Mechanics of Materials 7th Edition Beer Johnson Chapter 8

CHAPTER 8 P P PROBLEM 8.1 A W10  39 rolled-steel beam supports a load P as shown. Knowing that A D P  45 kips, a  10 in., and  all  18 ksi, determine (a) the maximum value of B C the normal stress  m in the beam, (b) the maximum value of the principal stress 10 ft  max at the junction of the flange and web, (c) whether the specified shape is a a acceptable as far as these two stresses are concerned. SOLUTION |V |max  90 kips |M |max  (45)(10)  450 kip  in. For W10  39 rolled steel section, d  9.92 in., b f  7.99 in., t f  0.530 in., tw  0.315 in., I x  209 in 4 S x  42.1 in 3 1 c d  4.96 in. yb  c  t f  4.43 in. 2 |M |max 450 (a) m    m  10.69 ksi  Sx 42.1 yb  4.43  b  m    (10.69)  9.55 ksi c  4.96  A f  b f t f  4.2347 in 2 1 yf  (c  yb )  4.695 in. 2 Qb  A f y f  19.8819 in 3 |V |maxQb (45)(19.8819)  xy    13.5898 ksi I xt w (209)(0.315) 2  b  2 R     xy  14.4043 ksi  2  b (b)  max   R  19.18 ksi  max  19.18 ksi  2 (c) Since  max >  all ( 18 ksi), W10  39 is not acceptable.  PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part. 1241 PROBLEM 8.2 Solve Prob. 8.1, assuming that P  22.5 kips and a  20 in. PROBLEM 8.1 A W10  39 rolled-steel beam supports a load P as shown. Knowing that P  45 kips, a  10 in., and  all  18 ksi, determine (a) the maximum value of the normal stress  m in the beam, (b) the maximum value of the principal stress  max at the junction of the flange and web, (c) whether the specified shape is acceptable as far as these two stresses are concerned. SOLUTION |V |max  22.5 kips |M |max  (22.5)(20)  450 kip  in. For W10  39 rolled steel section, d  9.92 in., b f  7.99 in., t f  0.530 in., tw  0.315 in., I x  209 in 4 , S x  42.1 in 3 1 c d  4.96 in. yb  c  t f  4.43 in. 2 |M |max 450 (a) m    m  10.69 ksi  Sx 42.1 yb  4.43  b  m    (10.69)  9.55 ksi c  4.96  A f  b f t f  4.2347 in 2 1 yf  (c  yb )  4.695 in. 2 Qb  A f y f  19.8819 in 3 |V |max Qb (22.5)(19.8819)  xy    6.7949 ksi I x tw (209)(0.315) 2   R   b    xy 2  8.3049 ksi  2  b (b)  max   R  13.08 ksi  max  13.08 ksi  2 (c) Since  max <  all ( 18 ksi), W10  39 is acceptable.  PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part. 1242 P PROBLEM 8.3 A C An overhanging W920  449 rolled-steel beam supports a load P as shown. Knowing that P  700 kN, a  2.5 m, and  all  100 MPa, determine (a) the B maximum value of the normal stress  m in the beam, (b) the maximum value of the principal stress  max at the junction of the flange and web, (c) whether a a the specified shape is acceptable as far as these two stresses are concerned. SOLUTION |V |max  700 kN  700  103 N |M |max  (700  103 )(2.5)  1.75  106 N  m For W920  449 rolled steel beam, d  947 mm, b f  424 mm, t f  42.7 mm, tw  24.0 mm, I x  8780  106 mm 4 , S x  18,500  103 mm3 1 c d  473.5 mm, yb  c  t f  430.8 mm 2  m  94.595 MPa |M |max 1.75  106 (a) m    m  94.6 MPa  Sx 18,500  106 yb 430.8 b  m  (94.595)  86.064 MPa c 473.5 A f  b f tt  18.1048  103 mm 2 1 yf  (c  yb )  452.15 mm 2 Qb  A f y f  8186.1  103 mm3  8186.1  106 m3 |V |max Qb (700  103 )(8186.1  106 )  xy    27.194 MPa I x tw (8780  106 )(24.0  103 ) 2 2    86.064  R   b    xy 2     27.1942  50.904 MPa  2   2  b (b)  max   R  93.9 MPa  max  93.9 MPa  2 (c) Since 94.6 MPa   all ( 100 MPa), W920  449 is acceptable.  PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part. 1243 PROBLEM 8.4 Solve Prob. 8.3, assuming that P  850 kN and a  2.0 m. PROBLEM 8.3 An overhanging W920  449 rolled-steel beam supports a load P as shown. Knowing that P  700 kN, a  2.5 m, and  all  100 MPa, determine (a) the maximum value of the normal stress  m in the beam, (b) the maximum value of the principal stress  max at the junction of the flange and web, (c) whether the specified shape is acceptable as far as these two stresses are concerned. SOLUTION |V |max  850 kN  850  103 N |M |max  (850  103 )(2.0)  1.70  106 N  m For W920  449 rolled steel section, d  947 mm, b f  424 mm, t f  42.7 mm, tw  24.0 mm, I x  8780  106 mm 4 , S x  18,500  103 mm3 1 c d  473.5 mm yb  c  t f  430.8 mm 2  m  91.892 MPa |M |max 1.70  106 (a) m    m  91.9 MPa  Sx 18,500  106 yb 430.8 b  m  (91.892)  83.605 MPa c 473.5 A f  b f t f  18.1048  103 mm 2 1 yf  (c  yb )  452.15 mm 2 Qb  A f y f  8186.1  103 mm3  8186.1  106 m3 |V |max Qb (850  103 )(8186.1  106 )  xy    33.021 MPa I x tw (8780  106 )(24.0  103 ) 2 2    83.605  R   b    xy 2   2   33.021  53.271 MPa  2   2  b (b)  max   R  95.1 MPa  max  95.1 MPa  2 (c) Since 95.1 MPa <  all ( 100 MPa), W920  449 is acceptable.  PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part. 1244 40 kN PROBLEM 8.5 2.2 kN/m (a) Knowing that  all  160 MPa and  all  100 MPa, select the most A C economical metric wide-flange shape that should be used to support the B loading shown. (b) Determine the values to be expected for  m ,  m , and the principal stress  max at the junction of a flange and the web of the 4.5 m 2.7 m selected beam. SOLUTION  M C  0: 7.2RA  (2.2)(7.2)(3.6)  (40)(2.7)  0 RA  22.92 kN VA  RA  22.92 kN VB  22.92  (2.2)(4.5)  13.02 kN VB  13.02  40  26.98 kN VC  26.98  (2.2)(2.7)  32.92 kN MA  0 1 MB  0  (22.92  13.02)(4.5)  80.865 kN  m 2 MC  0 M 80.865  103 S min  max   490  106 m3  all 165  106  490  103 mm3 Shape S (103 mm3) W360  39 578 W310  38.7 547  W250  44.8 531 W200  52 511 (a) Use W310  38.7.  d  310 mm t f  9.65 mm   tw  5.84 mm PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part. 1245 PROBLEM 8.5 (Continued) MB 80.865  103 (b) m   6  147.834  106 Pa S 547  10  m  147.8 MPa  V 32.92  103 m  max   18.1838  106 Pa dtw (310  103 )(5.84  103 )  m  18.18 MPa  1 c d  155 mm yb  c  t f  155  9.65  145.35 mm 2 y  145.35  b  b m    (147.834)  138.630 MPa c  155  V (26.98  103 ) At point B, w    14.9028 MPa dtw (310  103 )(5.84  103 ) 2  b     w  2 R (69.315) 2  (14.9028)2  70.899 MPa  2  b  max   R  69.315  70.899  max  140.2 MPa  2 PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part. 1246 275 kN PROBLEM 8.6 B C (a) Knowing that  all  160 MPa and  all  100 MPa, select the most A D economical metric wide-flange shape that should be used to support the loading shown. (b) Determine the values to be expected for 275 kN  m ,  m , and the principal stress  max at the junction of a flange and 3.6 m 1.5 m 1.5 m the web of the selected beam. SOLUTION RB  504.17 kN  RC  504.17  V max  275 kN M max  412.5 kN  m M 412.5  102 Smin  max   2578  106 m3  all 160  10 6  2578  103 mm3 Shape Sx (103 mm3) W760  147 4410  W690  125 3490 W530  150 3720 W460  158 3340 W360  216 3800 (a) Use W690  125.   d  678 mm t f  16.3 mm tw  11.7 mm  M 412.5  103  (b) m  max   118.195  106 Pa   m  118.2 MPa  S 3490  106 V V 275  103  m  max  max   34.667  106 Pa   m  34.7 MPa  Aw dtw (678  103 )(11.7  103 ) 1 67.8  c d   339 mm, t f  16.3 mm, yb  c  t f  339  16.3  322.7 mm  2 2 yb  322.7  b  m    (118.195)  112.512 MPa c  339  2  b     m  2  R (56.256) 2  (34.667)2  66.080 MPa   2  b   max   R  56.256  66.080   max  122.3 MPa  2 PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part. 1247 20 kips 20 kips PROBLEM 8.7 2 kips/ft B C (a) Knowing that  all  24 ksi and  all  14.5 ksi, select the most A D economical wide-flange shape that should be used to support the loading shown. (b) Determine the values to be expected for  m ,  m , and the principal stress  max at the junction of a flange and the web of the 10 ft 30 ft 10 ft selected beam. SOLUTION RA  50 kips  RD  50 kips  V  30 kips M  200 kip  ft  2400 kip  in. max max M 2400 S min  max   100 in 3  all 24 Shape S (in 3 ) W24  68 154  W21  62 127 W18  76 146 W16  77 134 W12  96 131 W10  112 126 (a) Use W21  62.  d  21.0 in. t f  0.615 in. tw  0.400 in. M 2400 (b) m  max   18.8976 ksi  m  18.9 ksi  S 127 V 30 m  max   3.5714 ksi  m  3.57 ksi  dtw (21.0)(0.400) 1 21.0 c d   10.50 in. yb  c  t f  10.50  0.615  9.8850 in. 2 2 yb  9.8850  b  m    (18.8976)  17.7907 ksi c  10.50  2  b     m  2 R (8.8954) 2  (3.5714) 2  9.5856 ksi  2 b  max   R  8.8954  9.5856  18.4810 ksi  max  18.48 ksi ◄ 2 PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part. 1248 1.5 kips/ft PROBLEM 8.8 (a) Knowing that  all  24 ksi and  all  14.5 ksi, select the most A C economical wide-flange shape that should be used to support the loading B shown. (b) Determine the values to be expected for  m ,  m , and the principal stress  max at the junction of a flange and the web of the selected 12 ft 6 ft beam. SOLUTION  M B  0: 12 RA  (1.5)(18)(3)  0 RA  6.75 kips   M A  0: 12 RB  (1.5)(18)(9)  0 RB  20.25 kips  V max  11.25 kips M max  27 kip  ft  324 kip  in. M 324 Smin  max   13.5 in 3  all 24 Shape S (in 3 ) W12  16 17.1  W10  15 13.8 W8  18 15.2 W6  20 13.4 (a) Use W10  15.  d  10.0 in.  t f  0.270 in.  tw  0.230 in. M 324 (b) m  max   23.478 ksi S 13.8  m  23.5 ksi  V 11.25 m  max   4.8913 ksi dtw (10.0)(0.230)  m  4.89 ksi  1 10.0  c  d   5.00 in.  2 2 PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part. 1249 PROBLEM 8.8 (Continued) yb  c  t f  5.00  0.270  4.73 in. yb  4.73  b  m    (23.478)  22.210 ksi c  5.00  2 2  b   22.210     m  2 2 R    (4.8913)  12.1345 ksi  2  2  b 22.210  max  R  12.1345  max  23.2 ksi  2 2 PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part. 1250 PROBLEM 8.9 The following problem refers to a rolled-steel shape selected in a problem of Chap. 5 to support a given loading at a minimal cost while satisfying the requirement  m   all . For the selected design (use the loading of Prob. 5.73 and selected W530  92 shape), determine (a) the actual value of  m in the beam, (b) the maximum value of the principal stress  max at the junction of a flange and the web. SOLUTION Reactions: RA  97.5 kN  RD  97.5 kN  |V |max  97.5 kN |M |max  286 kN  m For W530  92 rolled-steel section, d  533 mm, b f  209 mm, t f  15.6 mm, 1 tw  10.2 mm, c d  266.5 mm 2 I  554  106 mm 4 S  2080  103 mm3 | M |max 286  103 (a) m    137.5  106 Pa S 2080  106  m  137.5 MPa  yb  c  t f  250.9 mm A f  b f t f  3260.4 mm 2 1 y  (c  yb )  258.7 mm 2 Q  A f y  843.47  103 mm3 At midspan: V 0 b  0 yb 250.9 b  m  (137.5)  129.5 MPa  max  129.5 MPa  c 266.5 PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part. 1251 PROBLEM 8.9 (Continued) At sections B and C: M 270  103 m    129.808 MPa S 2080  106 y 250.9 b  b m  (129.808)  122.209 MPa c 266.5 VQ VA f y (82.5  103 )(3260.4  106 )(258.7  103 ) b    It Itw (554  106 )(10.2  103 )  12.3143 MPa 2  b  2 R     b  62.333 MPa  2  b  max  R  max  123.4 MPa  2 PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part. 1252 PROBLEM 8.10 The following problem refers to a rolled-steel shape selected in a problem of Chap. 5 to support a given loading at a minimal cost while satisfying the requirement  m   all . For the selected design (use the loading of Prob. 5.74 and selected W250  28.4 shape), determine (a) the actual value of  m in the beam, (b) the maximum value of the principal stress  max at the junction of a flange and the web. SOLUTION From Problem 5.74,  all  160 MPa M max  48 kN  m at section E, which lies 1.6 m to the right of B. V 0 For W250  28.4 rolled-steel section, d  259 mm b f  102 mm t f  10.0 mm tw  6.35 mm I  40.1  106 mm 4 S  308  103 mm3 1 c d  129.5 mm 2 M 48  103 (a) m  max  S 308  106  m  155.8 MPa ◄ yb  c  t f  119.5 mm 1 Af  b f t f  1020 mm 2 2 1 y  (c  yb )  124.5 mm 2 Q  A f y  126.99  103 mm3  126.99  106 m3 At section E, V 0 M  M max yb b   m  143.8 MPa c VQ b  0 Itw (b)  max   b  max  143.8 MPa ◄ PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part. 1253 PROBLEM 8.10 (Continued) At section C, V  40 kN M  32 kN  m Myb (32  103 )(119.5  103 ) b    95.6 MPa I 40.0  106 VQ (40  103 )(126.99  106 ) b    19.95 MPa Itw (40.1  106 )(6.35  103 ) 2  b     b  2 R 47.82  19.952  51.08 MPa  2 b  max   R  99.6 MPa (less than value at section E) 2 PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part. 1254 PROBLEM 8.11 The following problem refers to a rolled-steel shape selected in a problem of Chap. 5 to support a given loading at a minimal cost while satisfying the requirement  m   all . For the selected design (use the loading of Prob. 5.75 and selected S12  31.8 shape), determine (a) the actual value of  m in the beam, (b) the maximum value of the principal stress  max at the junction of a flange and the web. SOLUTION From Prob. 5.75,  all  24 ksi |M |max  54 kip  ft  648 kip  in. at C. At C, |V |  18 kips For S12  31.8 rolled-steel shape, d  12.0 in., b f  5.00 in., t f  0.544 in., tw  0.350 in. I z  217 in 4 , S z  36.2 in 3 1 c d  6.00 in. 2 |M | 648 m    17.9006 ksi (a)  m  17.90 ksi  S z 36.2 yb  c  t f  5.456 in. yb b b   m  16.2776 ksi  8.1388 ksi c 2 A f  b f t f  2.72 in 2 1 y (c  yb )  5.728 in. 2 Q  A f y  15.5802 in 3 VQ (18)(15.5802) b    3.6925 ksi I z tw (217)(0.350) 2   R   b    b2  8.13882  3.69252  8.9373 ksi  2  b  max   R  8.1388  8.9373 (b)  max  17.08 ksi  2 PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part. 1255 PROBLEM 8.12 The following problem refers to a rolled-steel shape selected in a problem of Chap. 5 to support a given loading at a minimal cost while satisfying the requirement  m   all . For the selected design (use the loading of Prob. 5.76 and selected S15  42.9 shape), determine (a) the actual value of  m in the beam, (b) the maximum value of the principal stress  max at the junction of a flange and the web. SOLUTION From Prob. 5.76,  all  24 ksi |M |max  96 kip  ft  1152 kip  in. at D. At D, |V |  38.4 kips For S15  42.9 shape, d  15.0 in., b f  5.50 in. t f  0.622 in., tw  0.411 in., I z  446 in 4 , S z  59.4 in 3 1 c d  7.5 in. 2 |M | 1152 m    19.3939 ksi (a)  m  19.39 ksi  S 59.4 yb  c  t f  6.878 in. yb b b   m  17.7855 ksi  8.8928 ksi c 2 A f  b f t f  3.421 in 2 1 y (c  yb )  7.189 in. 2 Q  A f y  24.594 in 3  VQ (57.6)(24.594) b    7.7281 ksi I z tw (446)(0.411) 2   R   b    b2  8.89282  7.72812  11.7816 ksi   2  b   max   R  8.8928  11.7816  (b)  max  20.7 ksi  2 PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part. 1256 PROBLEM 8.13 The following problem refers to a rolled-steel shape selected in a problem of Chap. 5 to support a given loading at a minimal cost while satisfying the requirement  m   all . For the selected design (use the loading of Prob. 5.77 and selected S510  98.2 shape), determine (a) the actual value of  m in the beam, (b) the maximum value of the principal stress  max at the junction of a flange and the web. SOLUTION From Problem 5.77,  all  160 MPa M  256 kN  m at point B max V  360 kN at B For S510  98.2 rolled-steel section, d  508 mm, b f  159 mm, t f  20.2 mm tw  12.8 mm, I x  495  106 mm 4 , S x  1950  103 mm3 1 c d  254 mm 2 M 256  103 (a) m  max   131.3 MPa  Sx 1950  106 yb  c  t f  233.8 yb b b   m  120.9 MPa  60. 45 MPa c 2 A f  b f t f  3212 mm 2 1 y  (c  yb )  243.9 mm 2 Q  A f y  783.4  103 mm3 VQ (360  103 )(783.4  106 ) b    44.5 MPa Itw (495  106 )(12.8  103 ) 2  b     b  2 R 60.452  44.52  75.06 MPa  2 b (b)  max   R  60.45  75.06  135.5 MPa 2  max  135.5 MPa ◄ PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part. 1257 PROBLEM 8.14 The following problem refers to a rolled-steel shape selected in a problem of Chap. 5 to support a given loading at a minimal cost while satisfying the requirement  m   all . For the selected design (use the loading of Prob. 5.78 and selected S460  81.4 shape), determine (a) the actual value of  m in the beam, (b) the maximum value of the principal stress  max at the junction of a flange and the web. SOLUTION Reactions: RA  65 kN  RD  35 kN  |V |max  65 kN |M |max  175 kN  m For S460  81.4 rolled-steel section, d  457 mm, b f  152 mm, t f  17.6 mm, 1 tw  11.7 mm, c  d  228.5 mm 2 I  333  106 mm 4 S  1460  103 mm3 M 175  103 (a) m  max   119.863  106 Pa S 1460  106  m  119.9 MPa  (b) yb  c  t f  210.9 mm A f  b f t f  2675.2 mm 2 1 y  (c  yb )  219.7 mm 2 Q  A f y  587.74  103 mm3 yb 210.9 At section C, b  m  (119.863)  110.631 MPa c 228.5 VQ VA f y (35  103 )(2675.2  106 )(219.7  103 ) b     5.2799 MPa It Itw (333  106 )(11.7  103 ) 2  b      b  55.567 MPa 2 R  2 b  max  R  max  110.9 MPa  2 PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part. 1258 PROBLEM 8.14 (Continued) M 162.5  103 At section B, m    111.301 MPa S 1460  106 y 210.9 b  b m  (111.301)  102.728 MPa c 228.5 VQ VA f y (65  103 )(2675.2  106 )(219.7  103 ) b     9.8055 MPa It Itw (333  106 )(11.7  103 ) 2  b      b  52.292 MPa 2 R  2  b  max  R  max  103.7 MPa  2 PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part. 1259 PROBLEM 8.15 A Determine the smallest allowable diameter of the solid shaft ABCD, knowing that  all  60 MPa and that the radius of disk B is r  80 mm. r B P 150 mm C 150 mm D T ⫽ 600 N · m SOLUTION T 600 M axis  0: T  Pr  0 P    7.5  103 N r 80  103 1 RA  RC  P 2  3.75  103 N M B  (3.75  103 )(150  103 )  562.5 N  m Bending moment: (See sketch). Torque: (See sketch). Critical section lies at point B. M  562.5 N  m, T  600 N  m J   c3   M2  T2  max c 2  all 2 M2  T2 2 (562.5) 2  (600) 2 c3     all  60  106  8.726  106 m3 c  20.58  103 m d  2c  41.2  103 m d  41.2 mm  PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part. 1260 PROBLEM 8.16 A Determine the smallest allowable diameter of the solid shaft ABCD, knowing that  all  60 MPa and that the radius of disk B is r  120 mm. r B P 150 mm C 150 mm D T ⫽ 600 N · m SOLUTION T 600 M AD  0, T  Pr  0 P  3  5  103 N r 120  10 1 RA  RC  P 2  2.5  103 N M B  (2.5  103 )(0.150  103 )  375 N  m Bending moment: (See sketch). Torque: (See sketch). Critical section lies at point B. M  375 N  m, T  600 N  m J   c3   M2  T2  max c 2  all 2 M2  T2 2 3752  6002 c3    7.5073  106 m3   all  60  106 c  19.5807  103 m d  2c  39.2  103 m  39.2 mm  PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part. 1261 PROBLEM 8.17 Using the notation of Sec. 8.2 and neglecting the effect of shearing stresses caused by transverse loads, show that the maximum normal stress in a circular shaft can be expressed as follows: c 2     1/2 1/2   max  M y  M z2  M y2  M z2  T 2 J   max SOLUTION Maximum bending stress: |M | c M y2  M z2 c m   I I Maximum torsional stress: Tc m  J m M y2  M z2 c c   M y2  M z2 2 2I J Using Mohr’s circle, 2  m  c2 T 2c 2 R   2  2   m  J2 M  2 y  M 2 z  J2  c  M y2  M z2  T 2 J m c c  max  R M y2  M z2  M y2  M z2  T 2 2 J J c 2     1/2 1/2   M y  M z2  M y2  M z2  T 2  J   PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part. 1262 y PROBLEM 8.18 The 4-kN force is parallel to the x axis, and the force Q is parallel to the A z axis. The shaft AD is hollow. Knowing that the inner diameter is half the 60 mm outer diameter and that  all  60 MPa, determine the smallest permissible Q outer diameter of the shaft. B 90 mm 100 mm C 4 kN 80 mm D 140 mm z x SOLUTION My  0: 60  103 Q  (90  103 )(4  103 )  0 Q  6  103 N  6 kN Bending moment and torque diagrams. In xy plane: ( M z )max  315 N  m at C. In yz plane: ( M x )max  412.5 N  m at B. About z axis: Tmax  360 N  m between B and C. PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part. 1263 PROBLEM 8.18 (Continued) At B,  100  Mz    (315)  175 N  m  180  M x2  M z2  T 2  1752  412.52  3602  574.79 N  m At C,  140  Mx    (412.5)  262.5 N  m  220  M x2  M z2  T 2  3152  262.52  3602  545.65 N  m Largest value is 574.79 N  m. max M x2  M z2  T 2 c  max  J J max M x2  M z2  T 2 574.79   c  max 60  106  9.5798  106 m3  9.5798  103 mm3   4 4  J 2 co  ci   c4   co3 1  i4   3   1 4    co 1     c co 2  co  2   2    1.47262co3 1.47262co3  9.5798  103 co  18.67 mm  d o  2co  d o  37.3 mm  PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part. 1264 6 in. P2 PROBLEM 8.19 The vertical force P1 and the horizontal force P2 are applied as shown 8 in. A B to disks welded to the solid shaft AD. Knowing that the diameter of the shaft is 1.75 in. and that  all  8 ksi, determine the largest permissible P1 C magnitude of the force P2. D 3 in. 10 in. 10 in. SOLUTION Let P2 be in kips. 4  M shaft  0: 6 P1  8P2  0 P1  P2 3 Torque over portion ABC: T  8P2 1 Bending in horizontal plane: M Cy  10  P2  5P2 2 Bending in vertical plane: M Bz  3P1 4  3 P2 3  4 P1 Critical point is just to the left of point C. T  8P2 M y  5P2 M z  2 P2 1 d  1.75 in. c d  0.875 in. 2  J  (0.875) 4  0.92077 in 4 2 c  all  T 2  M y2  M z2 J 0.875 8 (8P2 ) 2  (5P2 ) 2  (2 P2 ) 2  9.164 P2 0.92077 P2  0.873 kips P2  873 lb  PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part. 1265 y PROBLEM 8.20 7 in. 7 in. 7 in. The two 500-lb forces are vertical and the force P is parallel 4 in. 7 in. to the z axis. Knowing that  all  8 ksi, determine the smallest P A permissible diameter of the solid shaft AE. B 4 in. C z B E 6 in. D 500 lb x 500 lb SOLUTION Forces in vertical plane: M x  0: (4)(500)  6 P  (4)(500)  0 P  666.67 lb Torques: AB : T  0 BC : T  (4)(500)  2000 lb  in. CD : T  4(500)  2000 lb  in. DE : T  0 Critical sections are on either side of disk C. T  2000 lb  in. M z  3500 lb  in. M y  4667 lb  in. c  all  M y2  M z2  T 2 J Forces in horizontal plane: J  3  c c 2 M y2  M z2  T 2   all 4667 2  35002  20002  8  103  0.77083 in 3 c  0.7888 in. d  2c d  1.578 in.  PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part. 1266 PROBLEM 8.21 90⬚ M It was stated in Sec. 8.2 that the shearing stresses produced in a shaft by the H transverse loads are usually much smaller than those produced by the torques. In the preceding problems, their effect was ignored and it was assumed that the O maximum shearing stress in a given section occurred at point H (Fig. P8.21a) and was equal to the expression obtained in Eq. (8.5), namely, T c (a) H  M2 T2 J V Show that the maximum shearing stress at point K (Fig. P8.21b), where the effect M of the shear V is greatest, can be expressed as ␤ 2 c 2  O K  ( M cos  ) 2   cV  T  90⬚ J 3  K T where  is the angle between the vectors V and M. It is clear that the effect of the (b) shear V cannot be ignored when  K   H . (Hint: Only the component of M along V contributes to the shearing stress at K.) SOLUTION Shearing stress at point K. 2 3 Due to V: For a semicircle, Q c 3 For a circle cut across its diameter, t  d  2c 1 For a circular section, I J 2  xy  VQ (V ) 3 c  1   2 3  2 Vc 2 It  2 J  (2c) 3 J Tc Due to T:  xy  J Since these shearing stresses have the same orientation, c2   xy  Vc  T  J  3  PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part. 1267 PROBLEM 8.21 (Continued) Mu 2 Mu Bending stress at point K: x   I J Where u is the distance between point K and the neutral axis,   u  c sin   c sin      c cos   2  2 Mc cos  x  J Using Mohr’s circle, 2    K  R   x    xy 2 Cross section  2   2 c 2   ( M cos  )2   Vc  T  J 3  PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part. 1268 6 in. P2 PROBLEM 8.22 Assuming that the magnitudes of the forces applied to disks A and C of A 8 in. Prob. 8.19 are, respectively, P1  1080 lb and P2  810 lb, and using B the expressions given in Prob. 8.21, determine the values  H and  K C in a section (a) just to the left of B, (b) just to the left of C. P1 D 3 in. 10 in. 10 in. SOLUTION From Prob. 8.19, shaft diameter  1.75 in. 1 c d  0.875 in. 2  J  c 4  0.92077 2 Torque over portion ABC: T  (6)(1080)  (8)(810)  6480 lb  in. (a) Just to the left of point B: V  1080 lb M  3240 lb  in.   90 T  6480 lb  in. c 0.875 H  M2  T2  (3240)2  (6480)2 J 0.92077  H  6880 psi  c c 2   32 VC  T  2 K  (M cos  ) 2    VC  T  J J 3  0.875  2       (1080)(0.875)  6480   K  6760 psi  0.92077  3   PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part. 1269 PROBLEM 8.22 (Continued) (b) Just to the left of point C: V  (162)2  (405) 2  436.2 lb  162   tan 1  21.80 405 M  (1620) 2  (4050)2  4362 lb  in. 1620   tan 1  21.80 4050   90  21.8  21.8  46.4 0.875  H  (6480)2  (4362) 2   H  7420 psi  0.92077 2 2 VC  T    (436.2)(0.875)  6480  6734 lb  in.  3 3  M cos   4362 cos 46.4  3008 lb  in. 0.875  K  (3008)2  (6734)2   K  7010 psi  0.92077 PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part. 1270 PROBLEM 8.23 120 mm The solid shaft AB rotates at 600 rpm and transmits 160 mm 80 kW from the motor M to a machine tool M A connected to gear F. Knowing that  all  60 MPa, 120 mm determine the smallest permissible diameter of F shaft AB. C D E 80 mm B 60 mm SOLUTION 600 rpm f   10 Hz 60 sec /min P 80  103 T    1273.24 N  m 2 f (2 )(10) T Gear C: FC  rC 1273.24 FC  3  15.9155  103 N 80  10 T Gear D: FD  rD 1273.24 FD   21.221  103 N 60  103 Forces in vertical plane:  7  M Cz  (120  103 )  Fc   1336.90 N  m  10  120 M Dz  M Cz  572.96 N  m 280 Forces in horizontal plane:  7  M D y  (120  103 )  FD   1782.56 N.m  10  120 M Cy  M Dy  763.95 N  m 280 PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part. 1271 PROBLEM 8.23 (Continued) At C , M y2  M z2  T 2  1998.01 N  m At D, M y2  M z2  T 2  2264.3 N  m  all  c J  M y2  M z2  T 2  max J   c3   M y2  M z2  T 2  max  2264.3  37.738  106 m3 c 2  all 60  106 c  28.855  103 m d  2c  57.7  103 m d  57.7 mm ◄ PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part. 1272 PROBLEM 8.24 120 mm Solve Prob. 8.23, assuming that shaft AB rotates at 160 mm 720 rpm. M A 120 mm PROBLEM 8.23 The solid shaft AB rotates at 600 F rpm and transmits 80 kW from the motor M to a C machine tool connected to gear F. Knowing that  all  60 MPa, determine the smallest permissible D diameter of shaft AB. E 80 mm B 60 mm SOLUTION 720 rpm f   12 Hz 60 sec / min P 80  103 T    1061.03 N  m 2 f (2 )(12) I Gear C: FC  rC 1061.03 FC   13.2629  103 N 80  103 T Gear D: FD  rD 1061.03 FD  3  17.6838  103 N 60  10 Forces in vertical plane:  7  M C z  (120  103 )  FC   1114.08 N  m  10  120 M Dz  M Cz  477.46 N.m 280 Forces in horizontal plane:  7  M D y  (120  103 )  FD   1485.44 N  m  10  120 MC y  M D y  636.62 N  m 280 PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part. 1273 PROBLEM 8.24 (Continued) At C , M y2  M z2  T 2  1665.01 N  m At D, M y2  M z2  T 2  1886.87 N  m  all  c J  M y2  M z2  T 2  max J   c3   M y2  M z2  T2  max  1886.87  31.448  106 m3 c 2  all 60  106 3 c  27.153  10 m d  2c  54.3  103 d  54.3 mm ◄ PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part. 1274 8 in. PROBLEM 8.25 4 in. M 3.5 in. D The solid shafts ABC and DEF and the gears shown are used to transmit 20 hp from the motor M to a machine tool connected to shaft DEF. A B E Knowing that the motor rotates at 240 rpm and that  all  7.5 ksi, F determine the smallest permissible diameter of (a) shaft ABC, (b) shaft DEF. C 6 in. SOLUTION 20 hp  (20)(6600)  132  103 in.  lb/s 240 240 rpm   4 Hz 60 P 132  103 (a) Shaft ABC: T   5252 lb  in. 2 f (2 )(4) T 5252 Gear C: FCD    875.4 lb rC 6 Bending moment at B: M B  (8)(875.4)  7003 lb  in. c  all  M2 T2 J J  3 M2 T2 (5252) 2  (7003)2  c    1.1671 in 3 c 2  all 7500 c  0.9057 in. d  2c d  1.811 in.  (b) Shaft DEF: T  rD FCD  (3.5)(875.4)  3064 lb  in. Bending moment at E: M E  (4)(875.4)  3502 lb  in. c  all  M2 T2 J J  3 M2 T2 (3502) 2  (3064)2  c    0.6204 in 3 c 2  all 7500 c  0.7337 in. d  2c d  1.467 in.  PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part. 1275 8 in. PROBLEM 8.26 4 in. M 3.5 in. D Solve Prob. 8.25, assuming that the motor rotates at 360 rpm. A B E PROBLEM 8.25 The solid shafts ABC and DEF and the gears shown are F used to transmit 20 hp from the motor M to a machine tool connected to shaft DEF. Knowing that the motor rotates at 240 rpm and that C  all  7.5 ksi, determine the smallest permissible diameter of (a) shaft ABC, (b) shaft DEF. 6 in. SOLUTION 20 hp  (20)(6600)  132  103 in.  lb/s 360 360 rpm   6 Hz 60 P 132  103 (a) Shaft ABC : T   3501 lb  in. 2 f (2 )(6) T 3501 Gear C: FCD    583.6 lb rC 6 Bending moment at B: M B  (8)(583.6)  4669 lb  in. c  all  M2 T2 J J  3 M2 T2 46692  35012  c    0.77806 in 3 c 2  all 7500 c  0.791 in. d  2c d  1.582 in.  (b) Shaft DEF: T  rD FCD  (3.5)(583.6)  2043 lb  in. Bending moment at E: M E  (4)(583.6)  2334 lb  in. c  all  M2 T2 J J  3 M2 T2 23342  20433  c    0.41362 in 3 c 2  all 7500 c  0.6410 in. d  2c d  1.282 in.  PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part. 1276 100 mm PROBLEM 8.27 M C The solid shaft ABC and the gears shown are used to transmit 10 kW B from the motor M to a machine tool connected to gear D. Knowing that the motor rotates at 240 rpm and that  all  60 MPa, determine C A 90 mm the smallest permissible diameter of shaft ABC. D E SOLUTION 240 rpm f   4 Hz 60 sec/ min P 10  103 T   397.89 N  m 2 f (2 )(4) Gear A: FrA  T  0 T 397.89 F   4421 N rA 90  103 Bending moment at B: M B  LAB F  (100  103 )(4421)  442.1 N  m c  all  M2 T2 J J  3 M2 T2  c  c 2  all 2 M2 T2 (2) 442.12  397.892 c3    6.3108  106 m3   all  (60  10 ) 6 c  18.479  103 m d  2c  37.0  103 m d  37.0 mm  PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part. 1277 100 mm PROBLEM 8.28 M C Assuming that shaft ABC of Prob. 8.27 is hollow and has an outer B diameter of 50 mm, determine the largest permissible inner diameter of the shaft. C A 90 mm PROBLEM 8.27 The solid shaft ABC and the gears shown are used to transmit 10 kW from the motor M to a machine tool connected to gear D. Knowing that the motor rotates at 240 rpm and that D  all  60 MPa, determine the smallest permissible diameter of shaft ABC. E SOLUTION 240 rpm f   4 Hz 60 sec/min P 10  103 T   397.89 N  m 2 f (2 )(4) Gear A: FrA  T  0 T 397.89 F   4421 N rA 90  103 Bending moment at B: M B  LAB F  (100  103 )(4421)  442.1 N  m co 1  all  M2 T2 co  d o  25  103 m J 2 J  (co4  ci4 ) M2 T2   co 2 co  all 2co M 2  T 2 ci4  co4   all (2)(25  103 ) 442.12  397.892  (25  103 ) 4   (60  106 )  390.625  109  157.772  109  232.85  109 ci  21.967  103 m di  2ci  43.93  103 m d  43.9 mm  PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part. 1278 4 in. PROBLEM 8.29 M 6 in. The solid shaft AE rotates at 600 rpm and transmits 60 hp from the motor M to F machine tools connected to gears G and H. A 8 in. Knowing that  all  8 ksi and that 40 hp is BC taken off at gear G and 20 hp is taken off at 6 in. gear H, determine the smallest permissible H C diameter of shaft AE. 3 in. D G 4 in. E 4 in. SOLUTION 60 hp  (60)(6600)  396  103 in.  lb/ sec 600 rpm f   10 Hz 60 sec/min Torque on gear B: P 396  103 TB    6302.5 lb  in. 2 f 2 (10) Torques on gears C and D: 40 TC  TB  4201.7 lb  in. 60 20 TD  TB  2100.8 lb  in. 60 Shaft torques: AB : TAB 0 BC : TBC  6302.5 lb  in. CD : TCD  2100.8 lb  in. DE : TDE 0 Gear forces: TB 6302.5 FB    2100.8 lb rB 3 T 4201.7 FC  C   1050.4 lb rC 4 T 2100.8 FD  D   525.2 lb rD 4 PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part. 1279 PROBLEM 8.29 (Continued) Forces in vertical plane: Forces in horizontal plane: At B  , M z2  M y2  T 2  2450.92  6477.62  6302.52  9364 lb  in. At C  , M z2  M y2  T 2  6127.32  3589.22  6302.52  9495 lb  in. (maximum)  all  c J  M z2  M y2  T 2  max J  3  c   M z2  M y2 T2  c 2  all 9495   1.1868 in 3 8  103 c  0.911 in. d  2c d  1.822 in.  PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part. 1280 4 in. PROBLEM 8.30 M 6 in. Solve Prob. 8.29, assuming that 30 hp is taken off at gear G and 30 hp is taken off at F gear H. A 8 in. BC PROBLEM 8.29 The solid shaft AE rotates 6 in. at 600 rpm and transmits 60 hp from the H C motor M to machine tools connected to gears G and H. Knowing that  all  8 ksi and that 40 hp is taken off at gear G and 20 hp is 3 in. D taken off at gear H, determine the smallest G permissible diameter of shaft AE. 4 in. E 4 in. SOLUTION 60 hp  (60)(6600)  396  103 in.  lb/ sec 600 rpm f   10 Hz 60 sec/min Torque on gear B: P 396  103 TB    6302.5 lb  in. 2 f 2 (10) Torques on gears C and D: 30 TC  TB  3151.3 lb  in. 60 30 TD  TB  3151.3 lb  in. 60 Shaft torques: AB : TAB  0 BC : TBC  6302.5 lb  in. CD : TCD  3151.3 lb  in. DE : TDE  0 PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part. 1281 PROBLEM 8.30 (Continued) Gear forces: Forces in vertical plane: T 6302.5 FB  B   2100.8 lb rB 3 TC 3151.3 FC    787.8 lb rC 4 TD 3151.3 FD    787.8 lb rD 4 At B  , M z2  M y2  T 2  1838.22  6214.92  6302.52 Forces in horizontal plane:  9040.2 lb  in. (maximum) At C  , M z2  M y2  T 2  4595.52  2932.42  6302.52  8333.0 lb  in.  all  c J  M z2  M y2  T 2  max J  3  c   M z2  M y2 T2  c 2  all 9040.3   1.1300 in 3 8  103 c  0.8960 in. d  2c d  1.792 in.  PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part. 1282 12 in. PROBLEM 8.31 A 1.8 in. Two 1.2-kip forces are applied to an L-shaped b 1.2 kips a c machine element AB as shown. Determine the 6 in. normal and shearing stresses at (a) point a, 1.2 kips (b) point b, (c) point c. d 0.5 in. e 1.0 in. B f 0.5 in. 3.5 in. 1.0 in. SOLUTION Let  be the slope angle of line AB. 6 tan     26.565 12 Draw a free body sketch of the portion of the machine element lying above section abc. P  (1.2)sin   0.53666 kips V  1.2cos   1.07331 kips M  (1.8)(1.2cos  )  1.93196 kip  in. Section properties: A  (1.0)2  1.0 in 2 1 I  (1.0)(1.0)3  0.083333 in 4 12 c  0.5 in. P Mx 0.53666 (1.93196)(0.5) (a) Point a:      A I 1.0 0.083333   11.06 ksi ◄   0◄ (b) Point b: P 0.53666      0.537 ksi ◄ A 1.0 Q  (0.5)(1.0)(0.25)  0.125 in 3 VQ (1.07331)(0.125)      1.610 ksi ◄ It (0.083333)(1.0) P Mx 0.53660 (1.93196)(0.5) (c) Point c:        12.13 ksi ◄ A I 1.0 0.083333   0 ◄ PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part. 1283 12 in. PROBLEM 8.32 A 1.8 in. Two 1.2-kip forces are applied to an L-shaped b 1.2 kips a c machine element AB as shown. Determine the 6 in. normal and shearing stresses at (a) point d, 1.2 kips (b) point e, (c) point f. d 0.5 in. e 1.0 in. B f 0.5 in. 3.5 in. 1.0 in. SOLUTION Let  be the slope angle of line AB. 6 tan     26.535 12 Draw a free body sketch of the portion of the machine element lying to the right of section def. P  (1.2) cos   1.07331 kips V  1.2sin   0.53666 kips M  (3.5)(1.2sin  )  1.87831 kip  in. Section properties: A  (1.0)2  1.0 in 2 1 I  (1.0)(1.0)3  0.083333 in 4 12 c  0.5 in. P My 1.07331 (1.87831)(0.5) (a) Point d:      A I 1.0 0.083333   12.34 ksi ◄ 0 ◄ P 1.07331 (b) Point e:      1.073 ksi ◄ A 1.0 Q  (0.5)(1.0)(0.25)  0.125 in 3 VQ (0.53666)(0.125)      0.805 ksi ◄ It (0.083333)(1.0) P My 1.07331 (1.87831)(0.5) (c) Point f:        10.20 ksi ◄ A I 1.0 0.083333 0 ◄ PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part. 1284 a PROBLEM 8.33 100 mm c The cantilever beam AB has a rectangular cross section of 150  200 mm. 100 mm Knowing that the tension in the cable BD is 10.4 kN and neglecting the D b weight of the beam, determine the normal and shearing stresses at the 150 mm three points indicated. 0.75 m a c A B b E 200 mm 14 kN 0.9 m 0.3 m 0.6 m SOLUTION DB  0.752  1.82  1.95 m  0.75  Vertical component of TDB :   (10.4)  4 kN  1.95   1.8  Horizontal components of TDB :   10.4   9.6 kN  1.95  At section containing points a, b, and c, P  9.6 kN V  14  4  10 kN M  (1.5)(4)  (0.6)(14)  2.4 kN  m Section properties: A  (0.150)(0.200)  0.030 m 2 1 I  (0.150)(0.200)3  100  106 m 2 12 c  0.100 m P Mc 9.6  103 (2.4  103 )(0.100) At point a, x       2.08 MPa  x  2.08 MPa ◄ A I 0.030 100  106  xy  0 ◄ P Mc 9.6  103 (2.4  103 )(0.100) At point b, x       2.72 MPa  x  2.72 MPa ◄ A I 0.030 100  106  xy  0 ◄ PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part. 1285 PROBLEM 8.33 (Continued) P 9.6  103 At point c, x     0.320 MPa ◄ A 0.030 Q  (150)(100)(50)  750  103 mm3  750  106 m3 VQ (10  103 )(750  106 )  xy     0.500 MPa ◄ It (100  106 )(0.150)  PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part. 1286 PROBLEM 8.34 Member AB has a uniform rectangular cross section of 10  24 mm. For the 60 mm loading shown, determine the normal and shearing stress at (a) point H, (b) point K. A 9 kN 30⬚ G 60 mm H K 12 mm 12 mm B 40 mm SOLUTION Fx  0: Bx  0 MA  0: By (120 sin 30 )  9(60 sin 30)  0 By  4.5 kN At the section containing points H and K, P  4.5 cos 30  3.897 kN V  4.5 sin 30  2.25 kN M  (4.5  103 )(40  103 sin 30)  90 N  m A  10  24  240 mm 2  240  106 m 2 1 I  (10)(24)3  11.52  103 mm 4 12  11.52  109 m 4 P 3.897  103 (a) At point H, x      16.24 MPa  A 240  106 3 V 3 2.25  103  xy     14.06 MPa  2 A 2 240  106 P Mc 3.897  103 (90)(12  103 ) (b) At point K, x      A I 240  106 11.52  109   110.0 MPa    0  PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part. 1287 PROBLEM 8.35 60 mm A 9 kN Member AB has a uniform rectangular cross section of 10  24 mm. For the loading shown, determine the normal and shearing stress at (a) point H, 30⬚ G (b) point K. 60 mm H K 12 mm 12 mm B 40 mm SOLUTION M B  0: (120 cos 30) RA  (60sin 30)(9)  0 RA  2.598 kN Fy  0: By  9  0 By  9 kN  Fx  0: 2.598  Bx  0 Bx  2.598 kN  At the section containing points H and K, P  9 cos 30  2.598 sin 30  9.093 kN V  9 sin 30  2.598 cos 30  2.25 kN M  (9  103 )(40  103 sin 30)  (2.598  103 )(40  103 cos 30)  90 N  m A  10  240  240 mm 2  240  106 m 2 1 I (10)(24)3  11.52  103 mm 4  11.52  109 m 4 12 (a) At point H, P 9.093  103 x      37.9 MPa  A 240  106 3 V 3 2.25  103  xy     14.06 MPa  2 A 2 240  106 (b) At point K, P Mc x    A I 9.093  103 (90)(12  103 )   240  106 11.52  109   131.6 MPa    0  PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part. 1288 A PROBLEM 8.36 60 mm 9 kN Member AB has a uniform rectangular cross section of 10  24 mm. For the loading shown, determine the normal and shearing stress at (a) point H, G (b) point K. 60 mm 30⬚ H K 12 mm 12 mm B 40 mm SOLUTION M B  0: (9)(60sin 30)  120 RA  0 RA  2.25 kN Fx  0: 2.25cos 30  Bx  0 Bx  1.9486 kN  Fy  0: 2.25sin 30  9  By  0 By  7.875 kN  At the section containing points H and K, P  7.875cos 30  1.9486sin 30  7.794 kN V  7.875sin 30  1.9486 cos 30  2.25 kN M  (7.875  103 )(40  103 sin 30)  (1.9486  103 )(40  103 cos 30)  90 N  m A  10  24  240 mm 2  240  106 m 2 1 I (10)(24)3  11.52  103 mm 4  11.52  109 m 4 12 P 7.794  103 (a) At point H, x     x  32.5 MPa  A 240  106 3 V 3 2.25  103  xy    xy  14.06 MPa  2 A 2 240  106 P Mc 7.794  103 (90)(12  103 ) (b) At point K, x      A I 240  106 11.52  109  x  126.2 MPa   xy  0  PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part. 1289 9 kip · in. PROBLEM 8.37 C 1.5 kips A 1.5-kip force and a 9-kip · in. couple are applied at the top of the 2.5-in.- diameter cast-iron post shown. Determine the normal and shearing stresses at (a) point H, (b) point K. H K 9 in. SOLUTION Diameter  2.5 in. At the section containing points H and K, P0 V  1.5 kips T  9 kip  in. M  (1.5)(9)  13.5 kip  in. 1 d  2.5 in. c d  1.25 in. 2  A   c 2  4.909 in 2 I  c 4  1.9175 in 4 J  2I  3.835 in 4 4 2 3 For a semicircle, Q c  1.3021 in 3 3 (a) At point H, H  0 ◄ Tc VQ (9)(1.25) (1.5)(1.3021) H      2.934  0.407  3.34 ksi  H  3.34 ksi ◄ J It 3.835 (1.9175)(2.5) Mc (13.5)(1.25) (b) At point K, K     8.80 ksi  K   8.80 ksi ◄ I 1.9175 Tc (9)(1.25) K    2.93 ksi  K  2.93 ksi ◄ J 3.835 PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part. 1290 y PROBLEM 8.38 45 mm Two forces are applied to the pipe AB as shown. Knowing that the pipe has inner and outer diameters equal to 35 and 42 mm, 45 mm A respectively, determine the normal and shearing stresses at 1500 N (a) point a, (b) point b. 1200 N a b 75 mm B z 20 mm x SOLUTION co  do 2  21 mm, ci  di 2    17.5 mm A   co2  ci2  423.33 mm 2  J  2 c4 o  ci4  1  158.166  103 mm 4 I  J  79.083  103 mm 4 2 For semicircle with semicircular cutout, Q 2 3 3   co  ci3  2.6011  103 mm3 At the section containing points a and b, P  1500 N Vz  1200 N Vx  0 M z  (45  103 )(1500)  67.5 N  m M x  (75  103 )(1200)  90 N  m T  (90  103 )(1200)  108 N  m P M xc 1500 (90)(21  103 ) (a)        20.4 MPa  A I 423.33  106 79.083  109 Tc VxQ (108)(21  103 )     0   14.34 MPa  J It 158.166  109 P M zc 1500 (67.5)(21  103 ) (b)        21.5 MPa  A I 423.33  106 79.083  109 Tc Vz Q (108)(21  103 ) (1200)(2.6011  106 )        19.98 MPa  J It 158.166  109 (79.083  109 )(7  103 ) PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part. 1291 y PROBLEM 8.39 200 lb Several forces are applied to the pipe assembly shown. Knowing that the pipe has inner and outer diameters equal to 1.61 in. and 1.90 in., respectively, determine the D 150 lb normal and shearing stresses at (a) point H, (b) point K. H K z 10 in. 4 in. 4 in. 150 lb 6 in. 50 lb x SOLUTION Section properties: P  150 lb T  (200 lb)(10 in.)  2000 lb  in. M z  (150 lb)(10 in.)  1500 lb  in. M y  (200 lb  50 lb)(10 in.)  (150 lb)(4 in.)  900 lb  in. Vz  200  150  50  0 Vy  0 A   (0.952  0.8052 )  0.79946 in 2  (0.954  0.8054 )  0.30989 in 4 I 4 J  2 I  0.61979 in 4 P M c 150 lb (1500 lb  in.)(0.95 in.) (a) Point H: H   z  2  A I 0.79946 in 0.30989 in 4  187.6 psi  4593 psi  H  4.79 ksi  Tc (2000 lb  in.)(0.95 in.) H    3065.6 psi  H  3.07 ksi  J 0.61979 in 4 P M yc 150 lb (900 lb  in.)(0.95 in.) (b) Point K: K    2  A I 0.79946 in 0.30989 in 4  187.6 psi  2759 psi  K  2.57 ksi  Tc K   same as for  H  K  3.07 ksi  J PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part. 1292 y PROBLEM 8.40 50 mm 20 mm The steel pipe AB has a 100-mm outer diameter and an 8-mm wall t ⫽ 8 mm thickness. Knowing that the tension in the cable is 40 kN, determine A the normal and shearing stresses at point H. D 225 mm H 60⬚ E x B z SOLUTION Vertical force: 40cos 30  34.64 kN Horizontal force: 40sin 30  20 kN Point H lies on neutral axis of bending. Section properties: 1 d o  100 mm, co  d o  50 mm, ci  co  t  42 mm, 2 A   (co2  ci2 )  2.312  103 mm 2  2.312  103 m 2 P 34.64  103      14.98 MPa  A 2.312  106 V (2)(20  103 ) For thin pipe,  2    17.29 MPa  A 2.314  103 PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part. 1293 y PROBLEM 8.41 2 in. 6 kips 2 in. Three forces are applied to a 4-in.-diameter plate that is attached to the 6 kips solid 1.8-in. diameter shaft AB. At point H, determine (a) the principal stresses and principal planes, (b) the maximum shearing stress. A 2.5 kips 8 in. H B z x SOLUTION At the section containing point H, P  12 kips (compression) V  2.5 kips T  (2)(2.5)  5 kip  in. M  (8)(2.5)  20 kip  in. 1 d  1.8 in. c  d  0.9 in. 2 A   c 2  2.545 in 2  I  c 4  0.5153 in 4 4 J  2 I  1.0306 in 4 For a semicircle, 2 3 c  0.486 in 3 Q 3 Point H lies on neutral axis of bending. P 12 H    4.715 ksi A 2.545 Tc VQ (5)(0.9) (2.5)(0.486) H     J It 1.0306 (0.5153)(1.8)  5.676 ksi PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part. 1294 PROBLEM 8.41 (Continued) Use Mohr’s circle. 1  ave  (4.715) 2  2.3575 ksi 2  4.715  2 R    5.676  2   6.1461 ksi (a)  a   ave  R  a  3.79 ksi   b   ave  R  b  8.50 ksi  (2)(5.676) tan 2 p   2.408  a  33.7  4.715 b  123.7  (b)  max  R  max  6.15 ksi  PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part. 1295 y PROBLEM 8.42 3 kN The steel pipe AB has a 72-mm outer diameter B and a 5-mm wall thickness. Knowing that the arm CDE is rigidly attached to the pipe, C H D x determine the principal stresses, principal planes, and the maximum shearing stress at point H. 9 kN 120 mm A 150 mm z 120 mm E SOLUTION Replace the forces at C and E by an equivalent force-couple system at D. FD  9  3  6 kN TD  (9  103 )(120  103 )  (3  103 )(120  103 )  1440 N  m At the section containing point H, P0 V  6 kN T  1440 N  m M  (6  103 )(150  103 )  900 N  m 1 Section properties: d o  72 mm co  d o  36 mm ci  co  t  31 mm 2 A   (co2  ci2 )  1.0524  103 mm 2  1.0524  103 m 2  I  (co4  ci4 )  593.84  103 mm 4  593.84  109 m 4 4 J  2 I  1.1877  106 m 4 2 3 For half-pipe, (co  ci3 )  11.243  103 mm3  11.243  106 m3 Q 3 At point H, point H lies on the neutral axis of bending.  H  0. Tc VQ (1440)(36  103 ) (6  103 )(11.243  106 ) H      55.0 MPa J It 1.1877  106 (593.84  109 )(10  103 ) Use Mohr’s circle. c  0 R  55.0 MPa a  c  R  a  55.0 MPa ◄ b  c  R  b  55.0 MPa ◄  a  45, b  45 ◄  max  R  max  R  max  55.0 MPa ◄  PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part. 1296 y PROBLEM 8.43 B A 13-kN force is applied as shown to the 60-mm-diameter cast-iron D post ABD. At point H, determine (a) the principal stresses and principal planes, (b) the maximum shearing stress. 13 kN 300 mm H 100 mm A E z 125 mm 150 mm x SOLUTION DE  1252  3002  325 mm At point D, Fx  0  300  Fy     (13)  12 kN  325   125  Fz     (13)  5 kN  300  Moment of equivalent force-couple system at C, the centroid of the section containing point H:    i j k     M  0.150 0.200 0  1.00i  0.75 j  1.8k kN  m 0 12 5 Section properties: 1 d  60 mm c  d  30 mm 2 A   c 2  2.8274  103 mm 2  I  c 4  636.17  103 mm 4 4 J  2 I  1.2723  106 mm 4 2 3 For a semicircle, Q c  18.00  103 mm3 3 PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part. 1297 PROBLEM 8.43 (Continued) P Mc 12  103 (1.8  103 )(30  103 ) At point H, H       89.13 MPa A I 2.8274  103 636.17  109 Tc VQ (0.75  103 )(30  103 ) (5  103 )(18.00  106 ) H      20.04 MPa J It 1.2723  106 (636.17  109 )(60  103 ) H (a)  ave   44.565 MPa 2 2 H     H  48.863 MPa 2 R   2   a   ave  R  a  4.30 MPa    b   ave  R  b  93.4 MPa  2 H tan 2 p   0.4497 H  a  12.1, b  102.1  (b)  max  R  48.9 MPa  PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part. 1298 y PROBLEM 8.44 1 in. P A vertical force P of magnitude 60 lb is applied to the crank at point A. 2 in. 60° A Knowing that the shaft BDE has a diameter of 0.75 in., determine the principal stresses and the maximum shearing stress at point H located D E H at the top of the shaft, 2 in. to the right of support D. 8 in. z B 5 in. x SOLUTION Force-couple system at the centroid of the section containing point H: Fx  0, Vy  0.06 kips, Vz  0 M z  (5  2  1)(0.06)  0.24 kip  in. M x  (8sin 60)(0.06)  0.41569 kip  in. 1 d  0.75 in. c d  0.375 in. 2   I  c4  (0.375) 4  15.5316  103 in 4 4 4 J  2 I  31.063  103 in 4 At point H, Mzy (0.24)(0.375) H     5.7946 ksi Iz 15.5316  103 Tc (0.41569)(0.375) H    5.0183 ksi J 31.063  103 1 Use Mohr’s circle.  ave   H  2.8973 ksi 2 2 H  2 R     H  5.7946 ksi  2   a   ave  R  max  8.69 ksi   b   ave  R  min  2.90 ksi  2 H 2(5.0183) tan 2 p    1.7321 H 5.7946  a  30.0 b  120.0  max  R  max  5.79 ksi  PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part. 1299 50 kips PROBLEM 8.45 0.9 in. 2 kips 2.4 in. 0.9 in. C Three forces are applied to the bar shown. Determine the normal 2 in. and shearing stresses at (a) point a, (b) point b, (c) point c. 6 kips h ⫽ 10.5 in. 1.2 in. 1.2 in. a b c 4.8 in. 1.8 in. SOLUTION Calculate forces and couples at section containing points a, b, and c. Forces Couples h  10.5 in. P  50 kips Vx  6 kips Vz  2 kips M z  (10.5  2)(6)  51 kip  in. M x  (10.5)(2)  21 kip  in. Section properties. A  (1.8)(4.8)  8.64 in 2 1 Ix  (4.8)(1.8)3  2.3328 in 4 12 1 I z  (1.8)(4.8)3  6.5888 in 4 12 Stresses. P Mzx Mxz Vx Q      A Iz Ix Izt PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part. 1300 PROBLEM 8.45 (Continued) (a) Point a: x  0, z  0.9 in., Q  (1.8)(2.4)(1.2)  5.184 in 3 50 (21)(0.9)   0   2.31 ksi  8.64 2.3328 (6)(5.184)    1.042 ksi  (16.5888)(1.8) (b) Point b: x  1.2 in., z  0.9 in., Q  (1.8)(1.2)(1.8)  3.888 in 3 50 (51)(1.2) (21)(0.9)       6.00 ksi  8.64 16.5888 2.3328 (6)(3.888)   0.781 ksi   0.781 ksi  (16.5888)(1.8) (c) Point c: x  2.4 in., z  0.9 in., Q  0 50 (51)(2.4) (21)(0.9)       9.69 ksi  8.64 16.5888 2.3328   0  PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part. 1301 50 kips PROBLEM 8.46 0.9 in. 2 kips 2.4 in. 0.9 in. C Solve Prob. 8.45, assuming that h  12 in. 2 in. PROBLEM 8.45 Three forces are applied to the bar shown. Determine the normal and shearing stresses at (a) point a, (b) point b, (c) point c. 6 kips h ⫽ 10.5 in. 1.2 in. 1.2 in. a b c 4.8 in. 1.8 in. SOLUTION Calculate forces and couples at section containing points a, b, and c. Forces Couples h  12 in. P  50 kips Vx  6 kips Vz  2 kips M z  (12  2)(6)  60 kip  in. M x  (12)(2)  24 kip  in. Section properties. A  (1.8)(4.8)  8.64 in 2 1 Ix  (4.8)(1.8)3  2.3328 in 4 12 1 I z  (1.8)(4.8)3  6.5888 in 4 12 Stresses. P Mzx Mxz Vx Q      A Iz Ix Izt PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part. 1302 PROBLEM 8.46 (Continued) (a) Point a: x  0, z  0.9 in., Q  (1.8)(2.4)(1.2)  5.184 in 3 50 (24)(0.9)   0   3.47 ksi  8.64 2.3328 (6)(5.184)    1.042 ksi  (16.5888)(1.8) (b) Point b: x  1.2 in., z  0.9 in., Q  (1.8)(1.2)(1.8)  3.888 in 3 50 (60)(1.2) (24)(0.9)       7.81 ksi  8.64 16.5888 2.3328 (6)(3.888)    0.781 ksi  (16.5888)(1.8) (c) Point c: x  2.4 in., z  0.9 in., Q  0 50 (60)(2.4) (24)(0.9)       12.15 ksi  8.64 16.5888 2.3328   0  PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part. 1303 60 mm PROBLEM 8.47 24 mm Three forces are applied to the bar c shown. Determine the normal and b a 15 mm shearing stresses at (a) point a, 180 mm (b) point b, (c) point c. 40 mm 750 N 32 mm 16 mm 30 mm 500 N C 10 kN SOLUTION A  (60)(32)  1920 mm 2  1920  106 m 2 1 Iz  (60)(32)3  163.84  103 mm 4 12  163.84  109 m 4 1 I y  (32)(60)3 12  579  103 mm 4  576  109 m 4 At the section containing points a, b, and c, P  10 kN Vy  750 N, Vz  500 N M z  (180  103 )(750)  135 N  m Side View M y  (220  103 )(500)  110 N  m T 0 P Mz y Myz    A Iz Iy Vz Q  Top View I yt PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part. 1304 PROBLEM 8.47 (Continued) (a) Point a: y  16 mm, z  0, Q  Az  (32)(30)(15)  14.4  103 mm3 10  103 (135)(16  103 )   0   18.39 MPa  1920  106 163.84  109 (500)(14.4  106 )    0.391 MPa  (576  109 )(32  103 ) (b) Point b: y  16 mm, z  15 mm, Q  Az  (32)(15)(22.5)  10.8  103 mm3 10  103 (135)(16  103 ) (110)(15  103 )      21.3 MPa  1920  106 163.84  109 576  109 (500)(10.8  106 )    0.293 MPa  (576  109 )(32  103 ) (c) Point c: y  16 mm, z  30 mm, Q  0 10  103 (135)(16  103 ) (110)(30  106 )      24.1 MPa  1920  106 163.84  109 576  109  0  PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part. 1305 60 mm PROBLEM 8.48 24 mm Solve Prob. 8.47, assuming that the c 750-N force is directed vertically b a 15 mm upward. 180 mm 750 N 40 mm PROBLEM 8.47 Three forces are 32 mm applied to the bar shown. Determine 16 mm the normal and shearing stresses at 30 mm (a) point a, (b) point b, (c) point c. 500 N C 10 kN SOLUTION A  (60)(32)  1920 mm 2  1920  106 m 2 1 Iz  (60)(32)3 12  163.84  103 mm 4  163.84  109 m 4 1 Iy  (32)(60)3 12  576  103 mm 4  576  109 m 4 At the section containing points a, b, and c, P  10 kN T 0 Vy  750 N Vz  500 N M z  (180  103 )(750)  135 N  m Side View M y  (220  103 )(500)  110 N  m P M z y M yz     A Iz Iy Vz Q   Top View I yt PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part. 1306 PROBLEM 8.48 (Continued) (a) Point a: y  16 mm, z  0, Q  Az  (32)(30)(15)  14.4  103 mm3 10  103 (135)(16  103 )   0   7.98 MPa  1920  106 163.84  109 (500)(14.4  106 )    0.391 MPa  (163.84  109 )(32  103 ) (b) Point b: y  16 mm, z  15 mm, Q  Az  (32)(15)(22.5)  10.8  103 mm3 10  103 (135)(16  103 ) (110)(15  103 )      5.11 MPa  1920  106 163.84  109 576  109 (500)(10.8  106 )    0.293 MPa  (163.84  109 )(32  109 ) (c) Point c: y  16 mm, z  30 mm, Q  0 10  103 (135)(16  103 ) (110)(30  103 )      2.25 MPa  1920  106 163.84  109 576  109  0  PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part. 1307 y PROBLEM 8.49 Two forces are applied to the small post BD as shown. B Knowing that the vertical portion of the post has a cross section of 1.5  2.4 in., determine the principal stresses, principal planes, and maximum shearing stress at point H. 6000 lb 500 lb 1.5 in. 2.4 in. 4 in. H 6 in. D 1 in. z 3.25 in. x 1.75 in. SOLUTION Components of 500-lb force: (500)(1.75) Fx   140 lb 6.25 (500)(6) Fy    480 lb 6.25 Moment arm of 500-lb force:    r  3.25i  (6  1) j Moment of 500-lb force:    i j k   M  3.25 5 0  2260k lb  in. 140 480 0 At the section containing point H, P  480 lb Vx  140 lb Vz  6000 lb, M z  2260 lb  in., M x  (4)(6000)  24,000 lb  in. 1 A  (1.5)(2.4)  3.6 in 2 (2.4)(1.5)3  0.675 in 4 Iz  12 P M zx 480 (2260)(0.75) H      2644 psi A Iz 3.6 0.675 3 Vz 3 6000 H    2500 psi 2 A 2 3.6 PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part. 1308 PROBLEM 8.49 (Continued) Use Mohr’s circle. 2644  ave    1322 psi 2 2  2644  2 R    (2500)  2828 psi  2   a   ave  R  a  1506 psi   b   ave  R  b  4150 psi  2 H (2)(2500) tan 2 p    1.891 H 2644  a  31.1, b  121.1   max  R  max  2830 psi  PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part. 1309 y PROBLEM 8.50 Solve Prob. 8.49, assuming that the magnitude of the 6000-lb B force is reduced to 1500 lb. PROBLEM 8.49 Two forces are applied to the small post BD 6000 lb as shown. Knowing that the vertical portion of the post has a cross section of 1.5  2.4 in., determine the principal stresses, 500 lb principal planes, and maximum shearing stress at point H. 1.5 in. 2.4 in. 4 in. H 6 in. 1 in. D z 3.25 in. x 1.75 in. SOLUTION Components of 500-lb force: (500)(1.75) Fx   140 lb 6.25 (500)(6) Fy    480 lb 6.25 Moment arm of 500-lb force:    r  3.25i  (6  1) j    i j k   M  3.25 5 0  2260k lb  in. 140 480 0 At the section containing point H, P  480 lb Vx  140 lb Vz  1500 lb, M z  2260 lb  in., M x  (4)(1500)  6000 lb  in. 1 A  (1.5)(2.4)  3.6 in 2 (2.4)(1.5)3  0.675 in 4 Iz  12 P M zx 480 (2260)(0.75) H      2644 psi A Iz 3.6 0.675 3 Vz 3 1500 H    625 psi 2 A 2 3.6 PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part. 1310 PROBLEM 8.50 (Continued) Use Mohr’s circle. 1  ave   H  1322 psi 2 2  2644  2 R    (625)  1462.30 psi  2   a   ave  R  a  140.3 psi   b   ave  R  b  2780 psi  2 H (2)(625) tan 2 p    0.4728 H 2644  a  12.6 b  102.6   max  R  max  1462 psi  PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part. 1311 y 50 mm PROBLEM 8.51 150 mm Three forces are applied to the machine component ABD A 40 mm 0.5 kN as shown. Knowing that the cross section containing H point H is a 20  40-mm rectangle, determine the z principal stresses and the maximum shearing stress at B 20 mm point H. 3 kN D 160 mm x 2.5 kN SOLUTION Equivalent force-couple system at section containing point H: Fx  3 kN, Fy  0.5 kN, Fz  2.5 kN M x  0, M y  (0.150)(2500)  375 N  m M z  (0.150)(500)  75 N  m A  (20)(40)  800 mm 2  800  106 m 2 1 Iz  (40)(20)3 12  26.667  103 mm 4  26.667  109 m 4 P Mzy H   A Iz 3000 (75)(10  103 )   800  106 26.667  109  24.375 MPa 3 | Vz | 3 2500 H    2 A 2 800  106  4.6875 MPa PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part. 1312 PROBLEM 8.51 (Continued) Use Mohr’s circle. 1  ave  H 2  12.1875 MPa 2  24.375  2 R    (4.6875)  2   13.0579 MPa  a   ave  R  a  25.2 MPa   b   ave  R  b  0.87 MPa  2 H (2)(4.6875) tan 2 p    0.3846 H 24.375  a  10.5, b  100.5  max  R  max  13.06 MPa  PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part. 1313 y 50 mm PROBLEM 8.52 150 mm Solve Prob. 8.51, assuming that the magnitude of the A 40 mm 0.5 kN 2.5-kN force is increased to 10 kN. H z PROBLEM 8.51 Three forces are applied to the B 20 mm machine component ABD as shown. Knowing that the 3 kN cross section containing point H is a 20  40-mm 160 mm rectangle, determine the principal stresses and the D x maximum shearing stress at point H. 2.5 kN SOLUTION Equivalent force-couple system at section containing point H: Fx   3 kN, Fy   0.5 kN, Fz   10 kN M x  0, M y  (0.150)(10,000)  1500 N  m M z   (0.150)(500)  75 N  m A  (20)(40)  800 mm 2  800  106 m 2 1 Iz  (40)(20)3  26.667  103 mm 4  26.667  109 m 4 12 P Mzy 3000 (75)(10  103 ) H      24.375 MPa A Iz 800  106 26.667  109 3 Vz 3 10,000 H     18.75 MPa 2 A 2 800  106 1  c   H  12.1875 MPa 2 2  24.375  2 R    (18.75)  22.363 MPa  2  a  c  R  a  34.6 MPa ◄ b  c  R  b   10.18 MPa ◄ 2 H (2)(18.75) tan 2 p    1.5385 H 24.375  a  28.5, b  118.5  max  R  max  22.4 MPa ◄ PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part. 1314 PROBLEM 8.53 y e b d Three steel plates, each 13 mm a 60 mm 400 mm thick, are welded together to form 30 mm 75 mm a cantilever beam. For the 60 mm C x loading shown, determine the normal and shearing stresses at 9 kN 150 mm points a and b. t ⫽ 13 mm C 13 kN SOLUTION Equivalent force-couple system at section containing points a and b: Fx  9 kN, Fy  13 kN, Fz  0 M x  (0.400)(13  103 )  5200 N  m M y  (0.400)(9  103 )  3600 N  m Mz  0 A  (2)(150)(13)  (13)(75  26)  4537 mm 2  4537  106 m 2 1  1 I x  2  (150)(13)3  (150)(13)(37.5  6.5) 2   (13)(75  26)3  12  12  3.9303  106 mm 4  3.9303  106 m 4 1 1 I y  2  (13)(150)3  (75  26)(13)3 12 12  7.3215  10 mm  7.3215  106 mm 4 6 4 For point a, Qx  0, Qy  0 For point b, A*  (60)(13)  780 mm 2 x  45 mm y  31 mm Qx  A* y  24.18  103 mm3  24.18  106 m3 Qy  A* x  35.1  103 mm3  35.1  106 m3 PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part. 1315 PROBLEM 8.53 (Continued) Direction of shearing stress for horizontal and for vertical components of shear: At point a, Mx y Myx   Ix Iy (5200)(37.5  103 ) (3600)(75  103 )     86.5 MPa  3.9303  106 7.3215  106   0  At point b,  Mx y Myx     Ix Iy (5200)(37.5  103 ) (3600)(15  103 )       57.0 MPa  3.9303  106 7.3215  106 |Vx ||Qy | |Vy ||Qx |   I yt I xt (9  103 )(35.1  106 ) (13  103 )(24.18  106 )     (7.3215  106 )(13  103 ) (3.9303  106 )(13  103 )   3.32 MPa  6.15 MPa    9.47 MPa  PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part. 1316 PROBLEM 8.54 y e b d Three steel plates, each 13 mm a 400 mm thick, are welded together to 60 mm 30 mm 75 mm form a cantilever beam. For the 60 mm C x loading shown, determine the normal and shearing stresses at 9 kN 150 mm points d and e. t ⫽ 13 mm C 13 kN SOLUTION Equivalent force-couple system at section containing points d and e. Fx  9 kN, Fy  13 kN, Fz  0 M x  (0.400)(13  103 )  5200 N  m M y  (0.400)(9  103 )  3600 N  m Mz  0 A  (2)(150)(13)  (13)(75  26)  4537 mm 2  4537  106 m 2 1  1 I x  2  (150)(13)3  (150)(13)(37.5  6.5)2   (13)(75  26)3  12  12  3.9303  106 mm 4  3.9303  106 m 4 1  1 I y  2  (13)(150)3   (75  26)(13)3 12  12  7.3215  106 mm 4  7.3215  106 m 4 For point d, A*  (60)(13)  780 mm 2 x  45 mm y  31 mm Qx  A y  24.18  103 mm3  24.18  106 m3 * Qy  A* x  35.1  103 mm3 = 35.1  106 m3 For point e, Qx  0, Qy  0 PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part. 1317 PROBLEM 8.54 (Continued) At point d, M x y M yx    Ix Iy (5200)(37.5  10 3 ) (3600)(15  103 )     42.2 MPa  3.9303  106 7.3215  106 Due to Vx : |Vx ||Qy | (9000)(35.1  106 )   I yt (7.3215  106 )(13  103 )  3.32 MPa  Due to Vy : |Vy ||Qx | (13,000)(24.18  106 )    6.15 MPa  I xt (3.9303  106 )(13  103 ) By superposition, the net value is   2.83 MPa  At point e, M x y M y x (5200)(37.5  103 ) (3600)(75  103 )     Ix Iy 3.9303  106 7.3215  106   12.74 MPa     0  PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part. 1318 y PROBLEM 8.55 75 mm a Two forces P1 and P2 are applied as shown in directions a perpendicular to the longitudinal axis of a W310  60 beam. x Knowing that P1  25 kN and P2  24 kN, determine the principal stresses and the maximum shearing stress at point a. b b P2 1.2 m W310 ⫻ 60 0.6 m P1 SOLUTION At the section containing points a and b, M x  (1.8)(25)  45 kN  m M y  (1.2)(24)  28.8 kN  m Vx  24 kN Vy  25 kN For W310  60 rolled-steel section, d  302 mm, b f  203 mm, t f  13.1 mm, tw  7.49 mm I x  128  106 mm 4  128  106 m 4 , I y  18.4  106 mm 4  18.4  106 m 4 bf Normal stress at point a: x  75  26.5 mm 2 1 y  d  151 mm 2 M x y M y x (45  103 )(151  103 ) (28.8  103 )(26.5  103 ) z     Ix Iy 128  106 18.4  106  53.086 MPa  41.478 MPa  11.608 MPa Shearing stress at point a:  Vx A x Vy A y  xz    I yt f I xt f A  (75  103 )(13.1  103 )  982.5  106 m 2 bf 75 x    64 mm 2 2 d tf y    144.45 mm 2 2 (24  103 )(982.5  106 )(64  103 ) (25  103 )(982.5  106 )(144.45  103 )  xz    (18.4  106 )(13.1  103 ) (128  106 )(13.1  103 )  6.2609 MPa  2.1160 MPa  4.1449 MPa PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part. 1319 PROBLEM 8.55 (Continued) 11.608  ave   5.804 MPa 2 2  11.608  2 R    (4.1449)  7.1321 MPa  2   max   ave  R  max  12.94 MPa    min   ave  R   min  1.328 MPa    max  R    max  7.13 MPa  PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part. 1320 y PROBLEM 8.56 75 mm a Two forces P1 and P2 are applied as shown in directions a perpendicular to the longitudinal axis of a W310  60 beam. x Knowing that P1  25 kN and P2  24 kN, determine the principal stresses and the maximum shearing stress at point b. b b P2 1.2 m W310 ⫻ 60 0.6 m P1 SOLUTION At the section containing points a and b, M x  (1.8)(25)  45 kN  m M y  (1.2)(24)  28.8 kN  m Vx  24 kN, Vy  25 kN For W310  60 rolled-steel section, d  302 mm, b f  203 mm, t f  13.1 mm, tw  7.49 mm I x  128  106 mm 4  128  106 m 4 , I y  18.4  106 mm 4  18.4  106 m 4 1 Normal stress at point b: x  0, y   d  t f  137.9 mm 2 M x y M y x (45  103 )(137.9  103 ) z    0 Ix Iy 128  106  48.480 MPa Shearing stress at point b: Vy A y  yz   I xt w A  A f  b f t f  2659.3 mm 2  2659.3  106 m 2 1 1 x  0, y   d  t f  144.45 mm 2 2 (25  103 )(2659.3  106 )(144.45  103 )  yz    10.0169 MPa (128  106 )(7.49  103 ) PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part. 1321 PROBLEM 8.56 (Continued) 48.480  ave    24.240 MPa 2 2  48.48  2 R    (10.0169)  26.228 MPa  2   max   ave  R  max  1.988 MPa    min   ave  R   min  50.5 MPa    max  R   max  26.2 MPa  PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part. 1322 20 kips W8 3 28 PROBLEM 8.57 1.6 kips y 4 in. 1.6 kips Four forces are applied to a W8  28 rolled-steel beam as shown. Determine the principal stresses and maximum shearing stress at point a. 5 kips b x 20 in. a 3 in. b a SOLUTION Calculate forces and couples at section containing point of interest. P  20 kips Vx  3.2 kips Vy  5 kips M x  (20)(5)  (4.03)(20)  180.6 kip  in. M y  (20  4)(3.2)  76.8 kip  in. Section properties: A  8.24 in 4 d  8.06 in. b f  6.54 in. t f  0.465 in. tw  0.285 in. I x  98.0 in 4 I y  21.7 in 4 6.54 8.06 Point a: xa    3  0.27 in. ya    4.03 in. 2 2 P M y M x a    x a  y a A Ix Iy 20 (180.6)(4.03) (76.8)(0.27)    8.24 98.0 21.7   2.4272  7.4267  0.95558  4.0439 ksi PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part. 1323 PROBLEM 8.57 (Continued) Shearing stress at point a due to Vx : Shearing stress at point a due to Vy : A  (3)(0.465)  1.395 in 2 A  (3)(0.465)  1.395 in 2 6.54 3 8.06 0.465 x    1.77 in. y    3.7975 in. 2 2 2 2 Q   Ax  2.4692 in 3 Q   Ay  5.2975 in 3 Vx Q (3.2)(2.4692) V yQ (5)(5.2975)  xz    0.78306 ksi  xz    0.58125 ksi I yt (21.7)(0.465) I xt (98.0)(0.465) Combined shearing stress:  a  0.78306  0.58125  0.20181 ksi 4.0439  0  ave   2.0220 ksi 2 2  4.0439  0  2 R    (0.20181)  2.0320 ksi  2   max   ave  R  max  4.05 ksi ◄  min   ave  R  min   0.010 ksi ◄  max  R  max  2.03 ksi ◄ PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part. 1324 20 kips W8 3 28 PROBLEM 8.58 1.6 kips y 4 in. 1.6 kips Four forces are applied to a W8  28 rolled-steel beam as shown. Determine the principal stresses and maximum shearing stress at point b. 5 kips b x 20 in. a 3 in. b a SOLUTION Calculate forces and couples at section containing point of interest. P  20 kips Vx  3.2 kips Vy  5 kips M x  (20)(5)  (4.03)(20)  180.6 kip  in. M y  (20  4)(3.2)  76.8 kip  in. Section properties: A  8.24 in 4 d  8.06 in. b f  6.54 in. t f  0.465 in. tw  0.285 in. I x  98.0 in 4 I y  21.7 in 4 Point b: xb  0 yb  0 P M x yb M y xb b     A Ix Iy 20   0  0  2.4272 ksi 8.24 PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part. 1325 PROBLEM 8.58 (Continued) Shearing stress at point b due to Vx:  xz  0 Shearing stress at point b due to Vy: A(in 2 ) y (in.) Ay (in 3 )  3.0411 3.7975 11.5486  1.01603 1.7825 1.81107  13.3597 Q  Ay  13.3597 in 3 V yQ (5)(13.3597) b    2.3916 ksi I xt w (98.0)(0.285) 2.4272  0  ave   1.2136 ksi 2 2  2.4272  0  2 R    (2.3916)  2   2.6819 ksi  max   ave  R  max  1.468 ksi ◄  min   ave  R  min  3.90 ksi ◄  max  R  max  2.68 ksi ◄ PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part. 1326 B PROBLEM 8.59 a A force P is applied to a cantilever beam by means of a cable attached to a b A bolt located at the center of the free end of the beam. Knowing that P acts in a C h direction perpendicular to the longitudinal axis of the beam, determine (a) the l normal stress at point a in terms of P, b, h, l, and  , (b) the values of  for which the normal stress at a is zero. ␤ P SOLUTION 1 3 1 3 Ix  bh I y  hb 12 12 M (h/2) M y (b/2)   x  Ix Iy 6M x 6M y   bh 2 hb 2 P  P sin  i  P cos  j r  l k M  r  P  l k  ( P sin  i  P cos  j )  Pl cos  i  Pl sin  j M x  Pl cos  M y  Pl sin  6 Pl cos  6 Pl sin  6 Pl  cos  sin   (a)       bh 2 hb 2 bh  h b  cos  sin  b (b)  0  0 tan   h b h b    tan 1    h PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part. 1327 PROBLEM 8.60 l ⫽ 1.25 m a B A vertical force P is applied at the center of the free end of cantilever beam AB. (a) If the beam is installed with the web vertical (  0) and with its longitudinal axis AB horizontal, determine the magnitude of the A force P for which the normal stress at point a is 120 MPa. W250 ⫻ 44.8 (b) Solve part a, assuming that the beam is installed with   3. P ␤ SOLUTION For W250  44.8 rolled-steel section, S x  531  103 mm3  531  106 m3 S y  94.2  103 mm3  94.2  106 m3 At the section containing point a, M x  Pl cos  , M y  Pl sin  Stress at a: Mx My Pl cos  Pl sin       Sx Sy Sx Sy 1  all  cos  sin   Allowable load. Pall     l  S x S y  1 120  106  1  (a)   0: Pall   531  106  0   51.0  103 N 1.25   Pall  51.0 kN  1 120  106  cos 3 sin 3  (b)   3: Pall     39.4  103 N 1.25  531  10 6 94.2  106  Pall  39.4 kN  PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part. 1328 PROBLEM 8.61* B d A 5-kN force P is applied to a wire that is wrapped around bar AB as shown. a Knowing that the cross section of the bar is a square of side d  40 mm, d 2 determine the principal stresses and the maximum shearing stress at point a. A P SOLUTION Bending: Point a lies on the neutral axis.  0 T Torsion:   where a  b  d c1ab 2 and c1  0.208 for a square section. Pd P P Since T  , T  2  2.404 2 . 2 0.416d d Transverse shear: 1 4 V  P I  d 12 1 2 1 1 3 A d y  d Q  Ay  d 2 4 8 t d VQ P V   1.5 2 It d By superposition, P    T  V  3.904 d2 (3.904)(5  103 )   (40  103 )2  12.2  106 Pa  12.2 MPa By Mohr’s circle,  max  12.2 MPa   min  12.2 MPa    max  12.2 MPa  PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part. 1329 PROBLEM 8.62* 3 in. H Knowing that the structural tube shown has a uniform wall thickness of 6 in. 0.3 in., determine the principal stresses, principal planes, and maximum K shearing stress at (a) point H, (b) point K. 4 in. 2 in. 10 in. 0.15 in. 9 kips SOLUTION At the section containing points H and K, V  9 kips M  (9)(10)  90 kip  in. T  (9)(3  0.15)  25.65 kip  in. Torsion: Ꮽ  (5.7)(3.7)  21.09 in 2 T 25.65     2.027 ksi 2t Ꮽ (2)(0.3)(21.09) Transverse shear: QH  0 QK  (3)(2)(1)  (2.7)(1.7)(0.85)  2.0985 in 3 1 1 I  (6)(4)3  (5.4)(3.4)3  14.3132 in 4 12 12 VQK (9)(2.0985) H 0 K    4.398 ksi It (14.3132)(0.3) Bending: Mc (90)(2) H    12.576 ksi, K  0 I 14.3132 PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part. 1330 PROBLEM 8.62* (Continued) 12.576 (a) Point H:  ave   6.288 ksi 2 2  12.576  2 R    (2.027)  6.607 ksi  2   max   ave  R  max  12.90 ksi    min   ave  R   min  0.32 ksi  2  tan 2 p   0.3224   p  8.9, 81.1    max  R  6.61 ksi  max  6.61 ksi  (b) Point K:  0   2.027  4.398  6.425 ksi  max  6.43 ksi    min  6.43 ksi     45   max  6.43 ksi   PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part. 1331 3 in. PROBLEM 8.63* The structural tube shown has a uniform wall thickness of 0.3 in. a Knowing that the 15-kip load is applied 0.15 in. above the base of the tube, determine the shearing stress at (a) point a, (b) point b. 1.5 in. b 4 in. 2 in. 10 in. A 15 kips SOLUTION Calculate forces and couples at section containing points a and b. Vx  15 kips M z  (2  0.15)(15)  27.75 kip  in. M y  (10)(15)  150 kip  in. Shearing stresses due to torque T  M z : Ꮽ  [3  (2)(0.15)][4  (2)(0.15)]  9.99 in 2 Mz 27.75 q   1.3889 kip/in. 2Ꮽ (2)(9.99) q 1.3889 At point a, t  0.3 in.  a    4.630 ksi  t 0.3 q 1.3889 At point b, t  0.3 in.  b    4.630 ksi  t 0.3 PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part. 1332 PROBLEM 8.63* (Continued) Shearing stresses due to Vx : At point a, Part A(in 2 ) x (in.) Ax (in 3 )  0.45 0.75 0.3375  1.02 1.35 1.377  0.45 0.75 0.3375  2.052 Q  Ax  2.052 in 3 t  (2)(0.3)  0.6 in. 1 1 Iy  (4)(3)3  (3.4)(2.4)3  5.0832 12 12 VQ (15)(2.052)   x   10.092 ksi I yt (5.0832)(0.6) At point b, b  0 Combined shearing stresses. (a) At point a,  a  4.630  10.092   5.46 ksi   a  5.46 ksi  (b) At point b,  b  4.630   0  4.63 ksi   b  4.63 ksi  PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part. 1333 3 in. PROBLEM 8.64* For the tube and loading of Prob. 8.63, determine the principal stresses a and the maximum shearing stress at point b. PROBLEM 8.63* The structural tube shown has a uniform wall 1.5 in. b 4 in. thickness of 0.3 in. Knowing that the 15-kip load is applied 0.15 in. 2 in. above the base of the tube, determine the shearing stress at (a) point a, (b) point b. 10 in. A 15 kips SOLUTION Calculate forces and couples at section containing point b. Vx  15 kips M z  (2  0.15)(15)  27.75 kip  in. M y  (10)(15)  150 kip  in. Forces Couples 1 1 Iy  (4)(3)3  (3.4)(2.4)3  5.0832 in 4 12 12 M x (150)(1.5) b   y b    44.26 ksi Iy 5.0832 Shearing stress at point b due to torque M z: Ꮽ  [3  (2)(0.15)][4  (2) (0.15)]  9.99 in 2 Mz 27.75 q   1.38889 kip/in. 2Ꮽ (2)(9.99) q 1.38889     4.630 ksi t 0.3 Shearing at point b due to Vx :  0 PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part. 1334 PROBLEM 8.64* (Continued) Calculation of principal stresses and maximum shearing stress. 44.26  0  ave   22.13 ksi 2 2  44.26  0  2 R    (4.630)  22.61 ksi  2   max   ave  R  max  0.48 ksi    min   ave  R   min  44.7 ksi   max  R  max  22.6 ksi  PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part. 1335 2 kips/ft 12.5 kips PROBLEM 8.65 B C (a) Knowing that  all  24 ksi and  all  14.5 ksi, select the most A D economical wide-flange shape that should be used to support the loading shown. (b) Determine the values to be expected for  m ,  m , and the 9 ft principal stress  max at the junction of a flange and the web of the 3 ft 3 ft selected beam. SOLUTION M D  0:  15RA  (10.5)(9)(2)  (3)(12.5)  0 RA  15.1 kips  RD  15.4 kips  M (57.003  12 kip  in.) S min  max   28.502 in 3  all 24 ksi Shape S (in3) W16  26 38.4  W14  22 29.0 (a) Use W14  22.  W12  26 33.4 W10  30 32.4 W8  35 31.2 For W14  22, Aweb  dtw  (13.7)(0.230)  3.151 in 2 ME (57.003)(12) (b) Point E: m    23.587 ksi  m  23.6 ksi  S 29.0 1 c  d  6.85 in. yb  c  t f  6.515 in. 2 y  6.515  b  b m    (23.587)  22.433 ksi c  6.85  Since  m  0,  max   b  22.4 ksi  M (46.2)(12) Point C: m  C   19.1172 ksi  m  19.12 ksi S 29.0 V 15.4 m    4.8873 ksi  m  4.89 ksi  Aweb 3.151 2 yb  6.515   b  b  m    (19.1172)  18.1823 ksi R 2     m  10.3216 ksi c  6.85   2 b  max  R  max  19.41 ksi  2 For  max , point B controls;  max  22.4 ksi  PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part. 1336 200 mm PROBLEM 8.66 180 mm 160 mm Neglecting the effect of fillets and of stress concentrations, determine 500 N 1250 N the smallest permissible diameters of the solid rods BC and CD. Use  all  60 MPa. A D C B SOLUTION  all  60  106 Pa J  3 M2 T2  c  c 2  all 2 M2 T2 c3  d  2c   all Bending moments and torques. Just to the left of C: M  (500)(0.16)  80 N  m T  (500)(0.18)  90 N  m M 2  T 2  120.416 N  m Just to the left of D: T  90 N  m M  (500)(0.36)  (1250)(0.2)  430 N  m M 2  T 2  439.32 N  m Smallest permissible diameter d BC . (2)(120.416) c3   1.27765  106 m3  (60  106 ) c  0.01085 m  10.85 mm d BC  21.7 mm  Smallest permissible diameter dCD . (2)(439.32) c3   4.6613  106 m3  (60  106 ) c  0.01670 m  16.7 mm dCD  33.4 mm  PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part. 1337 200 mm PROBLEM 8.67 180 mm 160 mm Knowing that rods BC and CD are of diameter 24 mm and 36 mm, 500 N 1250 N respectively, determine the maximum shearing stress in each rod. Neglect the effect of fillets and of stress concentrations. A D C B SOLUTION 1 Over BC: c d  12 mm  0.012 m 2 1 Over CD: c d  18 mm  0.018 m 2 M 2 T2c 2 M2 T2  J  c3 Bending moments and torques. Just to the left of C: M  (500)(0.16)  80 N  m T  (500)(0.18)  90 N  m M 2  T 2  120.416 N  m Just to the left of D: T  90 N  m M  (500)(0.36)  (1250)(0.2)  430 N  m M 2  T 2  439.32 N  m Maximum shearing stress in portion BC. (2)(120.416)  max   44.36  106 Pa  max  44.4 MPa   (0.012) 3 Maximum shearing stress in portion CD. (2)(439.32)   max   47.96  106 Pa   max  48.0 MPa   (0.018)3 PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part. 1338 PROBLEM 8.68 150 mm F The solid shaft AB rotates at 450 rpm and transmits 225 mm 20 kW from the motor M to machine tools connected A to gears F and G. Knowing that  all  55 MPa and 225 mm assuming that 8 kW is taken off at gear F and 12 kW is taken off at gear G, determine the smallest 60 mm 150 mm permissible diameter of shaft AB. M D 100 mm 60 mm E G B SOLUTION 450 f   7.5 Hz 60 Torque applied at D: P 20  103 TD   2 f (2 )(7.5)  424.41 N  m Torques on gears C and E: 8 TC  TD  169.76 N  m 20 12 TE  TD  254.65 N  m 20 Forces on gears: TD 424.41 FD    4244 N rD 100  103 TC 169.76 FC    2829 N Forces in horizontal plane: rC 60  103 TE 254.65 FE    4244 N rE 60  103 Torques in various parts: AC : T 0 CD : T  169.76 N  m DE : T  254.65 N  m EB : T 0 PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part. 1339 PROBLEM 8.68 (Continued) Critical point lies just to the right of D. T  254.65 N  m M y  1007.9 N  m Forces in vertical plane: M z  318.3 N  m  M y2  M z2  T 2  max  1087.2 N  m  all  c J  M y2  M z2  T 2  max J   c3   M y2  M z2  T2  max c 2  all 1087.2  55  106  19.767  103 m3 c  23.26  103 m d  2c  46.5  103 m d  46.5 mm  PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part. 1340 8 in. 8 in. 6 kips PROBLEM 8.69 35⬚ A A 6-kip force is applied to the machine element AB as shown. Knowing that the uniform thickness of the 8 in. element is 0.8 in., determine the normal and shearing B a d stresses at (a) point a, (b) point b, (c) point c. 1.5 in. b e 1.5 in. c f SOLUTION Thickness  0.8 in. At the section containing points a, b, and c, P  6 cos35  4.9149 kips V  6sin 35  3.4415 kips M  (6sin 35)(16)  (6 cos35)(8)  15.744 kip  in. A  (0.8)(3.0)  2.4 in 2 1 I  (0.8)(3.0)3  1.80 in 4 12 P Mc 4.9149 (15.744)(1.5) (a) At point a, x      x  11.07 ksi  A I 2.4 1.80   xy  0  P 4.9149 (b) At point b, x    x  2.05 ksi  A 2.4 3V 3 3.4415    xy      xy  2.15 ksi  2A 2 2.4 P Mc 4.9149 (15.744)(1.5) (c) At point c, x      x  15.17 ksi  A I 2.4 1.80   xy  0  PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part. 1341 PROBLEM 8.70 l A thin strap is wrapped around a solid rod of radius c  20 mm as shown. H Knowing that l  100 mm and F  5 kN, determine the normal and shearing stresses at (a) point H, (b) point K. K c F SOLUTION At the section containing points H and K, T  Fc M  Fl V F   J c4 I c4 2 4 Mc Flc 4 Fl Point H:    I  c4 4  c2 Tc Fc 2 2F    J  c4 4  c2 Point K: Point K lies on the neutral axis.  0 Tc 2F Due to torque:   J  c2 2 3 Due to shear: For a semicircle, Q c , t  d  2c 3 VQ F 23 c3 4F    It  4 c (2c) 3 c 2 4 2F 4F 10 F Combined:    c 2 3 c 2 3 c 2 PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part. 1342 PROBLEM 8.70 (Continued) Data: F  5 kN  5  103 N, l  100 mm  0.100 m c  20 mm  0.020 m (4)(5  103 )(0.100) (a) Point H:    79.6 MPa   (0.020)3 (2)(5  103 )    7.96 MPa   (0.020)2 (b) Point K:  0  (10)(5  103 )    13.26 MPa  3 (0.020)2 PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part. 1343 P P PROBLEM 8.71 R R A close-coiled spring is made of a circular wire of radius r that is formed into a helix of radius R. Determine the maximum shearing stress produced by the two equal and opposite forces P and P. (Hint: First determine the shear V and the torque T in a transverse T cross section.) r V P' SOLUTION  Fy  0: P V  0 V  P  M C  0: T  PR  0 T  PR Shearing stress due to T. Tc 2T 2 PR T    J c 3  r3 Shearing stress due to V. 2 3 For semicircle, Q r , t  d  2r 3 1  For solid circular section, I  J  r3 2 4 V  VQ   4   V 23 r 3  4V  4P 2 It 4 r (2r ) 3 r 3 r 2 By superposition,  max   T   V  max  P(2R  4r/3)/ r 3  PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part. 1344 y PROBLEM 8.72 1 in. Forces are applied at points A and B of the solid cast-iron bracket shown. Knowing that the bracket has a diameter of 0.8 in., determine H the principal stresses and the maximum shearing stress at (a) point H, K x (b) point K. 2500 lb z B A 3.5 in. 2.5 in. 600 lb SOLUTION At the section containing points H and K, P  2500 lb (compression) Vy  600 lb Vx  0 M x  (3.5  1)(600)  1500 lb  in. My  0 M z  (2.5)(600)  1500 lb  in. Forces Couples 1 c d  0.4 in. 2 A   c 2  0.50265 in 2  I c 4  20.106  103 in 4 4 J  2 I  40.212  103 in 4 For semicircle, 2 3 Q c 3  42.667  103 in 3 PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part. 1345 PROBLEM 8.72 (Continued) (a) At point H, P Mc 2500 (1500)(0.4) H     A I 0.50265 20.106  103  24.87  103 psi Tc (1500)(0.4) H   3  14.92  103 psi J 40.212  10 24.87  ave   12.435 ksi 2 2  24.87  2 R    (14.92)  19.423 ksi  2   max   ave  R  max  31.9 ksi   min   ave  R  min  6.99 ksi  1  max  ( max   min )  max  19.42 ksi  2 (b) At point K, P 2500 K    4.974  103 psi A 0.50265 Tc VQ K   J It (1500)(0.4) (600)(42.667  103 )   40.212  103 (20.106  103 )(0.8)  16.512  103 psi 4.974  ave    2.487 ksi 2 2  4.974  R    (16.512) 2  16.698 ksi  2   max   ave  R  max  14.21 ksi   min   ave  R  min  19.18 ksi  1  max  ( max   min )  max  16.70 ksi  2 PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part. 1346 3 kips PROBLEM 8.73 30⬚ K Knowing that the bracket AB has a uniform thickness of 85 in., 2.5 in. A determine (a) the principal planes and principal stresses at B point K, (b) the maximum shearing stress at point K. 2 in. 5 in. SOLUTION Resolve the 3-kip force F at point A into x and y components. Fx   F cos 30  (3) cos 30  2.598 kips Fy  F sin 30  (3)sin 30  1.5 kips At the section containing points H and K, P   Fx  2.598 kips, V  Fy  1.5 kips M  (5)(1.5)  7.5 kip  in. 5 5 Section properties. t  in.  0.625 in., A    (2.5)  1.5625 in 2 , 8 8 1 I  (0.625)(2.5)3  0.8138 in 4 , c  1.25 in. 12 At point k, y  0.75 in. Q  (0.625)(0.50)(1.00)  0.3125 in 3 Stresses at point K. P My 2.598 (7.5)(0.75)       5.249 ksi A I 1.5625 0.8138 VQ (1.5)(0.3125)     0.9216 ksi It (0.8138)(0.625)  x  5.249 ksi,  y  0,  xy  0.9216 ksi Mohr’s circle. X : ( x ,  xy )  (5.249 ksi, 0.9216 ksi) Y : ( y ,  xy )  (0, 0.9216 ksi) C: ( ave , 0)  (2.6245 ksi, 0) x  y  2.6245 ksi 2 R (2.6245)2  (0.9216) 2  2.7816 ksi 0.9216 tan 2 p   0.35112 2 p  19.35 2.6245 (a)  max   ave  R  2.6245  2.7816  max  0.157 ksi at 80.3  (b)  min   ave  R  2.6245  2.7816  min  5.41 ksi at 9.7   max  R  2.7816 ksi  max  2.78 ksi  PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part. 1347 y PROBLEM 8.74 120 kN 50 mm 50 mm 75 mm 75 mm For the post and loading shown, determine the principal stresses, principal 50 kN planes, and maximum shearing stress at point H. C 30⬚ 375 mm H K z x SOLUTION Components of force at point C: Fx  50 cos 30  43.301 kN Fz  50 sin 30  25 kN Fy  120 kN Section forces and couples at the section containing points H and K: P  120 kN (compression) Vx  43.301 kN Vz  25 kN M x  (25)(0.375)  9.375 kN  m My  0 M z  (43.301)(0.375)  16.238 kN  m A  (100)(150)  15  103 mm 2  15  103 m 2 1 I x  (150)(100)3  12.5  106 mm 4  12.5  106 m 4 12 Forces Couples PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part. 1348 PROBLEM 8.74 (Continued) Stresses at point H: P Mxz (120  103 ) (9.375  103 )(50  103 ) H       29.5 MPa A Ix 15  103 12.5  106 3 Vx 3 43.301  103 H    4.33 MPa 2 A 2 15  103 Use Mohr’s circle. 1  ave   H  14.75 MPa 2 2  29.5  2 R    4.33  15.37 MPa  2   a   ave  R  a  30.1 MPa    b   ave  R   b  0.62 MPa  2 H tan 2 p   0.2936  H  a  8.2 b  81.8   max  R  max  15.37 MPa  PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part. 1349 3 in. 6 in. PROBLEM 8.75 600 lb Knowing that the structural tube shown has a uniform wall thickness 600 lb 1500 lb of 0.25 in., determine the normal and shearing stresses at the three 5 in. points indicated. 1500 lb 2.75 in. 3 in. 20 in. 0.25 in. a b c SOLUTION Bending moment Shear bo  6 in. bi  bo  2t  5.5 in. ho  3 in. hi  ho  2t  2.5 in. 1 Ix  (bo ho3  bi hi3 )  6.3385 in 4 12 1 Iz  (hobo3  hibi3 )  19.3385 in 4 12 Normal stresses.  (60)(3) (30)(1.5) At a :   a  16.41 ksi  19.3385 6.3385  M x M z (60)(2.75) (30)(1.5)   z  x At b :   b  15.63 ksi  Iz Ix  19.3385 6.3385  (60)(0) (30)(1.5) At c :   c  7.10 ksi   19.3385 6.3385 PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part. 1350 PROBLEM 8.75 (Continued) Shearing stresses. Point a is an outside corner.  a  0  At point b, Qzb  (1.5)(0.25)(2.875)  1.0781 in 3 VxQz (3)(1.0781)  b, v x    0.66899 ksi I zt (19.3385)(0.25) At point c,  2.75  3 Qzc  Qzb  (2.75)(0.25)    2.0234 in  2  VQ (3)(2.0234)  c,v x  x z   1.256 ksi I zt (19.3385)(0.25) At point b, Qxb  (2.75)(0.25)(1.375)  0.9453 in 3 Vz Q y (1.2)(0.9453)  b, v z    0.71585 ksi I xt (6.3385)(0.25) At point c, (symmetry axis)  c,vz  0 Net shearing stress at points b and c:  b  0.71585  0.66899  b  0.0469 ksi   c  1.256  c  1.256 ksi  PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part. 1351 PROBLEM 8.76 B a The cantilever beam AB will be installed so that the 60-mm side forms an 300 mm b angle  between 0 and 90° with the vertical. Knowing that the 600-N vertical force is applied at the center of the free end of the beam, 40 mm A determine the normal stress at point a when (a)   0, (b)   90. C (c) Also, determine the value of  for which the normal stress at point a is a maximum and the corresponding value of that stress. 60 mm ␤ 600 N SOLUTION 1 Sx  (40)(60) 2  24  103 mm3 6  24  106 m3 1 Sy  (60)(40)2  16  103 mm3 6  16  103 m3 M  Pl  (600)(300  103 )  180 N  m M x  M cos   180 cos  M y  M sin   180 sin  M x M y 180 cos  180 sin  a     Sx Sy 24  106 16  106  3   (7.5  106 )  cos   sin   Pa  2   3   7.5  cos   sin   MPa  2  (a)   0.  a  7.50 MPa   7.50 MPa  (b)   90.  a  11.25 MPa   11.25 MPa  d a  3  (c) Maximum.  7.5   sin   cos    0 d  2  3 3 sin   cos  tan     56.3  2 2  3     a  7.5  cos 56.3  sin 56.3     13.52 MPa   2  PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part. 1352 PROBLEM 8.C1 Let us assume that the shear V and the bending moment M have been determined in a given section of a rolled-steel beam. Write a computer program to calculate in that section, from the data available in Appendix C, (a) the maximum normal stress  m , (b) the principal stress  max at the junction of a flange and the web. Use this program to solve parts a and b of the following problems: (1) Prob. 8.1 (Use V ⫽ 45 kips and M  450 kip  in.) (2) Prob. 8.2 (Use V ⫽ 22.5 kips and M  450 kip  in.) (3) Prob. 8.3 (Use V ⫽ 700 kN and M  1750 kN  m.) (4) Prob. 8.4 (Use V ⫽ 850 kN and M  1700 kN  m.) SOLUTION We enter the given values of V and M and obtain from Appendix C the values of d , b f , t f , t w , I , and S for the given WF shape. d We compute c , yb  c  t f 2 1 M y  y  c  tf , a  , b  a  b  2 S  c  VQ Q  bf t f y, b  It w 1 From Mohr’s circle,  max   b  R 2 2 1 1   max   b    b    b2 2 2  Program Outputs Problem 8.1 Problem 8.2 Given Data: Given Data: V  45 kips, M  450 kip  in. V  22.5 kips, M  450 kip  in. d  9.92 in., b f  7.985 in. d  9.92 in., b f  7.985 in. t f  0.530 in., tw  0.315 in. t f  0.530 in., tw  0.315 in. I  209.00 in.4 , S  42.10 in.4 I  209.00 in.4 , S  42.10 in.4 Answers: Answers: (a)  A  10,688.84 psi (a)  A  10,688.84 psi (b)  m  19,169.08 psi (b)  m  13,073.82 psi PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part. 1353 PROBLEM 8.C1 (Continued) Program Outputs (Continued) Problem 8.3 Problem 8.4 Given Data: Given Data: V  700 kN, M  1750 kN  m V  850 kN, M  1700 kN  m d  9.30 mm, b f  423 mm d  930 mm, b f  423 mm t f  43 mm, tw  24 mm t f  43 mm, tw  24 mm I  8470 (106 mm 4 ) I  8470 (106 mm 4 ) S  18,200 (103 mm3 ) S  18,200 (103 mm3 ) Answers: Answers: (a)  A  96.15 MPa (a)  A  93.40 MPa (b)  m  95.39 MPa (b)  m  96.56 MPa  PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part. 1354 PROBLEM 8.C2 P A cantilever beam AB with a rectangular cross section of width b B A and depth 2c supports a single concentrated load P at its end A. K c Write a computer program to calculate, for any values of x/c and y/c, ␪p (a) the ratios  max / m and  min / m , where  max and  min are the y ␴max ␴min c principal stresses at Point K(x, y) and  m the maximum normal stress in the same transverse section, (b) the angle  p that the principal planes at K form with a transverse and a horizontal plane x through K. Use this program to check the values shown in Fig. 8.8 b and to verify that  max exceeds  m if x 艋 0.544c, as indicated in the second footnote on Page 560. SOLUTION  y Since the distribution of the normal stresses is linear, we have    m   (1) c MC Pxc where m   (2) I I 3 P y2  We use Equation (8.4), Page 498:  1  2  (3) 2 A c   2 y  3 I 1 c Dividing (3) by (2),   m 2 A xc  2 y 1 1 3 I 121 b(2c) 1  c or, since   c2 ,  (4) A b (2c) 3 m 2 x c x y Letting X  and Y  , Equations (1) and (4) yield c c    mY 1 Y 2   m 2X Using Mohr’s circle, we calculate 2 2 1  1 1Y 2  R       2   m Y 2    2  2  X   max 1  YR m 2  min 1  Y R  m 2 PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part. 1355 PROBLEM 8.C2 (Continued)  1 Y 2 1Y 2 tan 2 P     2 2 X ( 2y ) XY 1 1Y 2  P  tan 1    2  XY  Note: For y  0, the angle  P is , which is opposite to what was arbitrarily assumed in Figure P8.C2. Program Outputs x x For  2, For  8, c c x  min  max y  min  max   c m m c m m 1.0 0.000 1.000 0.00 1.0 0.000 1.000 0.00 0.8 –0.010 0.810 6.34 0.8 –0.001 0.801 1.61 0.6 –0.040 0.640 14.04 0.6 –0.003 0.603 3.80 0.4 –0.090 0.490 23.20 0.4 –0.007 0.407 7.35 0.2 –0.160 0.360 33.69 0.2 –0.017 0.217 15.48 0.0 –0.250 0.250 45.00 0.0 –0.062 0.063 45.00 –0.2 –0.360 0.160 –33.69 –0.2 –0.217 0.017 –15.48 –0.4 –0.490 0.090 –23.20 –0.4 –0.407 0.007 –7.35 –0.6 –0.640 0.040 –14.04 –0.6 –0.603 0.003 –3.80 –0.8 –0.810 0.010 –6.34 –0.8 –0.801 0.001 –1.61 –1.0 –1.000 0.000 –0.00 –1.0 –1.000 0.000 –0.00 To check that  max   m if x 艋 0.544c, we run the program for x c  0.544 and for x c  0.545 and observe  max y that m exceeds 1 for several values of c in the first case, but does not exceed 1 in the second case. x For  0.544, c y  min  max  c m m 0.30 –0.700 0.9997 39.92 0.31 –0.690 1.0001 39.72 0.32 –0.680 1.0004 39.51 0.33 –0.670 1.0005 39.30 0.34 –0.660 1.0005 39.09 0.35 –0.650 1.0003 38.88 0.36 –0.640 1.0000 38.66 0.37 –0.630 0.9996 38.44 0.38 –0.619 0.9990 38.21 0.39 –0.608 0.9983 37.98 0.40 –0.598 0.9975 37.74 PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part. 1356 PROBLEM 8.C2 (Continued) Program Outputs (Continued) x For  0.545, c y  min  max  c m m 0.30 –0.698 0.9982 39.91 0.31 –0.689 0.9986 39.71 0.32 –0.679 0.9989 39.50 0.33 –0.669 0.9990 39.29 0.34 –0.659 0.9990 39.08 0.35 –0.649 0.9988 38.87 0.36 –0.639 0.9986 38.65 0.37 –0.628 0.9982 38.42 0.38 –0.618 0.9976 38.20 0.39 –0.607 0.9970 37.96 0.40 –0.596 0.9962 37.73  PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part. 1357 PROBLEM 8.C3 y Disks D1 , D2 , . . . , Dn are attached as shown in Fig. 8.C3 to the solid shaft AB of length L, uniform diameter d, and ci P1 L allowable shearing stress  all . Forces P1, P2, . . . , Pn of known magnitude (except A for one of them) are applied to the disks, Pn either at the top or bottom of its vertical ri diameter, or at the left or right end of its z horizontal diameter. Denoting by ri the D1 radius of disk Di and by ci its distance from D2 B the support at A, write a computer program P2 Di x to calculate (a) the magnitude of the Pi Dn unknown force Pi, (b) the smallest permissible value of the diameter d of shaft AB. Use this program to solve Prob. 8.18. SOLUTION 1. Determine the unknown force Pi by equating to zero the sum of their torques Ti about the x axis. 2. Determine the components ( Fy )i and ( Fz )i of all forces. 3. Determine the components Ay and Az of reaction at A by summing moments about axes Bz  // z and By  // y : 1 M z   0: Ay L  ( Fy )i ( L  ci )  0, Ay   ( Fy )i ( L  ci ) L 1 M y  0: Az L  ( Fz )i ( L  ci )  0, Az   ( Fz )i ( L  ci ) L 4. Determine ( M y )i , ( M z )i and torque Ti just to the left of disk Di : ( M y )i  Az ci   ( F ) c  c 1 k z k i k ( M z )i   Ay ci   ( F ) c  c  k y k i k 1 Ti   T c  c  k k i k 0 where < > indicates a singularity function. 5. The minimum diameter d required to the left of Di is obtained by first computing ( J/c)i from Equation (8.7).  M y i   M z i2  Ti2 2 J c   all  i PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part. 1358 PROBLEM 8.C3 (Continued)  Jc i  12  ci3 , we have ci  2  Jc i and di  4  Jc  1/3 1/ 3 6. Recalling that J  12  c 4 and, thus, that  i This is the required diameter just to the left of disk Di . 7. The required diameter just to the right of disk Di is obtained by replacing Ti with Ti 1 in the above computation. 8. The smallest permissible value of the diameter of the shaft is the largest of the values obtained for Di . Program Output Problem 8.19 Length of shaft  28 in.  (ksi)  8 For Disk 1, Force  0.500 kips Radius of disk  4.0 in. Distance from A  7.0 in. For Disk 2, Force  0.000 kips Radius of disk  6.0 in. Distance from A  14.0 in. For Disk 3, Force  0.500 kips Radius of disk  4.0 in. Distance from A  21.0 in. Unknown force  –0.667 kips AY  0.500 kips, AZ  0.333 kips BY  0.500 kips, BZ  0.333 kips Just to the left of Disk 1: MY  2.3333 kip  in. MZ  –3.5000 kip  in. T  0.0000 kip  in. Diameter must be at least 1.389 in. Just to the right of Disk 1: T  2.00 kip  in. Diameter must be at least 1.437 in. PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part. 1359 PROBLEM 8.C3 (Continued) Program Output (Continued) Just to the left of Disk 2: MY  4.6667 kip  in. MZ  –3.5000 kip  in. T  2.0000 kip  in. Diameter must be at least 1.578 in.  Just to the right of Disk 2: T  –2.00 kip  in. Diameter must be at least 1.578 in. Just to the left of Disk 3: MY  2.3333 kip  in. MZ  –3.5000 kip  in. T  –2.0000 kip  in. Diameter must be at least 1.437 in. Just to the right of Disk 3: T  0.00 kip  in. Diameter must be at least 1.389 in. PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part. 1360 PROBLEM 8.C4 The solid shaft AB of length L, uniform diameter d, and allowable shearing y stress  all rotates at a given speed expressed in rpm (Fig. 8.C4). Gears G1, G2, . . . , Gn are attached to the shaft and L each of these gears meshes with another ci gear (not shown), either at the top or bottom of its vertical diameter, or at the A left or right end of its horizontal ri diameter. One of these gears is z connected to a motor and the rest of them to various machine tools. G1 Denoting by ri the radius of disk Gi, by G2 B ci its distance from the support at A, and Gi x by Pi the power transmitted to that gear Gn (sign) or taken of that gear (sign), write a computer program to calculate the smallest permissible value of the diameter d of shaft AB. Use this program to solve Probs 8.27 and 8.68. SOLUTION 1. Enter w in rpm and determine frequency f  w/60. 2. For each gear, determine the torque Ti  Pi /2 f , where Pi is the power input () or output (–) at the gear. 3. For each gear, determine the force Fi  Ti /ri exerted on the gear and its components ( Fy )i and ( Fz )i . 4. Determine the components Ay and Az of reaction at A by summing moments about axes Bz  // z and By  // y : 1 M z   0: Ay L  ( Fy )i ( L  ci )  0, Ay   ( Fy )i ( L  ci ) L 1 M y  0: Az L  ( Fz )i ( L  ci )  0, Az   ( Fz )i ( L  ci ) L 5. Determine ( M y )i ,( M z )i , and torque Ti just to the left of gear Gi :  ( F ) c  c  ( M y )i  Az ci  k z k i k 1 (M )   A c   ( F ) c  c  z i y i y k i k 1 k T   T c  c  i k i k 0 k where   indicates a singularity function. PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part. 1361 PROBLEM 8.C4 (Continued) 6. The minimum diameter d required to the left of Gi is obtained by first computing ( J/c)i from Equation (8.7).  M y i   M z i2  Ti2 2 J c   all  i 7. Recalling that J  12  c 4 and, thus, that  Jc i  12  ci3 , we have ci  2  Jc  and di  4  Jc  1/3 1/3  i i This is the required diameter just to the left of gear Gi . 8. The required diameter just to the right of gear Gi is obtained by replacing Ti with Ti 1 in the above computation. 9. The smallest permissible value of the diameter of the shaft is the largest of the values obtained for di . Program Outputs Problem 8.25 Problem 8.27   450 rpm   600 rpm Number of Gears: 3 Number of Gears: 3 Length of shaft  750 mm Length of shaft  24 in.   55 MPa   8 ksi For Gear 1, For Gear 1, Power input  –8.00 kW Power input  60.00 hp Radius of gear  60 mm Radius of gear  3.00 in. Distance from A in mm  150 Distance from A in inches  4.0 For Gear 2, FY  0 Power input  20.00 kW Fz  2.100845 Radius of gear  100 mm For Gear 2, Distance from A in mm  375 Power input  –40.00 hp For Gear 3, Radius of gear  4.00 in. Power input  –12.00 kW Distance from A in inches  10.0 Radius of gear  60 mm FY  1.050423 Distance from A in mm  600 FZ  0 AY  –0.849 kN, AZ  4.386 For Gear 3, BY  –3.395 kN, BZ  2.688 Power input  –20.00 hp Just to the left of Gear 1: Radius of gear  4.00 in. MY  657.84 Nm Distance from A in inches  18.0 MZ  127.32 Nm FY  0 T  0.00 Nm FZ  –0.5252113 PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part. 1362 PROBLEM 8.C4 (Continued) Diameter must be at least 39.59 mm. AY  –0.6127 kips, AZ  –1.6194 kips Just to the right of Gear 1: BY  –0.4377 kips, BZ  0.438 kips T  –169.77 N  m Just to the left of Gear 1: Diameter must be at least 40.00 mm. MY  –6.478 kip  in. Just to the left of Gear 2: MZ  2.451 kip  in. MY  1007.98 N  m T  0.000 kip  in. MZ  318.31 N  m Diameter must be at least 1.640 in. T  –169.77 N  m Just to the right of Gear 1: Diameter must be at least 46.28 mm. T  6.3025 kip  in. Just to the right of Gear 2: Diameter must be at least 1.813 in. T  254.65 N  m Just to the left of Gear 2: Diameter must be at least 46.52 mm.  MY  –3.589 kip  in. Just to the left of Gear 3: MZ  6.127 kip  in. MY  403.19 N  m T  6.303 kip  in. MZ  509.30 N  m Diameter must be at least 1.822 in.  T  254.65 N  m Just to the right of Gear 2: Diameter must be at least 40.13 mm. T  2.1008 kip  in. Just to the right of Gear 3: Diameter must be at least 1.677 in. T  0.00 N  m Just to the left of Gear 3: Diameter must be at least 39.18 mm. MY  0.263 kip  in. MZ  2.626 kip  in. T  2.101 kip  in. Diameter must be at least 1.290 in. Just to the right of Gear 3: T  0.000 kip  in. Diameter must be at least 1.189 in.  PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part. 1363 y PROBLEM 8.C5 b My Write a computer program that can be used to calculate the normal and shearing stresses at points with given coordinates y and z located on the surface of a machine part having a rectangular cross section. The Vy internal forces are known to be equivalent to the force-couple system h shown. Write the program so that the loads and dimensions can be C expressed in either SI or U.S. customary units. Use this program to P Vz solve (a) Prob. 8.45b, (b) Prob. 8.47a. x Mz z SOLUTION Enter: b and h b3 h bh3 Program: A  bh Iy  Iz  12 12 For point on surface, enter y and z. Note: y and z must satisfy one of following: h2 b2 y2  and z2  (1) 4 4 b2 h2 or z2  and y 2  (2) 4 4 If either (1) and (2) are satisfied, compute P Myz Mz y    A Iy Iz If z2  b2/4, then point is on vertical surface and h  h 1  h2 y 2  Qz  b   y   y   b     2  2 2  8 2  v y Qz  I zb 2 2 If y  h /4, the point is on horizontal surface, and b  b 1  b2 z 2  Qy  h   z   z   h    2  2 2  8 2  Vz Qy  I yh Force-couple system: P  50 kips Vz  6 kips Vy  2 kips M y  (6 kips) (8.5 in.)  51 kip  in. M z  (2 kips) (10.5 in.)  21 kip  in. PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part. 1364 PROBLEM 8.C5 (Continued) Problem 8.45b Force-Couple at Centroid P  50.00 kips M Y  51.00 kip  in. M Z  21.00 kip  in. VY  2.00 kips VZ  6.00 kips ++++++++++++++++++++++++++ At point of coordinates: y  0.90 in. z  1.20 in.   6.004 ksi   0.781 ksi Force-Couple System P  10 kN Vy  750 N Vz  500 N M y  (500 N)(220 mm)  110 N  m M z  (750 N)(180 mm)  135 N  m Problem 8.47a Force-Couple at Centroid P  10000.00 N M Y  110.00 N  m M Z  135.00 N  m VY  750.00 N VZ  500.00 N ++++++++++++++++++++++++++ At point of coordinates: y  16.00 mm x  0.00 mm   18.392 MPa   0.391 MPa PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part. 1365 PROBLEM 8.C6 A 9 kN d Member AB has a rectangular cross section of 10  24 mm. For the 120 mm loading shown, write a computer program that can be used to determine the normal and shearing stresses at Points H and K for values of d from 0 to 120 mm, using 15-mm increments. Use this program to solve 30⬚ K Prob. 8.35. H 12 mm 12 mm B 40 mm SOLUTION Cross section: Enter A  (0.010 m)(0.024 m)  240  106 m 2 I  (0.010 m)(0.024 m)3 /12  138.24  109 m 4 R  0.5(0.029 m)  12  103 m Compute reaction at A. M B  0: (9 kN)(120  d ) sin 30°  A (120) cos 30°  0 (120 mm  d ) A  (9 kN) tan 30 120 mm Free Body: From A to section containing Points H and K, Define: If d  80 mm, Then STP  1 Else STP  0 Program force-couple system. F   A sin 30  (9 kN) cos 30°(Step) V   A cos 30°  (9 kN) sin 30°(Step) M  A(80 mm) cos 30°  (9 kN)(80 mm  d )sin 30 (STP) At Point H,  H   F/ A 3  H  V/ A 2 At Point K,  K   F/A  Mc/I K  0 PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part. 1366 PROBLEM 8.C6 (Continued) Program Output Problem 8.35 Stresses in MPa d (mm) H H K K 0.0 43.30 0.00 43.30 0.00 15.0 41.95 3.52 65.39 0.00 30.0 40.59 7.03 87.47 0.00 45.0 39.24 10.55 109.55 0.00 60.0 37.89 14.06 131.64 0.00  75.0 36.54 17.58 153.72 0.00 90.0 2.71 7.03 96.46 0.00 105.0 1.35 3.52 48.23 0.00 120.0 0.00 0.00 0.00 0.00 PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part. 1367 y PROBLEM 8.C7* x The structural tube shown has a uniform wall thickness of 0.3 in. H 10 in. A 9-kip force is applied at a bar (not shown) that is welded to the d 3 in. end of the tube. Write a computer program that can be used to 3 in. 4 in. determine, for any given value of c, the principal stresses, principal planes, and maximum shearing stress at Point H for values of d from 3 in. to 3 in., using one-inch increments. Use z this program to solve Prob. 8.62a. 9 kips c SOLUTION Force-couple system. Enter: V  9 kips  M x  (9 kips)(10 in.)  90 kips  in. T  (9 kips)c Area enclosed. a  (a  t )(b  t ) T 9c T   2ta 2ta  T  Shearing stress due to torsion b t  Q  dt    2 2 I  ab3 /12  (a  2t )(b  2t )3 /12 VQ V  It  V  Shearing stress due to V  total   T   Y b Mx   Bending: H  2 I 2 1   Principal stresses:  ave   H ; R   H    total 2 2  2   max   ave  R;  min   ave  R; 2 1   total   H   P  tan 1   ;  max      total 2 2   ave   2  PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part. 1368 PROBLEM 8.C7* (Continued) Rectangular tube of uniform thickness t  0.3 in. Outside dimensions. Horizontal width: a  6 in. Vertical depth: b  4 in. Vertical load: P  9 kips; Line of action at x  c Find normal and shearing stresses at point H. ( x  d , y  b/2) Problem 8.62a Program output for value of c  2.85 in. d  V T  Total  max  min  max P in. ksi ksi ksi ksi ksi ksi ksi Degrees ----------------------------------------------------------------------------------------------------------------------------- 3.00 12.58 3.49 2.03 5.52 14.65 2.08 8.36 18.49 2.00 12.58 2.33 2.03 4.35 13.94 1.36 7.65 16.00 1.00 12.58 1.16 2.03 3.19 13.34 0.76 7.05 12.78  0.00 12.58 0.00 2.03 2.03 12.89 0.32 6.61 8.73  1.00 12.58 1.16 2.03 0.86 12.63 0.06 6.35 3.89 2.00 12.58 2.33 2.03 0.30 12.58 0.01 6.30 1.36 3.00 12.58 3.49 2.03 1.46 12.74 0.17 6.46 6.46  PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part. 1369