304CHAPTER 5 Stresses in Beams Design of Beams P Problem 5.6-1 The cross section of a narrow-gage railway bridge is shown in part (a) of the figure. The bridge is constructed with longitudinal steel girders that support the wood cross ties. The girders are restrained against lateral buckling by diagonal bracing, as indicated by the dashed lines. The spacing of the girders is s1 50 in. and the spacing of the rails is s2 30 in. The load transmitted by each rail to a single tie is P 1500 lb. The cross section of a tie, shown in part (b) of the figure, has width b 5.0 in. and depth d. Determine the minimum value of d based upon an allowable bending stress of 1125 psi in the wood tie. (Disregard the weight of the tie itself.) Solution 5.6-1 Railway cross tie P s2 P s2 Steel rail Wood tie d b Steel girder (b) s1 (a) P Steel rail Wood tie d b s1 (b) (a) s1 50 in. b 5.0 in. s2 30 in. d depth of tie P 1500 lb allow 1125 psi P(s1 s2 ) Mmax 15,000 lb-in. 2 bd 2 1 5d 2 S (50 in.)(d 2 ) d inches 6 6 6 Mmax sallow S15,000 (1125) ¢ Solving, d2 16.0 in. dmin 4.0 in. Note: Symbolic solution: d 2 Problem 5.6-2 A fiberglass bracket ABCD of solid circular cross section has the shape and dimensions shown in the figure. A vertical load P 36 N acts at the free end D. Determine the minimum permissible diameter dmin of the bracket if the allowable bending stress in the material is 30 MPa and b 35 mm. (Disregard the weight of the bracket itself.) 3P(s1 s2 ) bsallow 5b A B 2b D P Solution 5.6-2 Fiberglass bracket DATA P 36 N allow 30 MPa CROSS SECTION d = diameter I b 35 mm d 4 64 MAXIMUM BENDING MOMENT MINIMUM DIAMETER (96)(36 N)(35 mm) 96Pb d3 sallow (30 MPa) 1,283.4 mm3 Mmax P(3b) MAXIMUM BENDING STRESS Mmax c d 3Pbd 96 Pb smax c sallow I 2 2I d 3 5d 2 ≤ 6 dmin 10.9 mm C 2b 5 lb-ft 9450 lb-in. Appendix E. 305 .000 lb-ft 11.000 psi 15. W 8 21 P = 4000 lb Problem 5. Then select an 8-inch wide-flange beam (W shape) from Table E-1.2 in.000 psi q0 28 lb/ft S 24.6-4 q0 21 lb/ft 2 7.3 sallow 16.450 lb-in.2 in.3 in.250 lb-ft 4 8 26.000 psi.250 lb-ft 315.3 in.200 lb-in.3 2 M0 q0 L 787.3 sallow 16. Required S Mmax 227.000 psi use an 8-inch W shape TRIAL SECTION W 8 28 REQUIRED SECTION MODULUS PL qL2 Mmax 15.450 324. 2 Mmax 223.3 18. Solution 5.000 psi.6 Design of Beams P 2500 lb Problem 5. 20. S 14. and recalculate S taking into account the weight of the beam.600 lb-ft 223. Mmax 315. S 19.6-3 A cantilever beam of length L 6 ft supports a uniform load of intensity q 200 lb/ft and a concentrated load P 2500 lb (see figure). 15. Appendix E. q 200 lb/ft L = 6 ft Solution 5.3 in. calculate the required section modulus S.000 lb-ft 3.3 Use W 8 28 Beam is satisfactory. Select a new beam size if necessary.200 4.3 M0 q0 L 378 lb-ft 4536 lb-in.000 lb-in. Calculate the required section modulus S if allow 15.2 in.3 sallow 15.88 in.450 lb-in. Select a new 8-inch beam if necessary.000 psi L 6 ft REQUIRED SECTION MODULUS qL2 Mmax PL 15.6-4 A simple beam of length L 15 ft carries a uniform load of intensity q 400 lb/ft and a concentrated load P 4000 lb (see figure).000 psi TRIAL SECTION W 8 21 S 18.3 24.3 sallow 15.2 in.5 ft q = 400 lb/ft L = 15 ft Simple beam P 4000 lb q 400 lb/ft L 15 ft allow 16. Assuming allow 16.000 psi 20.3 in.600 lb-ft 2 18.000 lb-in. 8 Mmax 315.3 Use Beam is satisfactory.700 lb-in.6-3 Cantilever beam P 2500 lb q 200 lb/ft allow 15.SECTION 5.536 227.000 9.700 lb-in.69 in. Then select a suitable wide-flange beam (W shape) from Table E-1. and recalculate S taking into account the weight of the beam.200 lb-in. Mmax 223. Required S Mmax 324. 400 lb-in.4 lb/ft 2 REQUIRED SECTION MODULUS M0 PL qL2 12. which are called chesses.0 m .4 S 24.7 in. 16. P q B A L — 4 L — 4 Solution 5.3 24. and q 400 lb/ft.3 sallow 15.6-5 q L — 4 L — 4 Simple beam P 2000 lb q 400 lb/ft allow 15.000 psi q0 L 1829 lb-ft 21. P 2000 lb.950 252.0 m 2.000 psi L 24 ft TRIAL SECTION S 10 25.000 psi .200 lb-ft 230.3 Beam is satisfactory.8 in.6-6 A pontoon bridge (see figure) is constructed of two longitudinal wood beams.400 21. known as balks. 15.) Also.0 m long and that the balks are simply supported with a span of 3.400 lb-in.000 psi. that span between adjacent pontoons and support the transverse floor beams.0 kPa ALLOWABLE STRESS: allow 16 MPa Lc length of chesses Balk Lb 3. Then select a suitable I-beam (S shape) from Table E-2.3 q0 25. If the balks have a square cross section. Select a new beam size if necessary.950 lb-in.0 kPa acts over the chesses. L 24 ft.3 sallow 15. Use S 10 25. 8 Mmax 230.36 in.4 Problem 5. The allowable bending stress in the wood is 16 MPa.6-5 A simple beam AB is loaded as shown in the figure on the next page. assume that the chesses are 2.300 lb-in.8 in. Appendix E. and recalculate S taking into account the weight of the beam. assume that a uniform floor load of 8.6-6 Chess Pontoon Balk Pontoon bridge Chess LR 2.0 m Lb length of balks 3. Mmax S Mmax 230. Calculate the required section modulus S if allow 15.0 m.200 lb-ft 4 32 19.306 CHAPTER 5 Stresses in Beams Problem 5. (This load includes an allowance for the weights of the chesses and balks.000 lb-ft 7. Required S Mmax 252.300 lb-in.7 in. 16. what is their minimum required width bmin? Solution 5. For purposes of design.0 m Pontoon FLOOR LOAD: w 8. Solution 5.0 m S ∴ W total load wLb Lc Design of Beams 307 b3 6 qL2b (8. assuming that each joist may be represented as a simple beam carrying a uniform load. The span length L of each joist is 10.) 2 26.333 lb/in.5 106 m3 sallow 16 MPa b3 562. w floor load 120 lb/ft2 0. Mmax 26. measured from center to center (see figure).6-7 A floor system in a small building consists of wood planks supported by 2 in.)(126 in.000 N m 8 8 ˇ ˇ Mmax 9. 19.SECTION 5. q ws 13.0 m) 2 9.0 kNm)(3.6 LOADING DIAGRAM FOR ONE BALK Section modulus S KN q 8.460 lbin..0 kPa)(2.3 sallow 1350 psi From Appendix F: Select 2 10 in. and the allowable bending stress in the wood is 1350 psi.5 ft. bmin 0.0 kN/m Problem 5.333 lbin.150 m 150 mm wLc W q 2Lb 2 (8.6-7 Planks s s L Joists s Floor joists q Mmax qL2 1 (13. which includes an allowance for the weight of the floor system itself. joists .000 N m 562.0 m) 2 8. The uniform floor load is 120 lb/ft2. the spacing s of the joists is 16 in.8333 lb/in. (nominal width) joists spaced at distance s. 8 8 Required S L 10.5 106 m3andb3 3375 106 m3 6 Solving.0 m b Mmax b Lb 3.460 lb-in.5 ft allow 1350 psi L 10. Calculate the required section modulus S for the joists.5 ft 126 in. and then select a suitable joist size (surfaced lumber) from Appendix F.2 s spacing of joists 16 in.6 in. 6-8 Spacing of floor joists Planks h = 180 mm s s L Joists b = 40 mm s L 4.6-8 The wood joists supporting a plank floor (see figure) are 40 mm 180 mm in cross section (actual dimensions) and have a span length L 4. (Assume that each joist may be represented as a simple beam carrying a uniform load.308 CHAPTER 5 Stresses in Beams Problem 5. which includes the weight of the joists and the floor.0 m w floor load 3.450 m 450 mm .6 kPa) (4.0 m) 2 0.0 m q ws S bh2 6 Mmax S SPACING OF JOISTS qL2 wsL2 8 8 Mmax wsL2 bh2 sallow 8sallow 6 smax 4 bh2sallow 3wL2 Substitute numerical values: 4(40 mm)(180 mm) 2 (15 MPa) smax 3(3. Calculate the maximum permissible spacing s of the joists if the allowable bending stress is 15 MPa. The floor load is 3.0 m.) Solution 5.6 kPa s spacing of joists allow 15 MPa q L 4.6 kPa. (q q0 )L2 2Mallow qallow q0 2 L2 qallow 2Mallow 2(29.1 m long and has a cross section in the shape of a regular octagon. Mallow 29. Determine the allowable uniform load qallow if the distance L equals 3.94 in.) 2 Mmax occurs at support B.5 lbin. respectively. 10. Tension governs. The allowable stresses in tension and compression are 18 ksi and 12 ksi. C B L C 2. The design load is 1.033 in. (Assume that the ends of the bar are simply supported and that the weight of the bar is negligible.850 lb-in. qallow (43. The beam supports its own weight (30 lb/ft) plus a uniform load of intensity q acting on the overhang.4 ) 72.750 lb-in. c2 0.94 in. 0. st I (18 ksi)(3. Solution 5.384 in. Allowable bending moment based upon compression q load on overhang Mc L length of overhang 3.0 ft. The bar is 2.41)(12) 521 lb/ft Problem 5.41 lb/in.4 ) 29. and the allowable bending stress is 200 MPa.6 Problem 5. Determine the minimum height h of the bar.) q0 2.750 lb-in.384 in.384 in. 2 L (36 in. c1 2.6-10 A so-called “trapeze bar” in a hospital room provides a means for patients to exercise while in bed (see figure).2 kN applied at the midpoint of the bar.6-9 A beam ABC with an overhang from B to C is constructed of a C 10 30 channel section (see figure).SECTION 5.649 in.649 in. compression on bottom.5 lb/in.0 in. Mmax 45. Mt c2 0.) C h .94 in.5 43.0 ft = 36 in. I 3.6-9 Beam with an overhang DATA C 10 30 channel section Allowable bending moment based upon tension c1 2.91 2.649 in. Design of Beams 309 q A L 3. Allowable bending moment ALLOWABLE STRESSES t 18 ksi c 12 ksi sc I (12 ksi)(3.4 (from Table E-3) q0 weight of beam ABC 30 lb/ft 2. Allowable uniform load q Maximum bending moment Mmax (q q0 )L2 2 Tension on top.750 lb-in. and the wheelbase of the carriage is 5 ft. Appendix E. The weight of the beam itself may be disregarded. (b) Select a suitable I-beam (S shape) from Table E-2.643 mm Problem 5.6-10 Trapeze bar (regular octagon) P L 2 C h b L P 1.310 CHAPTER 5 Stresses in Beams Solution 5.1 m allow 200 MPa Determine minimum height h.2 kN L 2.85948b4 192 ˇ IC ¢ S¢ 11 82 4 42 5 4 ≤b ¢ ≤h 12 12 42 5 3 3PL ≤ h h3 6 2(42 5)sallow Substitute numerical values: h3 28.41421h)4 0. 4000 lb 5 ft A 2000 lb B 16 ft .41421) [3(2.7735 106 m3 h 0.030643 m s ˇ b C 2 h 2 b length of one side 360 360 45 n 8 b b tan (from triangle) b 2 h 2 b h h cot 2 b 2 b hmin 30. The load transmitted to the beam from the front axle is 2000 lb and from the rear axle is 4000 lb. with n 8 Minimum height h M M S s S 630N m 0.0 ksi.15 106 m3 200 MPa h3 28.41421h PL (1.41421 h 2 h 45 cot 2. (a) Determine the minimum required section modulus S for the beam if the allowable bending stress is 15. the length of the beam is 16 ft.109476h3 3.1 m) 630 N m 4 4 ˇ S IC 0.6-11 A two-axle carriage that is part of an overhead traveling crane in a testing laboratory moves slowly across a simple beam AB (see figure).85948(0.41421 b 2 Moment of inertia IC b b nb4 ¢ cot ≤ ¢ 3 cot2 1 ≤ 192 2 2 IC 8b4 (2. IC 1.054738h4 0.7735 106 m3 hmin 30.6 mm Alternative solution (n 8) M b b PL b 45tan 2 1cot 2 1 4 2 2 b ( 2 1)hh ( 2 1)b C For 45º: b 45 tan 0. Case 25.41421) 2 1] 1.2 kN)(2.054738h4 Section modulus Maximum bending moment Mmax b 0.109476h3 h2 h2 ˇ Properties of the cross section Use Appendix D. M lb-ft) Maximum bending moment dM Set equal to zero and solve for x xm.45m)2d 2 32 (400 N)(0. 260 lb-ft 231.2 in. B d P L Minimum diameter Mmax allow S PL gd 2L2 d 3 sallow ¢ ≤ 8 32 Rearrange the equation: 32 PL 0 (Cubic equation with diameter d as unknown.833465 0 Solve the equation numerically: d 0.41 in.000d 3 62.) sallow d 3 4g L2 d 2 Maximum bending moment Mmax PL Design of Beams qL2 gd 3L2 PL 2 8 Section modulus S d 32 3 Substitute numerical values (d meters): (60 106 N/m2)d 3 4(77.1667 ft dx 6 RA (x ft.6 Solution 5.6-12 A cantilever beam AB of circular cross section and length L 450 mm supports a load P 400 N acting at the free end (see figure).37d 2 1.6-12 Cantilever beam DATA L 450 mm P 400 N allow 60 MPa weight density of steel 77.000 psi (b) Select on I-beam (S shape) Table E-2.3 sallow 15. dx dM 43 125(43 6x) 0x xm 7.031614 m dmin 31. considering the effect of the beam’s own weight.61 mm . The beam is made of steel with an allowable bending stress of 60 MPa. Determine the required diameter dmin of the beam. 15. (a) Minimum section modulus Smin Mmax 231.45 m) 0 60.SECTION 5.130 lb-in.6-11 311 Moving carriage P2 P1 Bending moment under larger load P2 d x M RA x 125(43x 3x2) A B L P1 load on front axle 2000 lb P2 load on rear axle 4000 lb L 16 ft d 5 ft allow 15 ksi x distance from support A to the larger load P2 (feet) Lx Lxd RA P2 ¢ ≤ P1¢ ≤ L L x x 5 (4000 lb) ¢ 1 ≤ (2000 lb) ¢ 1 ≤ 16 16 16 125(43 3x) (x ft.130 lb-in. RA lb) Mmax (M) xxm 125 B (43) ¢ Solution 5.000 N/m3)(0.0 kN/m3 Weight of beam per unit length d 2 q g¢ ≤ 4 43 43 2 ≤ 3¢ ≤ R 6 6 19. Select S 8 23 (S 16.3) A Problem 5. Each beam has length L1 2. 5(72 in. (This load accounts for all loads except the weights of the cantilever beams. ∴ q w¢ L2 2. and D and has a splice (represented by the pin connection) at point C. Assuming that the middle cantilever supports 50% of the load and each outer cantilever supports 25% of the load.5 kPa acting over the entire floor area. B. The design load is 5.6-14 A small balcony constructed of wood is supported by three identical cantilever beams (see figure).5 kPa) ¢ ≤ 6875 Nm 2 2 b L1 2.800 psi. 2(10.5 m.5 kPa 5. Solution 5. width b.312 CHAPTER 5 Stresses in Beams Problem 5. The distance a 6.) 2 922 lb/ft qallow 922 lb/ft 57 lb/ft 865 lb/ft 2a 5qa2 2 Problem 5.2 in.3 ) 76.3 Allowable uniform load 121 qa2 128 B qmax 5q a2 sallow S 2 Data: a 6 ft 72 in.0 ft and the beam is a W 16 57 wide-flange shape with an allowable bending stress of 10.833 lbin.800 psi)(92.1 m L 2 2. Find the allowable uniform load qallow that may be placed on top of the beam.5 m ≤ (5. Solution 5.5 kNm2 )b2 3 3 7333b2 (N/m) (b meters) .800 psi S 92.) The allowable bending stress in the cantilevers is 15 MPa.5 m Floor dimensions: L 1 L 2 Design load w 5. and height h 4b/3.5 kN/m3. with L2 2.6-13 A compound beam ABCD (see figure) is supported at points A. taking into account the weight of the beam itself.5 kN/m3 (weight density of wood beam) allow 15 MPa Weight of beam q0 gbh 4gb2 4 (5.2 in.6-13 A B 4a C a 11qa RA 8 RB D 4a C D a 4a 45qa 8 C qmax D 2a qallow qmax (weight of beam) W 16 57 RD 2qa 2qa2 11a 8 2sallow S 5a2 allow 10. The dimensions of the balcony floor are L1 L2. which have a weight density 5.6-14q Cantilever beam for a balcony 4b h= — 3 L1 B Pin Mmax Pin A A Compound beam q M q 4b h= — 3 L2 b L1 Middle beam supports 50% of the load. determine the required dimensions b and h. 4a Pin connection at point C.1 m. 590b2 409. 16 in. B 1.1517 m S h 4b 0.25)(1.2 C A2 B A1 1.5 b 10.6 Maximum bending moment Rearrange the equation: (q 1 (6875 Nm 7333b2 )(2.25 in.75)(24) 130.5)(15) (0.6-15 A beam having a cross section in the form of an unsymmetric wide-flange shape (see figure) is subjected to a negative bending moment acting about the z axis.375b (in.170b2 (15 106 Nm2 ) ¢ ≤ 27 b 152 mm Problem 5.5b (in.375b)(7) 82. z C 12 in.2) b C1 1.42857 in.25) 15 in.5) 24 in.5 21.375b 45 in. 1.5 21.159 16. A 39 1. h 202 mm y b 1.2023 m 3 Required dimensions 8b3 15.5b 7 c1 4 sbottom c2 3 c2 4 60 c1 h 8.5 in.300 0 Solve numerically for dimension b b 0. 7 7 Solve for b stop 3 45 c2 h 6.5b)(45) (130.57143 in.3) Distance c2 from line B-B to the centroid C QBB 130. respectively. 1.SECTION 5. Locate centroid A A1 A2 A3 39 1.6-15 Unsymmetric wide-flange beam y Areas of the cross section (in.5 in.5b) (7.5b (39 1.125b 841.25 in.5 in. 16 in.170b2 (N m) q0 )L21 Mmax Design of Beams bh2 8b3 6 27 Mmax allow S (120 106)b3 436.25 in.5 in. 313 . Solution 5. 7 7 A2 (12)(1.2 A3 (16)(1. Find b (inches) h height of beam 15 in.1 m) 2 2 2 15. A1 z 12 in.159 16. Determine the width b of the top flange in order that the stresses at the top and bottom of the beam will be in the ratio 4:3.2) First moment of the cross-sectional area about the lower edge B-B QBB a yi Ai (14.5 21. Stresses at top and bottom are in the ratio 4:3. C2 A3 1. Calculate the thickness t of the channel in order that the bending stresses at the top and bottom of the beam will be in the ratio 7:3. respectively.6-16 A beam having a cross section in the form of a channel (see figure) is subjected to a bending moment acting about the z axis.314 CHAPTER 5 Stresses in Beams Problem 5. and (3) a circle (see figures).6-16 Channel beam Areas of the cross section (mm 2) A1 ht 50t A2 b1 t 120t 2t 2 A 2A1 A2 220t 2t 2 2t(110t) y z A1 A2 c1 c2 C A1 t B B t h 50 mm b1 t b 120 mm First moment of the cross-sectional area about the lower edge B-B t t QBB a yi Ai (2) ¢ ≤ (50 t) ¢ ≤ (b1 )(t) 2 2 t 2(25)(50t) ¢ ≤ (120 2t)(t) 2 t (2500 60t t 2) (t mm. (2) a square. are subjected to the same maximum bending moment. are made of the same material.6-17 Determine the ratios of the weights of three beams that have the same length. y t t C z t 50 mm 120 mm Solution 5. c2 Q BB t(2500 60t t 2 ) A 2t(110 t) 2500 60t t 2 15 mm 2(110 t) Determine the thickness t. Locate centroid stop c1 7 sbottom c2 3 c1 7 h 35 mm 10 c2 3 h 15 mm 10 Solve for t 2(110 t)(15) 2500 60t t 2 t 2 90t 800 0 t 10 mm Problem 5. Q mm3) t thickness (constant) (t is in millimeters) b1 b 2t 120 mm 2t Distance c2 from line B-B to the centroid C Stresses at the top and bottom are in the ratio 7:3. and have the same maximum bending stress if their cross sections are (1) a rectangle with height equal to twice the width. h = 2b b a a d . 6-18 A horizontal shelf AD of length L 900 mm.6905 S 23 4 4 (3) Circle: S Weights are proportional to the cross-sectional areas (since L and are the same in all 3 cases). The brackets are adjustable and may be placed in any desired positions between the ends of the shelf.3019S 2/3 (2) Square: S d 3 32S 13 d ¢ ≤ 32 d 2 32S 23 A3 ¢ ≤ 3. Mmax. and max are the same in all three beams. place the supports so that M1 ƒ M2 ƒ .6905 W1 : W2 : W3 1 : 1. Determine the maximum permissible value of the load q if the allowable bending stress in the shelf is allow 5. t A B D C b L (a) q A D B C L (b) Solution 5. and thickness t 20 mm is supported by brackets at B and C [see part (a) of the figure]. Let x length of overhang qL qx2 M1 (L 4x)ƒ M2 ƒ 8 2 2 qL qx ∴ (L 4x) 8 2 qL2 (3 212) 8 Mmax sallow S sallow ¢ Eq. which includes the weight of the shelf itself. (1) and (2) and solve for q: qmax 4bt 2sallow 3L2 (3 212) Substitute numerical values: qmax 5. (2) Equate Mmax from Eqs.3019 : 3. A uniform load of intensity q. width b 300 mm. bh2 2b3 3S 13 (1) Rectangle: S b ¢ ≤ 6 3 2 23 3S A1 2b2 2 ¢ ≤ 2. M S section modulus S s Since M and are the same. acts on the shelf [see part (b) of the figure].0 MPa and the position of the supports is adjusted for maximum load-carrying capacity. the section moduli must be the same. .6-17 Ratio of weights of three beams Beam 1: Rectangle (h 2b) Beam 2: Square (a side dimension) Beam 3: Circle (d diameter) L.6207 : 3.260 : 1.6 Solution 5.76 kN/m .6207S 23 2 315 Design of Beams a3 a (6S) 13 6 A2 a2 (6S)2/3 3.6-18 Shelf with adjustable supports q t A D B C L (a) L ( 12 1) 2 Substitute x into the equation for either M1 or ƒ M2 ƒ : M1 Mmax x Solve for x: x C B A b M2 M2 (b) D x L 900 mm b 300 mm t 20 mm allow 5.SECTION 5.0 MPa For maximum load-carrying capacity. (1) bt 2 ≤ 6 Eq.408 Problem 5. W1 : W2 : W3 A1 : A2 : A3 A1 : A2 : A3 2. 6-19 A steel plate (called a cover plate) having crosssectional dimensions 4.057 in.)(4. Problem 5.6-20 A steel beam ABC is simply supported at A and B and has an overhang BC of length L 150 mm (see figure on the next page).00 in.5 in. The beam supports a uniform load of intensity q 3.3 c2 7. 0. The cross section of the beam is rectangular with width b and height 2b.0 in.5 y 5.25 in.316 CHAPTER 5 Stresses in Beams Problem 5.3 Q1 13.5 in.3 in. (b) Taking into account the weight of the beam.2 c1 6.6-19 4.5)(6. A0 10.4 (12. is welded along the full length of the top flange of a W 12 35 wide-flange beam (see figure.3 in.5 in.8 in. A1 12.5 in.4 Section modulus (Use the smaller of the two section moduli) Iz 355.307 in. calculate t he required width b of the rectangular cross section. cover plate c1 L W 12 35 Beam with cover plate y z 4. 0.85 in.25 in.307 in. which shows the beam cross section).6 A1 A0 (4.3 in.6 in. Beam with cover plate (z axis is centroidal axis) Increase in section modulus S1 48.8% First moment with respect to axis L-L: Q1 a yi Ai (6.5 in.8 in.) 13.4 Moment of inertia about z axis: ILL Iz A1 y 2Iz I11 A1 y 2 Iz 369.3 y 1. calculate the required width b.25 y 7. What is the percent increase in section modulus (as compared to the wide-flange beam alone)? Solution 5.25 0.0 0.068 S0 45.2 I0 2.25 L 6.0 in.69 1.69 in.693 in.5) 3 (4.0 0.2)(1.057 in.50 in.)2 355.5 in. The allowable bending stress in the steel is allow 60 MPa and its weight density is 77.25 All dimensions in inches.4 S1 48.) 12. cover plate –y C c2 6. q C A 2b B 2L L b .)(0.00 in.2 Percent increase 6.)(0.3 Moment of inertia about axis L-L: 1 ILL I0 (4.0 kN/m3.0)(0.0)(0.4 S0 45.25) 2 12 369.5 kN/m over its entire length of 450 mm.25 0.5 in. c2 6. Wide-flange beam alone (Axis L-L is centroidal axis) W 12 35 d 12.0 in.3 in. (a) Disregarding the weight of the beam. 395b2 236.5 kN/m allow 60 MPa 77.6-20 Design of Beams 317 Beam with an overhang q C A 2b B Substitute numerical values: 3(3.96 mm 3 in. as shown in the figure.5 kNm)(150 mm) 2 b3 0. diam. calculate the maximum permissible spacing s of the piles. 12 in.00995 m 9. Top view p2 = 400 lb/ft2 Side view .SECTION 5. The lateral earth pressure is p1 100 lb/ft2 at the top of the wall and p2 400 lb/ft2 at the bottom.95 mm qL2 2b3 sallow ¢ ≤ 2 3 3qL2 4sallow Problem 5.00996 m 9.) q0 weight of beam per unit length q0 (b)(2b) 2b 2 Mmax S 2b3 3 (q q0 )L2 1 (q 2g b2 )L2 2 2 Mmax allow S 1 2b3 (q 2g b2 ) L2 sallow ¢ ≤ 2 3 Rearrange the equation: 4allow b 3 6L2 b2 3qL2 0 Substitute numerical values: (240 106)b3 10.98438 106 m3 4(60 MPa) b 2L L 9qL RB 4 3qL RA 4 (b) Include the weight of the beam RB M 9qL2 32 C B 0 A qL2 2 Mmax L 150 mm q 3. Assuming that the allowable stress in the wood is 1200 psi. To be on the safe side.6 Solution 5. p1 = 100 lb/ft2 12 in. (Hint: Observe that the spacing of the piles may be governed by the load-carrying capacity of either the planks or the piles. diameter (actual dimension).6-21 A retaining wall 5 ft high is constructed of horizontal wood planks 3 in.0 kN/m3 qL2 bh2 2b3 S 2 6 3 (a) Disregard the weight of the beam Mmax allow S b3 b 0.25 0 (b meters) Solve the equation: b 0. diam. thick (actual dimension) that are supported by vertical wood piles of 12 in. and consider the planks to act as simple beams between the piles. s 5 ft 3 in. Consider the piles to act as cantilever beams subjected to a trapezoidal distribution of load. assume that the pressure on the bottom plank is uniform and equal to the maximum pressure. even though the area of the cross section is reduced. 16h2 (2p1 p2 ) Plank governs smax 72. we can increase the section modulus and obtain a stronger beam. (a) Determine the ratio defining the areas that should be removed in order to obtain the strongest cross section in bending.318 CHAPTER 5 Stresses in Beams Solution 5.778 lb/in.2 q1 p1 s q2 p2 s d diameter of pile 12 in. B 3p2 q2 Divide the trapezoidal load into two triangles (see dashed line).4 in. (b) By what percent is the section modulus increased when the areas are removed? y a a z C a a . By removing a small amount of material at the top and bottom corners.6944 lb/in. Problem 5.6-21 Retaining wall q1 t (1) Plank at the bottom of the dam t thickness of plank 3 in. b width of plank (perpendicular to the plane of the figure) p2 maximum soil pressure 400 lb/ft 2 2.0 in. (2) Vertical pile h 5 ft 60 in. as shown by the shaded triangles in the figure.6-22 A beam of square cross section (a length of each side) is bent in the plane of a diagonal (see figure). 1 2h 1 h sh2 Mmax (q1 )(h) ¢ ≤ (q2 )(h) ¢ ≤ (2p1 p2 ) 2 3 2 3 6 d 3 S Mmax allow S or 32 sh2 d 3 (2p1 p2 ) sallow ¢ ≤ 6 32 Solve for s: 3 sallow d 3 s 81.0 in. p1 soil pressure at the top 100 lb/ft2 0.2 s spacing of piles q p2 b allow 1200 psi S section modulus Mmax qs2 p2 bs2 8 8 Mmax allow S or S bt 2 6 p2 bs2 bt 2 sallow ¢ ≤ 8 6 Solve for s: s h q s 4 sallow t 2 72. (1).1 . Reduced cross section (Area qmm. m. it is planned to add structural projections of width b/9 and height d to the top and bottom of the beam [see part (b) of the figure]. (1) a P P1 a Eq.6 Solution 5.10 h1 (S) S0 max 1.3 Entire cross section (Area D) I0 I0 a3 12 a4 a c0 S0 c0 12 12 12 Square mnpq (Area 1) I1 (1 b) 4a4 12 Parallelogram mm.00 0 a length of each side a amount removed Beam is bent about the z axis. For what values of d is the bending-moment capacity of the beam increased? For what values is it decreased? b — 9 d h b (a) h d b — 9 (b) . (1) (a) Value of for a maximum value of S/S0 d S ¢ ≤0 db S0 Take the derivative and solve this equation for .6-22 removed Beam of square cross section with corners y a (1 ) a am m 1 Ratio of section moduli S (1 3b)(1 b) 2 S0 q z 319 Design of Beams h C 1.0535 The section modulus is increased by 5. p. (1) 1.50 0.0535 Eq.35% when the triangular areas are removed. n. n (Area 2) 1 I2 (base)(height)3 3 (1 b)a 3 ba4 1 I2 (ba12) B R (1 b) 3 3 6 12 Graph of EQ. 0. pq) a4 I I1 2I2 (1 3b)(1 b) 3 12 c (1 b)a 12 I 12 a3 S (1 3b)(1 b) 2 c 12 Problem 5. (S/S0)max 1. 1 b 9 (b) MAXIMUM VALUE OF S/S0 Substitute 1/9 into Eq.SECTION 5.6-23 The cross section of a rectangular beam having width b and height h is shown in part (a) of the figure. For reasons unknown to the beam designer.2 19 0. 6861 and 0 h 3 S2 d 1 41 is minimum when 0.320 CHAPTER 5 Solution 5.000 0.6861 and h d 0 h d h S2 S1 0 0.0 d h Moment capacity is increased when d 7 0.2963 S2 S1 1.2937 0 0.75 1.50 0.0500 1.25 0.0 0.6861 h Notes: S2 2d 3 2d 1 when ¢ 1 ≤ 9 ¢ 1 ≤ 8 0 S1 h h d or 0.2937 S1 h 2 ¢ S2 ≤ 0.5 1.5 0.00 1.6861 0.8889 1. S1 h d 0.8399 S1 min .6861 h Moment capacity is decreased when d 6 0.8426 0.6-23 Stresses in Beams Beam with projections S2 d versus S1 h Graph of d 1 2 h b h d b — 9 (a) (b) (1) Original beam I1 bh2 bh3 h I1 c1 S1 c1 12 2 6 (2) Beam with projections 1 8b 1 b I2 ¢ ≤ h3 ¢ ≤ (h 2d) 3 12 9 12 9 b [8h3 (h 2d) 3 ] 108 h 1 c2 d (h 2d) 2 2 I2 b [8h3 (h 2d) 3 ] S2 c2 54(h 2d) Ratio of section moduli 2d 3 ≤ S2 b[8h (h 2d) ] h S1 2d 9(h 2d)(bh2 ) 9 ¢1 ≤ h 3 3 8 ¢1 Equal section moduli S2 d Set 1 and solve numerically for .