IS 800 - 19841. A 10 mm thick Gusset plate is connected to 6 mm angle section by Lap Joint. Find the rivet value of 16 mm dia of power drivern Rivets Givert Dia – 16mm Dia of rivet hole – 16 + 1.5 = 17.5 mm Permissible Stresses For Power driven rivet (Table 8.1 Page 95. IS 800 – 1984) τ vf = 100 N/mm 2 bt = 300 N/mm 2 (i) Strength in shearing = 4 2 πd x vf τ = 4 17.5 x x 100 2 π = 240252.82 N = 24.052 KN. (ii) Strength in bear = τ bft x d x t = 300 x 17.5 x 10 = 52. 5 KN (iii) Rivet Value (R) = Least of the strength in shearing (or) bearing Rivet value (R) = 24.052 KN 2. Find the value of the 16 mm power driven rivets connected a pair of double angle section consisting of ISA 75 x 75 x 6 mm through 10 mm thick Gusset plate. Find the Rivet value. Given (D) Dia = 16 mm (d) Dia of rivet hole = 16 + 1.5 = 17.5 mm vf = 100 N/mm 2 τ bf = 300 N/m [Refer Table 8.1 Page 95 – IS 800 – 1984] (ii) Rivet value (i) Strength of the rivet in double shearing = ] ] ] 4 πd x τ 2 2 vf 1 IS 800 - 1984 = ] ] ] 4 17.5 x π x 100 2 2 = 48.104 KN (ii) Strength of the rivet in bearing = τ bt x d x t = 300 x 17.5 x 6 = 31.50 KN (iii) Rivet value = 31.50 N (Least one) 3. Find the efficiency of the joint in a boiler. Shell connected using 16 mm dia of the rivet at a pitch of 60 mm C/C in a single riveted Lap Joint thickness of the plate is 8 mm. The rivets are power driven shop rivet Given (1) D = 16 mm d = 16 + 1.5 mm = 17.5 mm (2) Pitch Distance = 60 mm C/C (3) PDS - Rivets - vs = 100 N/mm 2 bt = 300 N/mm 2 τ at = 100 N/mm 2 (4) Thickness of plate = 8mm (i) plate solid the of Strength Tearig , Bearing , shearing the of Least · η (i) Strength in shearing = vf 4 17.5 x π x 100 4 πd x 2 2 · = 24.052 KN (ii) Strength in bearing = bt x d x t = 300 x 17.5 x 8 mm = 42.00 KN (iii) Strength in Tearing = σ at (P- d) t = 100 (60 – 17.5) x 8 = 42.5 KN x 8 = 34.00 KN plate solid of Strength 24.052 η · (iv) Strength of the solid plate = at x P x t = 100 x 60 x 8 2 IS 800 - 1984 = 48 KN (v) % 1 . 50 100 x 48 24.052 η · · 4. A tie member ISA 90 x 90 x 6 mm carning an axial tension of 40 KN is connected to a Gusset plate 10 mm thick design the Joint & sketch the arrangement of rivet. Given Angle section = ISA 90 x 90 x 6 mm Load (P) = 40 KN Thickness of plate = 10 mm Solution Step 1 Assume dia of rivet Assume Take Dia (D) = 12 mm Dia of hole = 12 + 1.5 = 13.5 mm Step 2 Find the value of rivet Assume Hand driven rivet vf = 80 N/mm 2 Refer table 8.1 τ bf = 250 N/mm 2 Page 95 tf = 80 N/mm2 IS 800 - 1984 (i) Strength in shearing = τ vf x 4 πd 2 = 80 x 4 13.5 x π 2 = 11.45 KN (ii) Strength in bearing = bf x d x t = 250 x 13.5 x 10 = 33. 75 KN (iii) Rivet value = Least value of shearing & bearing Rivet value (R) = 11.45 KN Step 3 Number of Rivet No of Rivet = value) (Rivet R load) (6 P 3 IS 800 - 1984 = Nos 4 3.49 11.45 40 ≅ · Step 4 Arrangement of rivet Edge distance (d) (i) d = 13.5 => Edge Distance = 19 mm [Refer Table 8.2] Pitch Distance (ii) Min = 2.5 x 12 (D) = 300 mm ≅ 50 mm (iii) Maxi = 16 t (or) 200 (whichever is less) = 16 x 6 = 96 mm (or) 200 (Take whichever is less) Maxi = 96 mm ≅ 100mm Step 5: plate solid the of Strength bearing & bearing , Shearing the of Least · η 100 x p d P − · 100 x 50 5 . 13 50 − · η = 73% 6. Two plates 6 mm tk are Jointed by 14 mm dia of the rivet in a triple straggled rivet Lap Joint as Shown in diagram in what way the Joint will failed. If allowable tensile stress 150 mpa, Allowable shering stress 90 mpa, Allowable bearing 270 mpa. Also find the efficiency of Joint. Step 1 : Dia of rivets & holes Nominal dia (D) = 14 mm Dia of rivet (d) = 15.5 mm Step 2 : Rivet value (i) Strength in shearing = 4 πd x τ 2 vf = 4 15.5 x π x 90 2 4 IS 800 - 1984 = 16.982 KN (ii) Strength in bearing = τ bf x d x t = 270 x 15.5 x 6 = 25.11 KN (iii) Rivet value = 16.98 KN (iv) Strength of Joint on the basis of rivet value = n x R.V = 7 x 16.982 = 118.874 KN Plate Failure (Consider Sec (1) – (1), (2) – (2), (3) – (3) for plate A Sec (3) – (3), (2) – (2), (3) – (3) for plate B) at (P – d) t Strength of the plate ‘A’ (a) section (1) – (1) = σ at (L – 2d) t = 150 (130 – 2 x 15.5) 6 = 89.1 KN Strength of plate ‘A’ (a) section (2) – (2) = Teaching strength (a) (2) – (2) + Strength Rivet Sec (1) – (1) = 150 (130 – 3(15.5)) x 6 + 2(R.V) = 150 [(130 – 3 (15.5)] x 6 + 2 (16, 982). =109. 114 KN Plate ‘A’ at section (2) – (2) can fail only it rivets at section (1) – (1) also fail. In the strength of het rivet at sec (1) – (1) will act along with the tearing of the plate (2) – (2) section Strength of the plat ‘A’ (a) sec (3) – (3) = tearing strength of the section (2) – (2) + Rivet value of (1) – (1) + Rivet value of (2) – (2) = 150 (130 – 2(15.5)) x 6 + 3 (16982) + 2(16982) = 174.01 KN Passable Failures (i) Combined Failure of rivet = 118.87 KN (ii) Failure of plate ‘A’ at section (1) – (1) = 89.10 KN (iii) Failure of plate ‘A’ at sec (2) – (2) = 109.114 KN 5 IS 800 - 1984 (iv) Failure of plate ‘A’ at sec (3) – (3) = 174. 01 KN The weakness critical section is (1) – (1) of plate ‘A’ strength of the Joint = 89.1 KN Strength of the solid plate = τ at x L x t (L = P) = 150 x 130 x 6 = 117.00 KN Efficiency = % 15 . 76 100 x 00 . 117 1 . 89 · Two plates 12 mm are joint by Double riveted double. Cover bult joint as shown in dia. Using 20 mm dia of the rivet design the pitch of the rivet. Take at = 150 Mpa also find the efficiency of the joint. Given at = 150 Mpa Dia = 20 mm Dia of the rivet hole = 20 + 1.5 = 21.5 mm Thickness of plate = 12 mm Assume PDS rivet, τ vf = 100 Mpa bf = 300 Mpa To find the rivet value (i) Strength of rivets in double shear = ] ] ] 4 πd x τvf 2 2 = ] ] ] 4 π(21.5) x 100 2 2 = 72.610 KN (ii) Strength of rivets in Bearing = bf. d.t = 300 x 21.5 x 20 = 77.4 KN (iii) Rivet value = least of shearing & bearing = 72. 61 KN For maximum efficiency of joint per pitch length, Strength of plate per pitch = 2 x Rivet value = 2 x 72.61 = 145. 22 KN σ at (P – d) x t = 145.22KN 6 IS 800 - 1984 150 (P – 21.5) x 12 = 145. 22 KN (or) 145220 P = 102.178 mm. Min pitch = 2.5 D = 2.5 x 20 = 50 mm Provide Pitch = 100mm 100 x P d P η − · 78.5% 100 x 100 21.5 - 100 · Two plates 12mm & 10 mm tk are jointed by trible riveted lab joint, In which the pitch of the centrel row of the rivet is half the pitch of rivet in outer row. Design the Joint & Find the efficiency Take: at = 150 N/mm 2 vf = 80 N/mm 2 bf = 250 N/mm 2 Assume 20mm dia Rivet hole = 20 + 1.5 = 21.5 mm To find the rivet value (i) Shearing = τ vf x π d2/4 = KN 04 . 29 4 5 . 21 x x 80 2 · π (ii) Bearing = bf x d x t = 250 x 21.5 x 10 = 53.75 KN (iii) Rivet value = least of shearing (or) bearing = 29.04 KN Strength of the plate (thinner) per pitch length along sec (1) – (1) = σ at (P – d) t = 1500 P – 32250 (1) Strength of plate per pitch length along sec (2) – (2) = σ at (P – 2d ) t + Rivet value = 150 (P x 21.5)10 + 29044 = 1500 P – 35456 (2) Sec (2) – (2) is weaker along which the strength of the plate is 1500 P – 35456 7 IS 800 - 1984 For maximum efficiency the strength of the per pitch length should be equal to strength of rivet per pitch length. 1500P – 35456 = 4 x R.V 1500 P = (4 x 29044) + 35456 P = 101. 088mm Min = 2.5D = 2.5 x 20 = 50mm Max = 32t (or) 300 whichever is lesser. = 32 (10) = 320 mm > 300 mm Max = 300 mm Outer row pitch = 120 mm Inner row pitch = 60 mm %) 64 ( %) 82 ( 100 x 60 5 . 21 60 ) or ( 100 x 120 5 . 21 120 d d P − − · − · η Note:- Strength of the plate = 1500 (60) = 35456 = 54, 544 N 42 = 4 x 29044N = 116, 76 N Take which value is user so take pitch, efficiency = sec (2) Design a bracket connection using two vertical lines of the rivet load carried by each plate is 120 KN the bracket plate of 10 mm tk are connected to 12mm tk flange plate. Assume pitch of 10 cm and horizontal distance between the vertical line is 12 cm. eccentriciting load is 25 cm. Given Load (P = 120 KN Thickness of the plate = 12 mm Thickness of the Flange = 12 mm Pitch (P) = 10 cm Gauge (G) = 12 cm Eccentricity (e) = 25 cm Soln:- Step: Assume 20 mm dia of rivet (PDS) D = 20mm d = 21.5 mm Step 2: Find the rivet value 8 IS 800 - 1984 (i) Strength in shear = Tvf x 4 2 21.5 x π x 100 4 2 πd · (ii) = 36305.03 N. Strength in bearing = bf x d x t = 300 x 21.5 x 10 = 64500 N Rivet value R = 36/305 KN Step 3: To find the no of the rivets vertical line Vertical line = 2 (given) M = P x Q = 120 x 25 = 3000 KN.cm M1 = KN.cm 150 2 3000 rivet vertical of. . no M · · n1= nos 5 4.979 10 x 36305 1500 x 6 R.P 6M 1 ≅ · · Step 4: Step 5: Check for the safety fo the joint q1 = qv1 + qv2 qv = .x r ε M n P 2 + (1) qn = .y r ε M 2 (2) r 2 = Σ x 2 + Σ y 2 = 0.(6) 2 + 4 (10 2 + 20 2 ) r 2 = 2360 cm qv = 6 . 2360 3000 10 120 + qv = 19.627 qn = max 2 y . r M ε 9 IS 800 - 1984 = 20 x 2320 3000 qn = 25.42 q = 2 2 2 2 42 . 25 627 . 19 qv qn + · + q = 32.118 KN < R = 36.305 K q < R hence safe Check the safety of the joint as shown in diagram Step 1:- Assume diameter = 16 mm Using PDS dia of the rivet hole = 16 + 1.5 = 17.5 mm Step 2: Shear stress due to Direct laod (Q) = p/n = 80/10 Q = 8 KN Step 3: Find ft max ft max = max 2 y . y M ε M = P x R = 80 x 16 = 1280 KN. Cm Ymax = 6 + 6 + 6 + 6 = 24 cm y 2 = 2(6 2 + 12 2 + 18 2 + 24 2 ) y 2 = 2160 cm Ft max = 24 x 2160 1280 Ftmax = 14.22KN Step 4 Find vf (cal) = 4 / d Q 2 π = 4 / 5 . 17 x 8 2 π = 0.083 = 83. 26 N/mm 2 10 IS 800 - 1984 Step 5 To find tf (cal) = 4 / d f 2 max t π = 4 / 5 . 17 22 . 14 2 π = 59.11 N/mm 2 Step 6 Check 4 . 1 ) cal ( ) cal ( tf tf vf vf ≤ σ σ + τ τ 4 . 1 fy 6 . 0 11 . 59 fy 4 . 0 26 . 33 ≤ + 4 x 4 . 0 26 . 33 A bracket plate of 10 mm thick is to be connected to the base of the flange using angles the load is applying through the bracket to the column. E = 10cm, P = 200KN. Design the connection between the angle & column Step 1 Assume 20mm dia of the rivet Using PDS rivet Dia of the rivet hoel = 20 + 1.5 = 21.5 mm Minimum pitch distance (D) = 2.5 = 2.5 x 20 = 50 mm Maximum pitch distance = 32t = 32 x 21.5 = 688 ≅ 690 mm Adopt pitch distance to 100 mm Step 2: To find rivet value Strength in shearing = vf x π d 2 / 4 = 100 x 4 5 . 21 x 2 π = 36.30 KN 11 IS 800 - 1984 Strength is bearing = σ bt x d x t = 300 x 21.5 x 10 = 64.5 KN Rivet value = 36.3 KN Step 3 No of rivet (n1) = 0.8 RP m 6 = 0.8 10 x 3 . 36 M x 6 M = P x e = 200 x 10 = 2000 KN.cm M1 = line rivet of no m = 2000/2 M1 = 1000 KN.cm n1 = 0.8 10 x 3 . 36 1000 x 6 = 3.25 4 Nos Adopt 4 nos of rivet each row Step 4 Arrangement of the rivet Step v To find ft max Ft max = max 2 y x y M ∑ M = 2000 KN.cm y 2 = 2(102 + 202 + 302) = 2800 Ymax = 10 + 10 + 10 = 30 cm Ft (max) = 30 x 2800 2000 Ft max = 21.42 KN 12 IS 800 - 1984 Q = 8 200 n P · Q = 25 KN Step VI τ vt (cal) = 4 / d Q 2 π = 4 / 5 . 21 x 25 2 π = 0.0688 KN/mm 2 = 68.8 N/mm 2 tf (cal)= 4 / d F 2 max t π = 4 / 5 . 21 x 42 . 21 2 π = 0.05 tf (cat) = 59 N/mm 2 Check 4 . 1 ) cal ( (cal) tf tf vf vf ≤ σ σ + τ τ 4 . 1 fy 6 . 0 9 . 5 fy 4 . 0 8 . 68 ≤ + 4 . 1 ? x 6 . 0 59 ? x 4 . 0 8 . 68 ≤ + Design the riveted connection between the column ISMB 300 & beam ISMB 350 transmitting the load of 35 KN/m over a span of 9m. Assume 20mm dia PDS rivet Given data:- Load = 35 KN/m Span l = 9m Solution: Step 1 The beam is connected to the column using angle. The size of the angle should not be less than 3d. 13 IS 800 - 1984 ∴ length of the angle = 3 x 21.5 = 64.5 mm Choose ISA 75 x 75 x 10mm Angles Step 2 Connection between the angle & web of the beam line Angle & flange of column line To find the rivet value: Strength of rivet in double shearing = 2 x τ vf x 4 d 2 π = 2 x 100 x 4 5 . 21 x 2 π = 7261 KN. Bearing for web of ISMB 350 = bE. d x t = 300 x 21.5 x 8.1 = 52. 24 KN Rivet value = 52.24 KN Number of rivets (n) = v . R P Load at the Joint (P) = Reaction from the beam = 5 . 157 2 9 x 35 2 WL · · n = 24 . 52 5 . 157 n = 3 Step 3 Connection between flange of the column of angle. To find the rivet value Strength in single shearing = τ vf x 4 d 2 π 14 IS 800 - 1984 = 100 x 4 5 . 21 x 2 π = 36. 30 KN Bearing for flange of ISMB = σ bf x d x t = 300 x 21.5 x 10 = 64.5 KN Rivet value = 36.3 KN Number of rivets n = V . R P = 3 . 36 5 . 157 n = 4.33 ≅ 5 nos Step 4 Arrangement of rivet Minimum pitch = 2.50 = 2.5 x 20 = 50 Maximum pitch = 32 t (or) 300 = 32 x 8.1 = 259.2 < 300 Max pitch distance = 260 mm Minimum edge distance = 29 mm [From IS 800 – 1984. Pg ∴ Provide 30 mm edge distance A tie bar 100 mm x 16 mm is to be welded to another plate 150 mm x 16 mm. find the minimum overlab length required if 8 mm fillet weld of used. Take σ at = 150 N/mm 2 . σ bt = 165 N/mm 2 , σ vf = 100 N/mm 2 Given data:- at = 150 N/mm 2 bt = 165 N/mm 2 vf = 100 N/mm 2 Size of the fillet welt = 8mm Load = the strength of the smaller plate Strength of the smaller plate = at x b x t 15 IS 800 - 1984 = 150 x 100 x 16mm P = 240 KN The value of the weld = ks. Fs = 0.78 x 8 x 100 R = 560 N/mm Length of the Weld = P/R = 560 10 x 2400 3 L = 428. 57 mm L ≅ 430 mm For minimum over lab in of the plate both end fillet weld & side fillet weld are provided. The length of the end fillet = 2 x 100 = 200 mm Length is to be provided by side Fillet = 430 – 200 Side = 230 mm The length shared by two side Length of the onside = 230 / 2 = 115 mm The weld Lab Joint is to be provided to connect two tie bar 150 x 16 mm stress in the plate is 150 N/mm 2 . To check the design if the size of the weld is 8mm & shear stress is taken as 100 N/mm 2 . Given data:- σ at = 150 N/mm 2 b = 150 mm t = 16 mm τ vf = 100 N/mm 2 S = 8 mm Solution To check the safety of the Joint should not be more than load at the joint. Load at the Joint = σ at x b x t = 130 x 150 x 16 = 360 KN Strength of the Joint = vf. K.S.L 16 IS 800 - 1984 L = 50 + 2 2 ] 80 50 2 2 + L = 477.38 mm Strength of the Joint = 100 x 0.707 x 8 x 4m = 266. 61 KN Hence the design is unsafe Load = 360 Strength = 267 KN Load 4 strength A 150 mm x 115 mm x 8mm angle section carries a tensile load of 200 KN it is to be connected gusset plate using 6 mm fillet weld at the extreame of the longer length (leg) Design the Joint along the shear stress 100 N/mm 2 . Angle section is unequal. The load is acting excentricity. We have to adopt Let x1 be the length of the weld at tob X2 be the length of the weld at bottom Total length = x1 + x2 L = weld the of Value Load Value of the weld = vf. K.S = 100 x 0.707 x 6 mm = 424.2 N/mm 2 Load at the Joint P = 200 KN L = 2 . 424 10 x 200 3 L= 471. 47 mm x1 + x2 – 150 = 471.48 – 150 = 321.48 mm Two unknowns S1 one equation to create another equation to find the either x1 (or) x2. Moment of the at the top = Moment of resistance of het bottom weld at top. Unequal section = 150 x 115 x 8 mm 17 IS 800 - 1984 Load acting at a distance lxx = 44.6 mm ( from steel table, Pg.) Moment of the load at top = 200 x 103 x 44.6 = 8.92 x 10 6 N/mm (1) Moment resistance of the bottom = 424.4 x2 x 130 = 63.66 x 10 3 x 2 x2 = 3 6 10 x 66 . 63 10 x 92 . 8 x2 = 140.119 mm x1 + x2 = 321.48 x1 + 140.119 = 321.48 x1 = 321.48 – 140.119 x1 = 181.36 mm Design a single angle section carring a axial load of 150 KN. Assume Fy. 250 N/mm 2 and dia of the rivet is 20mm. step 1 An = 2 3 at mm 1000 250 x 6 . 0 10 x 150 W · · σ Step 2: choose 70 x 70 x 10 mm in steel table (From steel table) L1 = L2 = 70mm, t = 10mm, d = 20 + 1.5 = 21.5 A = 1302 mm 2 Anet = A1 + A2 A1 = (L1 – t/2) t – d x t = [70 – 10/2] 10 – 25 x 10 A1 = 435 mm 2 A2 = [L2 – t/2]t = [70 – 10/2] 10 A2 = 650 mm 2 K = 2 1 1 A 3 A 3 A 3 + 18 IS 800 - 1984 = ) 650 ( ) 455 x 3 ( 55 x 3 + K = 0.67 Anet = 870.50 mm 2 Step 3 Load = Anet x σ at = 870.50 x 150 = 130.575 KN < 150 KN So unsafe Hence trial section choose ISA 100 x 75 x 10 Gross Area A = 1650 mm 2 L1 = 100, L2 – 75, t = 10 Anet = A1 + KA2 A1 = [L1 – t/2] t – (d x t) = [100 – 10/2]10 (21.5 x 10) A1 = 735 mm 2 A2 = [L2 – t/2] t = [75 – 10/2] x 10 A2 = 700 mm2 K = 76 . 0 700 ) 735 x 3 ( 735 x 3 A A 3 A 3 2 1 1 · + · + Anet = 735 + (0.76 x 700) Anet = 1266.32 mm 2 Load = Anet x σ at = 1266.32 x (0.6 x 250) Load (w) = 189.948 KN Design a tension member of roof truss carrings a axial tension of 250 KN using double angle section back to back of the Gusset plate (Opp side) dia of rivet is 20mm. Step 1 An = 150 10 x 250 W 3 at · σ An = 1666.66 mm 2 Step 2: Selected a section whose Gross area is 1.5 x An area = 1.5 x 1666.66 = 2500 mm 2 19 IS 800 - 1984 Take section ISA 150 x 115 x 12 mm L1 = 150 L2 = 115 t = 12 mm A = 3038mm 2 Anet = Ag – Area of Rivet holes. = 3038 – 2(21.5 x 10) Anet = 2608 mm 2 Load = 2606 x 0.6 x 250 Load = 391.2 KN Note: (i) for single angle section Ag = 1.35 to 1.5 times of Anet (ii) For Double angle section (a) angles on some side of the gusset plate Ag = 1.35 Anet (b) Angles on either side of the gusset plate Ag = 1.25 Anet (iii) (a) For chain riveting in plate section Anet = t (b – nd) (b) for zig – zag riveting (or) staggered riveting (i) Anet = t [(b – nd) 4 + m [s 2 /4g] (ii) Anet = t [(b – nd) + s 2 /4g 1 + s2 2 4g 2 )] Where b – breadth n – no of rivets d – dia of nivet hole m – no of zig. Zag line along the failure line s – Pitch g – guage Member under axial load and moment There will be axial tension due to axial force and bending stress due to bending moment. Direct stress due to axial tension = σ at (cal) = W/An Bending stress due to moment = bt (cal) = M/I. y The section is safe the following intraction formula is satisfied. bending uniaxial for 1 ) cal ( ) cal ( bt bt at at ≤ σ σ + σ σ for uniaxial bending 20 IS 800 - 1984 1 fy 66 . 0 ) cal ( fy 66 . 0 ) cal ( fy 6 . 0 ) cal ( bty btx at ≤ σ + σ + σ For biaxial bending A tension member made of two channels placed back to back carries a moment of 1900 N.m in addition to a direct tension of 450 KN. Design the section assume that f y = 250 N/mm 2 For the selection of the section assume that σ at = 30% to 40% of the preliminary stress at = (0.3 to 0.4) of 0.6 fy = 0.3 x 0.6 x 250 = 45 Area required = 250 x 6 . 0 x 3 . 0 1000 x 450 3 . 0 W at · σ = 10000 mm 2 This is offered by two channel section Area = 10000/2 = 5000 mm 2 Choose ISMC 400 Sectional properties Area = 6293 mm 2 Ixx = 15082 cm 4 tw = 8.6 mm Adopt 20mm dia Rivet for the connection An = Gross Area – area of Rivet hole = 2 x 6293 – 4 (21.5 x 8.6) = 11846.4 mm 2 at (cal) 2 3 n mm / N 99 . 37 4 . 11846 10 x 450 A W · · bt (cal) = 2 9 mm / N 60 . 12 2 400 x 10 x 15082 x 3 1000 x 19000 y . I M · · Check for Intraction formula 1 fy x 6 . 0 ) cal ( fy x 6 . 0 ) cal ( bt at ≤ σ + σ 21 IS 800 - 1984 1 250 x 66 . 0 60 . 12 250 x 6 . 0 99 . 37 ≤ + 0.253 + 0.07 < 1 0.33 < 1 Hence the section is safe A tie of roof truss consist of double angles ISA 100 x 75 x10 mm with it’s short leg back. To back and long leg connected to the same side of the gusset plate with 16 mm dia of the rivet determine the strength of the member take σ at = 150 N/mm 2 Step 1 Anet = A1 + KA2 K = 2 1 l A A 5 A 5 + K = 0.714 Anet = 650 + (0.714 x 1300) = 1578.2 mm 2 Strength = Anet x σ at = 1578.2 x 150 = 236730 N (or) 236.730 KN. A rolled steel is used as a column of height 5.5 m both ends are hinged. Design the column to carried axial toad of 600 KN. Solution: Both ends are hinged leff = L Leff = 3.5 m Load (P) = 600 KN Rolled steel section σ ac = 80 N/mm 2 Aread = 2 3 ac mm 7500 80 10 x 600 Load · · σ Choose ISHB 300 (1) 63.0 kg/m Area = 80.25 cm 2 = 8025 mm 2 rxx = 12.70 m ryy = 5.29 cm 22 IS 800 - 1984 Slanderness ratio λ = 052 . 0 5 . 3 r L min eff · λ = 66.16 IS 800 – 1984 Table 3.5 page 38 To find the bc permissible Fy = 250 (assume) λ = 66.16 60 122 70 112 x = 122 – ] ] ] − , ` . | − − ) 60 16 . 66 ( 70 60 112 122 x = 115.84 σ bc permissible = 115.84 N/mm 2 bc (assume) = 80 N/mm 2 σ bc Permissible > bc assume => Hence safe σ bc (act) = 76 . 74 8025 10 x 600 area W 3 · · Design a single angle discontinuous structs connected by 2 rivets to a gusset plate length 2.5m, applied load 150 KN. [Refer Is 800 – 1984 -> CL 5.2 Pg 46] Effective length = 0.85L = 2.125 m σ ac = 60 N/mm 2 Areqd = 2 3 ac mm 2500 60 10 x 150 W · · σ Choose the ISA 150 x 150 x 20 A = 2903 mm 2 rxx = 46.3mm ryy = 46.3mm λ 80 . 45 3 . 46 10 x 125 . 2 r L 3 min eff · · · [Refer IS 800 – 1984 Table 5.1 Pg.30] 23 IS 800 - 1984 fy = 250 λ = 45.89 40 139 50 132 45.89 134.88 σ bc Permissible = 134.88 N/mm 2 σ bc (assume) < bc permissible Hence safe Where σ bc (act) = 67 . 51 2903 10 x 150 area W 3 · · · Design a double angle strut continuous to a load of 250 KN/m3 length 3m. Given Load = 250 KN L = 3, If Double angle continuous member Leff = 0.7L to L Solution Leff = 0.85l = 0.85 x 3m Leff = 2.55 m Assume double angle σ bc = 80 N/mm 2 Area required = bc W σ = 80 10 x 250 3 = 3125 mm 2 Single angle area = 3125 / 2 = 1562.5 mm 2 Select ISA 90 x 90 x 100 mm @ 13.41 kg/m A = 1703 mm 2 Ixx = 126.7 x 10 4 mm 4 Iyy = 126.7 x 10 4 mm 4 Lyy = 25.9 mm (lyy – centrid distance) 24 IS 800 - 1984 To calculate rmin: rmin = a 2 L 2 A I R xx xx xx · · 1703 10 x 7 . 126 4 · rxx = 27.27 mm Ryy a 2 ) 2 / t iyy ( a iyy 2 2 + + = 2 1703 x 2 2 10 2 104 x 7 . 126 2 1703 2 10 x 7 . 126 2 2 4 , ` . | + + , ` . | Ryy = 447.9 x 10 3 mm λ = min eff r l = 27 . 27 10 x 55 . 2 3 λ = 93.50 To find the σ bc (Permissible) [Refer IS 800 – 1984 Pg 39 Table 5.1] 90 90 100 80 93.5 86.5 bc (Permissible) = 86.5 σ bc (act) = ) 1703 ( 2 10 x 250 3 bc (act) = 73.4 Design a compression member consist of two channels placed with toes facing each other subjected to load of 1300 KN. Eff ht of the column is 8m. Design the comp. member and also design a lacing system Solu: 25 IS 800 - 1984 Assume ac = 110 N/mm 2 Areq = 2 3 ac 1 mm 18 . 11818 110 10 x 1300 K · · σ This is offered two channel. Therefore Area of single channel = 09 . 5909 2 18 . 11818 · Select ISMC 400 @ 494 N/m Area = 6293 mm 2 ixx = 15082.8 x 10 4 mm 4 iyy = 504.8 x 10 4 mm 4 rxx = 154.8 mm ryy = 28.3 mm cyy = 24.2 mm Ixx = 2ixx = 39.656 x 10 6 mm 4 Iyy = 2[iyy + a (S – (yy) 2 ] = 2 [504.8 x 10 4 + (6293) [200 – 24.2) 2 ] = 399.074 x 10 6 mm 4 Rxx = rxx = 154.8 mm ryy = a 2 / Iyy ) or ( mm 07 . 178 6293 x 2 10 x 074 . 399 A Iyy 6 · · λ = 68 . 51 8 . 154 10 x 8 r L 3 min eff · · λ = 51.68 [Refer Table 5.1 Pg. 39 Is 800 = 1984 50 132 60 122 51.68 130.32 bc = 130.32 σ bc (dct) = 29 . 103 6293 x 2 10 x 1300 Area Load 3 · · σ ac (act) = 103.29 26 IS 800 - 1984 bc (per) = 130.32 > σ bc (act) = 103.29 The design is safe Design of Lacing Assume that the connection to the lacking bar in mode at the centre of the flange width Connection are at 50mm from the edge. Distance C/C of rivet across = 400 – 50 – 50 = 300 mm Assume the angle of inclination of lacing bar = 450 C = 2 x 300 = 600 mm [For angle 450 = 25 desare equal] Check min r C < 0.7 λ < 50 3 . 28 600 < 0.7 x 51.68 < 50 21.20 < 36.176 < 50 Hence ‘C’ is ok. Size of the Lacing bar:- Assume 20mm dia of rivet for connections. Width of the bar = 3 = 3 x 20 = 60 mm Thickness of bar ‘t’ = l1 / 40 for single Lacing l1 = length of the lacking bar l1 = 2 2 300 300 + l1 = 424.26 mm ‘t’ = mm 61 . 10 40 26 . 424 · T ≅ 12 mm Size of the bar = 60 x 12 mm Check for:- (i) Slenderness ratio λ > 145 27 IS 800 - 1984 rL = 46 . 3 ) 12 x 60 ( 12 12 x 60 A I 3 L L · , ` . | · 122.62 > 145 Hence O.K (ii) Check for compressive stress Compression leading in the lacing bar = Q Sin n V V = 25% of the load = 1300 x 100 5 . 2 V = 32.5 KN Comp. Load = KN 98 . 22 45 sin x 2 10 x 5 . 32 0 3 · 2 3 bc mm / N 92 . 31 12 x 60 10 x 98 . 22 A P ) cal ( · · · σ bc (Per) [For lacing bar] => λ = 122.62 fy = 250 120 64 130 57 from Table 5.1 122.62 62.17 in IS 800 – 1984 bc (Per) = 62.17 N/mm 2 (62.17) σ bc (Per) > σ bc (cal) (31.92) Hence safe (iii) Check for tensile stresses:- P = KN 98 . 22 Q Sin n V · σ at (cal) = net A P Anet = Agross – Area of rivet hole = (60 x 12) – (12 x 21.5) 28 IS 800 - 1984 = 462 mm 2 σ at (cal) = 74 . 49 462 98 . 22 · at (Per) = 0.6 fy = 0.6 x 250 = 150 KN [150] at (Per) > at (Cal) Hence safe in tensile stress. Design a simple slab base resting on a concrete slab for the following data Load from the column = 750 KN Size of the column = ISHB 400 σ cc = 4 N/mm 2 SBC = 100 KN/m 2 Design the slab base. Soln:- Bearing Area = 2 3 3 cc mm 10 x 5 . 187 4 10 x 750 Load · · σ Assume square base length of one side (l) L = mm 012 . 433 10 x 5 . 187 A l 3 · · · Provide 450 x 450 mm Thickness t = ) 4 / b a ( w 3 2 2 bs − σ a = mm 100 2 250 450 · − b = 2500 2 400 450 · − W = 2 2 3 mm / N 7 . 3 450 10 x 750 Area Load · · 29 IS 800 - 1984 t = ) 4 50 100 ( 185 ) 7 . 3 ( x 3 2 2 − t = 24.3 mm ≅ 25 mm Design of pedestile Size of the pedestil is design such that pressure on the soil is with in the safe bearing capacity of soil. Add 10% of the self wt Total Load = , ` . | + 750 x 100 10 750 Base area of the pedestil:- Area = 2 m 25 . 8 100 825 SBC Load · · Adopt square base, length of the one side (1) L = 2 m 25 . 8 25 . 8 A · · Provide 3m x 3m of the pedestil Area of pedestile = 3 x 3 = 9m 2 Depth of the pedestile Assume 450 despersion projection of the pedestile beyond the base plate = m 3 . 1 275 . 1 2 45 . 0 3 ≅ · − Adopt depth = 1.3 m Size of the pedestal = 3 x 3 x 1.3 m Size of the base plate = 450 x 450 x 25 mm Design of gusseted plate A builtup steel column compressing 2ISWB 400 RSJ section with their webs spaced at 325 mm and connections by 10mm thick battens. It transmit and axial load of 2000KN. SBC of soil at site is 300 KN/m 2 . The safe permissible stresses of concrete base is 4 N/mm 2 . Design the gusseted base. Grillage foundation. Load = 2000 KN 30 IS 800 - 1984 SBC = 300 KN/m 2 Area required = 4 10 x 2000 e permissibl . con Load 3 · = 500 x 10 3 mm 2 It is shared by two angle = 2 3 mm 2 10 x 500 Adopt angle section 150 x 75 x 12 mm gusset angles on flange side width 75 mm long horizontal Allow 30 mm projection on either side in the direction of parallel to web. Length base plate parallel to the web L reqd = 400 x 2(10) + 2(12) + 2(75) + 2(30) = 654 mm Provide length of base plate = 700 mm Breath of the plate = 700 10 x 500 L A 3 reqd reqd · = 714.29 mm Provide square plate = 750 mm x 750 mm Cantilever projection of the plate from face of the gusset angle is checked for bending stress due to the concrete below. Intensity of pressure below the plate = Area load W = 2 3 mm / N 56 . 3 750 x 750 10 x 2000 · Moment in the cantilever portion: 2 Wl M 2 · Where l = [750 – 400 – 2(10) – 2(12)] / 2 l = 153 mm w = load per ‘m’ length = 3.56 N/mm for 1 mm width bs = M/Z 31 IS 800 - 1984 185 = ) 2 / t ( / 12 t x 1 10 x 67 . 41 2 / t 12 / bt ) or ( Y T 10 x 67 . 41 3 3 3 3 , ` . | · ] ] ] , ` . | 6 / t 10 x 67 . 41 185 2 3 · 185t 2 = 41.67 x 10 3 t = 36.76 mm thickness of the base = 36.76 mm – 12 (thickness of the angle leg) = 24.76 mm Provide 25 mm plate thickness, size of Gusset base plate = 750 x 750 x 25 mm CONNECTIONS: Outstanding length of the each side = 2 400 750 − Load on each connection = 3.56 x 750 x 175 = 467.250 KN Using 20mm rivet (DDS) To find the rivet value Strength in shearing = 4 d x x 2 v π τ = 4 5 . 20 x x 100 2 π = 36.305 x 10 3 Strength in bearing = bc x b x t = 300 x 21.5 x 10 = 64.5 x 10 3 Rivet value = 30.305 KN No of rivet = 87 . 12 36305 250 . 467 V . R Load · · = 13 nos (or) 14 nos Pitch: Min pitch = 2.5 x D = 2.5 x 20 = 40 mm 32 IS 800 - 1984 Max pitch = 12 x t = 12 x 10 = 120 mm Provide 60 mm pitch edge distance code book = 30 mm A beam supporting a floor glab carries a distributed load of 20 KN/m span for the beam is 6m design suitable I – section for the beam Step I Assume 3% adding as a selt wb of section Total load = 20 + , ` . | 100 3 x 20 = 20.6 B.M = m KN 7 . 92 r WL 2 · Step 2: Z = 3 5 6 bt mm 103 x 81 . 561 10 10 x 27 . 0 M · · σ Choose ISLB 325 @ 431 N/m Area = 5490 m 2 Depth = 325 mm bf = 165 mm tf = 9.8 mm tw = 7.0 mm Zxx = 607.7 x 10 3 mm 3 Ixx = 9874.6 x 10 4 mm 4 Iyy = 510.8 x 10 4 x mm 4 Step 3 check for shear Shear is calculated at a distance of ‘d’ from the support V = w (l/2 – d) W = adi + self wt (of section) = 20 + 0.481 W = 20.481 V = 20.431 (6/2 = 0.2) V = 54.65 KN τ av (cal) = portion web of Area V 33 IS 800 - 1984 = 2 mm / N 56 . 25 7 x ] 007 ( 2 325 [ 65 . 54 · − τ av (Per) = 0.45 fy = 0.45 x 250 = 112.5 N/mm 2 av (Per) > τ av (cal) 112.5 N/mm 2 > 25.56 N/mm 2 Step 4 Check for deflection Ymax = xx 2 I E 384 WL 5 = 4 5 4 10 x 6 . 9874 x ) 10 x 1 . 2 ( x 384 6000 x 43 . 20 x 5 Ymax = 16.62 mm Permissible deflection = mm 46 . 18 325 6000 325 · · λ Ymax < yper 16.62 < 18.46 Hence the section is safe in deflection. A s/s beam of span 6m carring a point lead low Joist at Mid span and at support load applied at Midspan is 150 KN Design the beam, assuming fy = 250 N/mm 2 the beam developes B.M, S.F and check for shear and deflection Step 1 Assuming 3% adding as a self wt of the section Total load = 150 + (150 x 3/100) = 154.5 mm B.M = m . KN 75 . 231 4 6 x 5 . 154 · Z = M/σ at = 3 6 6 6 mm 10 x 55 . 1404 10 x 404 . 1 fy 66 . 0 10 x 75 . 231 · · Step 2 Take ISLB 500 at 750 N/m W = 750 Ixx = 38570 x 10 4 mm 4 A = 9550 Iyy = 1063.9 x 10 4 mm 4 D = 500 Zxx = 1543.2 x 10 3 mm 3 34 IS 800 - 1984 bf = 180 ryy = 33.4 mm tf = 14.1 tw = 9.2 Step 3 Check for shear Shear is calculated at a distance of ‘d’ from the support V = 2 W W = P.L + Selt wt = 150 + (0.750 x 6) τ av (cal) 2 3 mm / N 80 . 17 ) 2 . 9 ( )] 1 . 14 ( 2 500 [ 10 x 25 . 77 tw . dw V · − · τ av (Per) = 0.45 fy = 0.45 x 250 = 112.5 N/mm 2 av (cal) < τ av (Per) Hence safe in shearing Step 4 Check for deflection Ymax = mm 58 . 3 10 x 38579 x 10 x 1 . 2 x 48 6000 x 10 x 5 . 154 EI 48 WL 4 5 3 3 3 · · Yper = L/325 = 18.46 mm Ymax < y per Hence in deflection In the above problem the beam is laterly un support between the own beam Assume bc = 120 N/mm 2 (120 to 130 N/mm 2 ) M = 231.75 KN.m Z = 3 m 93125 . 1 120 75 . 231 · Z = 1031.25 mm 3 Choose ISLB 550 at 863 N/m A = 12669 mm 2 Zxx = 1933.2 x 10 3 mm 4 Ixx = 53161.6 x 10 4 mm 4 tf = 15 mm tw 9.9 mm 35 IS 800 - 1984 ryy = 34.8 mm To find τ bc permissible for the selected section effective length of the compressive flange distance between the cross beam. ∴ L = 6/2 = 3m 207 . 86 8 . 34 3000 ry l · · 67 . 36 15 550 t d T D f 1 · · · 53 . 52 99 ) 15 ( 2 550 tw dw · − · [Refer IS 800 – 1984 => Table 6.1 B => Page 58 5 . 1 9 . 9 15 tw tf t T · · · B5 36.67 40 85 131 130 86.21 130.01 129.71 129.04 90 127 120 04 . 130 2 . 1 x 80 N 85 127 131 131 · , ` . | − − · 04 . 129 2 . 1 x 85 N 80 126 N 130 130 · , ` . | − · 71 . 129 40 35 09 . 129 N 04 . 130 04 . 130 · , ` . | − − · bc (cal) = 2 3 6 xx mm / N 88 . 119 10 x 2 . 1933 10 x 75 . 231 Z M · · · bc (Per) = 129.71 Hence the section is safe bc (Per) > bc (cal) Check for shear: 36 IS 800 - 1984 Max shear at the support V = 2 1 2 W ) or ( 2 WL 2 W + + · 2 ) 6 x 863 . 0 ( 2 150 + · V = 75.59 KN. 37