ch11 - SOLUTION (11.1) Known: Two steel plates with Sy =...

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Unformatted text preview: SOLUTION (11.1) Known: Two steel plates with Sy = 425 MPa are butt welded together with E70 series welding rods. The weld length is known. Find: Determine the maximum tensile load that can be applied to the joint with a safety factor of 3. Schematic and Given Data: Weld length = 90 m S y = 425 MPa SF = 3 E70 series weld Assumption: The weld efficiency is 100%. Analysis: 1.From Section 11.4, the weld rod has a yield strength of sy = (70 - 12) ksi(§-8—is—1ivl—B§) = 399.6 MPa Since 399.6 < 425 the weld should yield first. 2. With the assumed 100% efficiency, the maximum tensile load that can be applied 18 6 2 F=§§=W=2038m I 11-1 SOLUTION (11.2) Known: Two steel plates are butt welded using E60 series welding rods. The safety factor is 3. Find: Determine the tensile load that can be applied to the plates per inch of welded plate width. Schematic and Given Data: Sy = 52.5 ksi (plate) SF = 3 E60 series weld Assumption: The weld efficiency is 100%. Analysis: 1 . From Section 11.4 the weld rod has a yield strength of Sy = (60 - 12) ksi = 48 ksi. Since 48 ksi is less than Sy of the plates, the weld will yield first. 2. With the assumed 100% efficiency: F __ _SLA; _ (48,000 psi)(0.50 in.)(L) )— SF _ 3 3. The maximum tensile load per inch of weld is F/L = 8000 lb/in. I Comment: The load that the plate can carry per inch of length is given by F = SyA = (52,500)(0.50) = 26,250 1b which significantly exceeds that of the weld. 11—2 SOLUTION (11.3) Known: Two steel plates are welded with a convex fillet weld. Find: Estimate the static load, F, that can be carried by the joint. Schematic and Given Data: 3),: 350 MPa (plate) h: 5 mm SF: 3 Assumptions: 1. The plates themselves do not fail; shear failure occurs in the weld throat area. 2. The throat length is given by t = 0.707 h. 3. The weld efficiency is 100%. Analysis: 1. With the assumed throat length, t = 0.707 h, the weld throat area is (0.707)(5)(100) = 353.5 mm2 for the two welds. 2. The weld throat area is stressed in shear. Using the distortion-energy theory, 85)! = Sy = 3. F = SsyA/SF = (203)(353.5)/3 = 23,920 N, 01‘ 23.9 kN I Comment: If the top plate has a cross—sectional area A = (40)(7) = 280 Ian, then the load capacity of the plate is F = SyA = 350(280) = 98 kN which significantly exceeds that of the weld; i.e., the plate will not fail. 11—3 SOLUTION (11.4) Known: Two steel plates with Sy = 400 MPa are butt welded together with E70 series welding rods. The weld length is known. Find: Determine the maximum tensile load that can be applied to the joint with a safety factor of 3. Schematic and Given Data: Weld length = 90 m S y = 400 MPa SF = 3 E70 series weld Assumption: The weld efficiency is 100%. Analysis: 1. From Section 11.4, the weld rod has a yield strength of sy = (70 - 12) ksi (28%;11392) = 399.6 MPa Since 399.6 < 400 the weld should yield first. 2. With the assumed 100% efficiency, the maximum tensile load that can be applied is _ SyA _ (399.6 x 106 N/m2)(0.015 m)(0.09 m) _ F__SF__——————————————3 —179.8kN I 11-4 SOLUTION (11.5) Known: Two steel plates are butt welded .using E60 series welding rods. The safety factor is 3. Find: Determine the tensile load that can be applied to the plates per inch of welded plate width. Schematic and Given Data: S}. = 50 ksi (plate) SF = 3 E60 series weld Assumption: The weld efficiency is 100%. Analysis: 1. From Section 11.4 the weld rod has a yield strength of Sy = (60 — 12) ksi = 48 ksi. Since 48 ksi is less than Sy of the plates. the weld will yield first. 2. With the assumed 100% efficiency: _ SyA _ (48,000 psi)(3/8 in.)(L) ’ SF 3 3. The maximum tensile load per inch of weld is F/L = 6000 lb/in. I F Comment: The load that the plate can carry per inch of length is given by F = SyA = (50,000)(.375) = 18,750 lb which significantly exceeds that of the weld. 11—5 SOLUTION (11.6) Known: Two steel plates are welded with a convex fillet weld. Find: Estimate the static load, F, that can be carried by the joint. Schematic and Given Data: Sy = 350 MPa (plate) 11 = 5 mm SF = 3 Assumptions: 1. The plates themselves do not fail; shear failure occurs in the weld throat area. 2. The throat length is given by t = 0.707 h. 3. The weld efficiency is 100%. Analysis: ‘1. With the assumed throat length, t = 0.707 h, the weld throat area is (0.707)(5)(100) = 353.5 mm2 for the two welds. 2. The weld throat area is stressed in shear. Using the distortion-energy theory, 83): = Sy = 3. F = SsyA/SF = (203)(353.5)/3 = 23,920 N, or 23.9 kN I Comment: If the top plate has a cross-sectional area A = (40)(8) = 320 m2, then the load capacity of the plate is F = SyA = 350(320) = 112 kN which significantly exceeds that of the weld; i.e., the plate will not fail. 11—6 SOLUTION (11.7) Known: Two steel plates are joined using 3/8 in. parallel-loaded fillet welds. The yield strength, Sy, and the length of the welds are known. The safety factor is 3. Find: Determine the maximum tensile load that can be applied. Schematic and Given Data: E60 series welding rods Sy = 50 ksi (plates) h = 0.375 in. SF = 3 Assumptions: 1. The throat length is given by t = 0.707 h. 2. The weld efficiency is 100%. Analysis: 1. With t = 0.707 h, the throat area = (0.707)(3/8)(6) = 1.59 in.2. 2. From Section 11.4, the yield strength of the weld material is Sy = 60 - 12 = 48 ksi. 3. Using the distortion energy theory, SSy = 0.58 Sy = O.58(48) = 27.8 ksi. 4. Thus, F = SsyA/SF = (27,800)(1.59)/3 = 14,700 lb. I 11-7 SOLUTION (11.8) Known: Two steel plates are joined using transverse-loaded fillet welds. Each weld is 100 mm long. A force of 150 kN is applied The safety factor is 3.5. Find: Determine the minimum leg length that must be used. Schematic and Given Data: E70 series welding rods F = 150 kN . Sy = 400 MPa (plates) SF = 3.5 h:? mm Assumptions: 1. The throat length is given by t = 0.707 h. 2. The weld efficiency is 100%. 3. The critical stress is at the minimum throat section, where the area is tL. This cross section carries the entire load F in shear. Analysis: 1. With throat length, t = 0.707 h, the throat area, A = (0.707)(200)h = 141.4h mm2 2. From Section 11.4, an estimated yield strength of the weld is sy = 70 ksi — 12 = 58 ksi = 58 MM) = 399.6 MPa. ksi 3. From the distortion—energy theory, Sys = (399.6)(0.58) = 231.8 MPa. SA MP . 2 4_ F = SY or150,000 N =W_ Thus, h =16_0 m. I SF 3.5 Comment: Because of assumption 3, the solution is less rigorous. 11—8 SOLUTION (11.9D) Known: Two steel plates with given yield strength are butt welded together. Find: Design a butt welded joint that can transmit a tensile load of 6000 lb. Schematic and Given Data: SV 2 50 ksi (plate) SF :2 3 E60 series weld Decisions: 1 . Select two 3/8 in. thick steel plates with 1 in. width. 2. Employ a safety factor of 3. 3. Use E60 series welding rods. Assumption: The weld efficiency is 100%. Analysis: 1. From Section 11.4 the weld rod has a yield strength of Sy = (60 — 12) ksi = 48 ksi. Since 48 ksi is less than Sy of the plates. the weld will yield first. 2. With the assumed 100% efficiency: F’SF’ 3 3. The maximum tensile load for the weld is 6000 lb. I Comment: The load that the plate can carry per inch of width is given by F = SyA = (50,000)(.375) = 18,750 lb which significantly exceeds that of the weld. 11-9 SOLUTION (1 1 . 10D) Known: Two steel plates are joined using 3/8 in. parallel—loaded fillet welds. The yield strength, Sy, and the loads the welds can carry are known. Find: Design a parallel—loaded fillet weld joint that can transmit a load greater than 14,000 lb. Schematic and Given Data: E60 series welding rods Sy = 50 ksi (plates) h = 0.375 in. SF = 3 Decisions: 1. Select two plates at least 3/8 in. in thickness. 2. Employ a safety factor of 3. 3. Use E60 series welding rods. 4. Use a weld length of 3 in. (6 in. total). Assumptions: 1. The throat length is given by t = 0.707 h. 2. The weld efficiency is 100%. 3. The steel p1ate(s) does not fail in tension. Analysis: 1. With t = 0.707 h, the throat area = (0.707)(3/8)(6) = 1.59 in.2. 2. From Section 11.4, the yield strength of the weld material is Sy = 60 ~12 =48 ksi. 11-10 3. Using the distortion energy theory, Ssy = 0.58 Sy = 058(48) = 27.8 ksi. 4. The maximum tensile load that can be applied is F = SsyA/SF = (27,800)(1.59)/3 = 14,7001b. I Comments: 1. The capacity achieved is slightly greater than required. 2. This design is only one of several possible. SOLUTION (11.11) Known: A bracket supports a total load of 60 kN. E60 series welding rods are used with a safety factor of 3.0. - Find: Determine the weld size that should be specified. Schematic and Given Data: Vertical 100 mm weld on inside of both plates along the y-axes E60 series welding rod SF = 3.0 Note: Each plate has two 75 mm welds and one 100 mm weld. Assumptions: 1 . The direct (transverse) shear stresses are uniformly distributed over the length of all welds. 2. The parts being joined are completely rigid. 3. The throat length is t = 0.707 h. ll-ll Analysis: 1. T(52.5)/J = 1992/! 75mm 2. Calculating the center of gravity, G, of the weld—segment. 2A 250 t y = 50 mm (by symmetry) 3. The polar moment of inertia of the weld pattern about G is the polar moment of inertia of one vertical plus the two horizontal welds. J = Jv + 21h = “gt + 100 «22.52 + 0) + 715% + 751 (502 + 152)] J = 613,020t mm4 4. T = (30,000 N)(55 mm + 22.5 mm) = 2.325 X 106 N.mm Torsional stress components are shown in the above figure at A, B, C, and D. . X = 30,000 N 2 1E , Direct shear adds a downward vector of A ———-——~(250 mum t at each locanon. Thus, the highest resultant stress is at C. 5- Atc,r=-1t— 189.6 +(85.34+120) =280.9 t 6. From Section 11.4, Sy = Su — 12 ksi = 60 — 12 = 48 ksi. From the distortion— energy theory, sy, = 0.58 sy = 0.58(48) = 27.84 ksi = {27.84 ksi) (wgflli) = 191.9 MPa 7. 280.9 _ Sys_191.9 .tz4’39 mm _ t z . = h ‘ 0.707 0.707 6'21 mm Conservatively we use h = 7 m. I Comment: For calculating transverse shear and axial stress, the throat dimension t is assumed to be in a 45° orientation. But the throat dimension t is assumed to be in the 11-12 plane of the weld pattern when computing torsional stresses. Also, for calculating the weld dimension h, t is assumed to be in a 45 ° plane. This simplification although not theoretically correct is justified for most engineering applications. SOLUTION (11.12) Known: A bracket supports a 4000 lb load. A fillet weld extends for the full 4 in. length on both sides. Series E60 welding rod is used. The safety factor is 3.0. Find: Calculate the minimum weld size required. Schematic and Given Data: SF = 3.0 E60 series welding rod Assumptions: 1. The throat length is given by t = 0.707 h. 2. The weld efficiency is 100%. Analysis: 1. The stress due to direct shear is given by: = X = 4000 lb 2 fl A St t _ . , 2. The stress due to bending is given by: G = k where I M = (3 in.)(4000 lb) = 12,000 lb.in 2 L_3t : 4_3t = . 4_ :2. Th f :12,000(2)=2249.3 I 2(12) 2(12) 10.67t1n., c 1n. ere ore,O —10I67t t 3. Vectorally adding 0' and 1:: Resultant stress = %V5007 + 2249.32 = Agni 11-13 4. From Section 11.4 in the text, Sy = 60 - 12 = 48 ksi. Using the distortion—energy theory, Sys = 0.58 Sy = 058(48) = 27.84 ksi 2304 2 E2: 27,840 5. t SF 30 ;t=0.251n. ~ : = t = 0.25 = - 6. Since t 0.707 h, h —-——0.707 ———0.707 0.35 1n. I SOLUTION (11.13) Known: A part welded using a E60 welding rod has out—of—plane eccentric loading. The safety factor is 3. Find: Determine the weld size required if only the top of the joint is welded. Schematic and Given Data: Force applied 10 m at midpoint of cylindrical hole E60 series welding rod SF = 3.0 sy =345 MPa Assumptions: 1. The direct shear stresses are uniformly distributed over the length of all welds. 2. The parts being joined are completely rigid. 3. The throat length is t = 0.7 07 h. 4. The absence of weld CD in Fig. 11.9 will not alter the bending stresses; that is there is a weld at AB and CD —— see analysis 1. 5. The weld at CD is absent and bending occurs at axis X-X —- see analysis 2. 6. Bending occurs at CD rather than at axis X-X -- see analysis 3. Analysis 1: 1. With welds assumed at AB and CD, the moment of inertia of the welds is x z 21}; = 2(70) t (60)2 = 504,000 t 11—14 _ Mc _ 1,600,000(60) _ 190.5 2. B di tres eldAB' 6—————————- MP en ngs sonw IS I 504,000t t a Transverse stress on weld AB is ’C = X = = 1429 MPa [X 70t t As in Fig. 11%, the resultant stress 1R =%«/190.57 + 142.92 = % 3. Using the distortion-energy theory, Sys = 0.58 Sy = 058(345) = 200.1 MPa S 4_ = t S}: 3 ,t 35 nun 5. s :07 hh= t =3_-57__HH_n-h= .0 meet 07 , 0.707 0.707 , 5 5 m I Anabsk 2: 1 . In this analysis we assume that the neutral bending axis is at the center of the cross section. Therefore for weld AB, Ix = In = (70)(t)(60)2 = 252,000 t 2. Bending stress on weld AB is o = ¥ = = 3899 MPa Transverse stress on weld AB is ’t = X = 10’000 = 1429 MPa A 70t t As in Fig. 11.9b, the resultant stress 1R =%\/380.9§ + 142.97 = 4038 3. Using the distortion-energy theory, Sys = 0.58 Sy = 058(345) = 200.1 MPa S 4. 4068 = ys = 200.1. = _1 t SF 3 ,t 6 nun . ‘ =. h. = t =6-1mm;h=s. 5 Smcet 0707 , h 0.707 0707 63 m I Analysis 3: 1. In this analysis we assume that the bending axis is at the bottom edge CD of the cm$SKMmLThadMemrwddABszh=(Wm0fl2m2=LMBDMM 2. Bending stress on weld AB is o = ¥ = W = 199-5 MPa Transverse stress on weld AB is “c = X = = 1429 MPa 1A 70t t As in Fig. 11%, the resultant stress TR 2 %1/190.52 + 142.92 2 g 3. Using the distortion—energy theory, Sys = 0.58 Sy = 058(345) = 200.1 MPa 11-15 4. fl=§§= 200-1't=3.57mm t SF 3 ’ 5. ' =. , = t =————3-57mm;h= .05 Slncet 0707hh 0.707 0.707 5 m I Comments: 1. It is not conservative to assume that bending would tend to occur closer to CD than to X-X since this assumption would end in the calculation of a smaller weld (leg) size. However, in practice bending most probably would take place closer to CD than to X-X. 2. With the omission of the bottom weld CD, there is an increase in the direct shear stress carried by the upper weld whereas the bending stress is not affected significantly. Even in the absence of the lower weld, the compressive stress at the bottom is carried by the support. SOLUTION (11.14) Known: Two steel plates are butt welded together. Both the plate and welding materials have known strength properties. The imposed loading fluctuates rapidly between —20 kN and +60 kN. The safety factor is 2.5. Find: Estimate the length of weld required: (a) If the weld "reinforcement" is not removed. (b) If the excess weld material is carefully ground off so as to provide a smooth, continuous surface. Schematic and Given Data: 31] = 500 MPa sy : 400 MPa SF = 2.5 Assumptions: 1. If the the weld "reinforcement" is not removed, the weld metal has a rough , surface comparable to an as—forged surface. 2. The gradient factor, CG = 0.8 3. Infinite life (106 cycles) is required 11—16 Analysis (3): 1 . With the assumption that Cs for the weld metal corresponds to "as-forged", Cs = 0.52. Also CG = 0.8. 2. Using Eq. (8.1) and Table 8.1, Sn = Sn'CLCGCs = (500/2)(1)(0.8)(0.52) = 104 MPa 3. Nominal load values: Pm = 20, Pa = 40. Design overload (SF 2 2.5): Pm = 50 kN; Pa = 100 kN. 4. From the above diagram, 63 = KfPa/A = 1.2(100,000)/A = 91 MPa A=1319mm2=(20mm)L;L=66mm I Analysis (b): 1. With the assumption that Cs for the weld corresponds to a ground surface, CS = 0.9. 2. Sn = Sn'CLCGCs = (500/2)(1)(0.8)(0.9) = 180 MPa 11-17 3. From the above diagram, O'a = Kf Pa/A = 1.0(100,000)/A = 150 MPa. Therefore, A = 667 m2. Also, A = (20 mm)L. Hence, L = 33.33 mm. Conservatively, L = 34 m. I SOLUTION (11.15D) This problem is left for the student to solve. A reference that will provide an excellent start for a solution is given in "Properties and Uses of 43 Adhesives Classified According to Chemical Type," N.J. DeLollis, p. 28—31, Product Engineering Design Manual, edited by Douglas C. Greenwood, McGraw-Hill, New York, 1959. SOLUTION (1 1 . 16D) This problem is left for the student to solve. A reference that will provide an excellent start for a solution is given in "Which Adhesive For What," R.W. James and R.W. Gormly, p. 32-33, Product Engineering Design Manual, edited by Douglas C. Greenwood, McGraw—Hill, New York, 1959. 11-18 ...
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This note was uploaded on 03/27/2009 for the course MECHENG EN.530.215 taught by Professor Wang during the Spring '09 term at Johns Hopkins.

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ch11 - SOLUTION (11.1) Known: Two steel plates with Sy =...

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