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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 111 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 distortionenergy 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 114 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 distortionenergy 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 crosssectional 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. parallelloaded 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 117 SOLUTION (11.8) Known: Two steel plates are joined using transverseloaded 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. 119 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. 1110 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 yaxes 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. llll 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 1112 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 1113 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 XX — see analysis 2. 6. Bending occurs at CD rather than at axis XX  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 distortionenergy 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_nh= .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 distortionenergy 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 =61mm;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 = 1995 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 1115 4. ﬂ=§§= 2001't=3.57mm t SF 3 ’
5. ' =. , = t =————357mm;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 XX 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 XX. 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 "asforged",
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 1117 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, McGrawHill, 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. 3233, Product Engineering Design Manual, edited by Douglas C.
Greenwood, McGraw—Hill, New York, 1959. 1118 ...
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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.
 Spring '09
 Wang

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