Page 381 - Marks Calculation for Machine Design

P1: Sanjay
January 4, 2005
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Other welding electrodes have higher ultimate tensile strengths, and therefore higher
allowable weld strengths. 15:14 MACHINE ASSEMBLY 363
Consider the following example where the steps are basically the same as those for
Example 3, except that the weld size (t) will be determined using Eq. (8.90).
U.S. Customary SI/Metric
Example 5. For the fillet weld and loading Example 5. For the fillet weld and loading
configuration shown in Figs. 8.18 to 8.20 deter- configuration shown in Figs. 8.18 to 8.20 deter-
mine the required weld size (t), where mine the required weld size (t), where
P = 18,000 lb P = 81,000 N
L o = 10 in L o = 26 cm = 0.26 m
L 1 = 5in L 1 = 13 cm = 0.13 m
L 2 = 10 in L 2 = 26 cm = 0.26 m
S weld = 18.0 kpsi (E60xx electrode) S weld = 126.0 MPa (E60xx electrode)
solution solution
Step 1. Substitute the given information in Step 1. Substitute the given information in
Eq. (8.83) to determine (τ shear ) as Eq. (8.83) to determine (τ shear ) as
P P
τ shear = τ shear =
2 L 1 + L 2 2 L 1 + L 2
18,000 lb 81,000 N
= =
2 (5in) + (10 in) 2 (0.13 m) + (0.26 m)
18,000 lb in 81,000 N m
= × = ×
20 in in 0.52 m m
2
2
= 900 (lb/in ) · in = 156,000 (N/m ) · m
= 0.9 kpsi · in = 0.16 MPa · m
Step 2. Substitute the given information in Step 2. Substitute the given information in
Eq. (8.85) to determine the distance (D o ) as Eq. (8.85) to determine the distance (D o ) as
L 2 (5in) 2 L 2 (0.13 m) 2
D o = 1 = D o = 1 =
2 L 1 + L 2 2 (5in) + (10 in) 2 L 1 + L 2 2 (0.13 m) + (0.26 m)
25 in 2 0.0169 m 2
= = 1.25 in = = 0.0325 in
20 in 0.52 m
Step 3. Substitute the distance (D o ) from Step 3. Substitute the distance (D o ) from
step 2 and the given information in Eq. (8.85) step 2 and the given information in Eq. (8.85)
to determine the radial distance (r o ) as to determine the radial distance (r o ) as

2 2

L 2 2 L 2 2
r o = + (L 1 − D o ) r o = + (L 1 − D o )
2 2

2
10 in 2 0.26 m
2
= + (5in − 1.25 in) = + (0.13 − 0.0325 m) 2
2 2

= (25 + 14.06) in 2 = (0.0169 + 0.0095) m 2
2
2
= 39.06 in = 6.25 in = 0.0264 m = 0.1625 m