Page 188 - Marks Calculation for Machine Design

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January 4, 2005
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a solid circular cross section. If other cross sections are of interest, then these expressions
should be modified accordingly. STRENGTH OF MACHINES
Also notice that the second term in the third line of Eq. (4.13) is four-thirds the direct
shear stress, which is the shear force (V ) divided by the cross-sectional area (A).
U.S. Customary SI/Metric
Example 5. Determine the maximum stresses Example 5. Determine the maximum stresses
on the top element of a solid circular crank arm, on the top element of a solid circular crank arm,
like the one shown in Fig. 4.16, subjected to a like the one shown in Fig. 4.16, subjected to a
downward applied force (P), where downward applied force (P), where
P = 500 lb P = 2,250 N
L AB = 2ft = 24 in L AB = 0.8 m
L BC = 1ft = 12 in L BC = 0.4 m
R = 1.0 in R = 2.5 cm = 0.025 m
solution solution
Step 1. Using Eq. (4.6), calculate the bending Step 1. Using Eq. (4.6), calculate the bending
moment (M A ). moment (M A ).
M A = P × L AB = (500 lb)(24.0in) M A = P × L AB = (2,250 N)(0.80 m)
= 12,000 in · lb = 1,800 N · m
Step 2. Using Eq. (4.5), calculate the torque Step 2. Using Eq. (4.5), calculate the torque
(T AB ). (T AB ).
T AB = P × L BC = (500 lb)(12.0in) T AB = P × L BC = (2,250 N)(0.40 m)
= 6,000 in · lb = 900 N · m
Step 3. Substitute the bending moment (M A ) Step 3. Substitute the bending moment (M A )
and the radius (R) in Eq. (4.10) to give the and the radius (R) in Eq. (4.10) to give the
normal stress (σ xx ). normal stress (σ xx ).
4M A 4 (12,000 in · lb) 4M A 4 (1,800 N · m)
σ xx = 3 = 3 σ xx = 3 = 3
πR π(1.0in) πR π(0.025 m)
48,000 in · lb 7,200 N · m
= 3 = 3
3.14 in 0.0000491 m
= 15,279 lb/in 2 = 146,680,000 N/m 2
= 15.3 kpsi = 146.7MPa
Step 4. Substitute the torque (T AB ) and the Step 4. Substitute the torque (T AB ) and the
radius (R) in Eq. (4.11) to give the shear stress radius (R) in Eq. (4.11) to give the shear stress
(τ xy ). (τ xy ).
2 T AB 2 (6,000 in · lb) 2 T AB 2 (900 N · m)
τ xy = = τ xy = =
πR 3 π(1.0in) 3 πR 3 π(0.025 m) 3
12,000 in · lb 1,800 N · m
= 2 = 3
3.14 in 0.0000491 m
= 3,820 lb/in 2 = 36,670,000 N/m 2
= 3.8 kpsi = 36.7MPa