Page 41 - Marks Calculation for Machine Design

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January 4, 2005
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Brown.cls
Brown˙C01
U.S. Customary FUNDAMENTAL LOADINGS SI/Metric 23
Example 5. Calculate shear stress (τ) and the Example 5. Calculate shear stress (τ) and the
angle of twist (φ) for a thin-walled rectangular angle of twist (φ) for a thin-walled rectangular
tube, similar to that shown in Fig. 1.22, where tube, similar to that shown in Fig. 1.22, where
T = 4,000 ft · lb = 48,000 in · lb T = 6,000 N · m
b = 4in b = 10 cm = 0.1 m
h = 8in h = 20 cm = 0.2 m
t 1 = 0.25 in t 1 = 0.6 cm = 0.006 m
t 2 = 0.5 in t 2 = 1.2 cm = 0.012 m
L = 2.5 ft = 30 in L = 0.8 m
6
2
9
2
G = 11.7 × 10 lb/in (steel) G = 80.8 × 10 N/m (steel)
solution solution
Step 1. Calculate the area (A m ) enclosed by Step 1. Calculate the area (A m ) enclosed by
the median using Eq. (1.33). the median using Eq. (1.33).
A m = bh = (4in)(8in) = 32 in 2 A m = bh = (0.1m)(0.2m) = 0.02 m 2
Step 2. Substitute the torque (T ), area (A m ), Step 2. Substitute the torque (T ), area (A m ),
and the thickness (t) in Eq. (1.31) to give the and the thickness (t) in Eq. (1.31) to give the
shear stress (τ) as shear stress (τ) as
T 48,000 in · lb T 6,000 N · m
τ = = 2 τ = = 2
2 A m t 1 2(32 in )(0.25 in) 2 A m t 1 2 (0.02 m )(0.006 m)
48,000 in · lb 2 6,000 N · m 2
= 3 = 3,000 lb/in = 3 = 25,000,000 N/m
(16 in ) (0.00024 m )
= 3 kpsi = 25 MPa
Step 3. Calculate the polar moment of inertia Step 3. Calculate the polar moment of inertia
(J) for the rectangular tube using Eq. (1.34). (J) for the rectangular tube using Eq. (1.34).
2 2 2 2
2 b h t 1 t 2 2 b h t 1 t 2
J = J =
bt 1 + ht 2 bt 1 + ht 2
2
2
2
2
2 (4in) (8in) (0.25 in)(0.5in) 2 (0.1m) (0.2m) (.006 m)(.012 m)
= =
(4in)(0.25 in) + (8in)(0.5in) (0.1m)(.006 m) + (0.2m)(.012 m)
256 in 6 4 5.76 × 10 −8 m 6 −5 4
= = 51.2in = = 1.92 × 10 m
5in 2 0.003 m 2
Step 4. Substitute the torque (T ), length (L), Step 4. Substitute the torque (T ), length (L),
shear modulus of elasticity (G), and the po- shear modulus of elasticity (G), and the po-
lar moment of inertia (J) just calculated into lar moment of inertia (J) just calculated, in
Eq. (1.32) to find the angle of twist (φ). Eq. (1.32) to find the angle of twist φ.
TL TL
φ = φ =
GJ GJ
(48,000 in · lb)(30 in) (6,000 N · m)(0.8m)
= 2 4 = 9 2 −5 4
6
(11.7 × 10 lb/in )(51.2in ) (80.8 × 10 N/m )(1.92 × 10 m )
864,000 lb · in 2 4,800 N · m 2
= = 2
599,000,000 lb · in 2 1,551,360 N · m
= 0.0024 rad = 0.0031 rad