Page 39 - Marks Calculation for Machine Design

P1: Shibu
January 4, 2005
12:26
Brown.cls
Brown˙C01
U.S. Customary FUNDAMENTAL LOADINGS SI/Metric 21
Step 3. Using this value for the polar moment Step 3. Using this value for the polar moment
ofinertia(J),andthemaximumallowableshear ofinertia(J),andthemaximumallowableshear
stress (τ max ), substitute in the equation devel- stress (τ max ), substitute into the equation deter-
oped in Step 1. mined in Step 1.
τ max J τ max J
T max = T max =
R R
4
4
2
8
2
(60,000 lb/in )(127.2in ) (4.1 × 10 N/m )(4.97 × 10 −5 m )
= =
(3in) (0.075 m)
7,632,000 lb · in 2 20,377 N · m 2
= =
3in 0.075 m
= 2,544,000 in · lb = 271,693 N · m
= 212 ft · kip = 272 kN · m
Step 4. For a maximum angle of twist (φ max ), Step 4. For a maximum angle of twist (φ max ),
solve for the maximum torque (T max ) in Eq. solve for the maximum torque (T max ) in Eq.
(1.29) to give (1.29) to give
φ max GJ φ max φ max GJ φ max
T max = = × GJ T max = = × GJ
L L L L
where the quantity (GJ) is called the torsional where the quantity (GJ) is called the torsional
stiffness. stiffness.
Step 5. Using the polar moment of inertia (J) Step 5. Using the polar moment of inertia (J)
calculated earlier and the shear modulus of elas- calculated earlier and the shear modulus of elas-
ticity (G), calculate the torsional stiffness. ticity (G), calculate the torsional stiffness.
6
4
9
2
2
4
GJ = (4.1 × 10 lb/in )(127.2in ) GJ = (26.7 × 10 N/m )(4.97 × 10 −5 m )
8
6
= 5.2 × 10 lb · in 2 = 1.3 × 10 N · m 2
Step 6. Substitute the given maximum allow- Step 6. Substitute the given maximum allow-
able angle of twist (φ max ), the torsional stiffness able angle of twist (φ max ), the torsional stiffness
(GJ) just calculated, and the length (L) in the (GJ) just calculated, and the length (L) into the
equation of Step 4 to give equation of Step 4 to give
φ max φ max
T max = × GJ T max = × GJ
L L
(0.026 rad) 8 2 (0.026 rad) 6 2
= (5.2 × 10 lb · in ) = (1.3 × 10 N · m )
(36 in) (1m)
1 1
6
2
2
8
= 7.2 × 10 −4 (5.2 × 10 lb · in ) = 2.6 × 10 −2 (1.3 × 10 N · m )
in m
= 374,400 in · lb = 33,800 N · m
= 31 ft · kip = 34 kN · m
Step 7. As the maximum torque (T max ) associ- Step 7. As the maximum torque (T max ) associ-
ated with the maximum angle of twist (φ max ) is ated with the maximum angle of twist (φ max ) is
smaller than that for the maximum shear stress smaller than that for the maximum shear stress
(τ max ), angle of twist governs, so (τ max ), angle of twist governs, so
T max = 31 ft · kip T max = 34 kN · m