## Strength of Materials |

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Page 252

The integral of dT over the whole cross section equals the applied torque, and

pdl = 2dA, so T = J pqdl = q J pdl = 2q J dA = 2qA where A is the total area within

the center line of the

...

The integral of dT over the whole cross section equals the applied torque, and

pdl = 2dA, so T = J pqdl = q J pdl = 2q J dA = 2qA where A is the total area within

the center line of the

**tube**wall. Assuming that the shear stress does not vary over...

Page 260

What is the maximum torque on a

made from a circular

with an equilateral triangular cross section. 9-42. Redo Prob. 9-40 for a

a ...

What is the maximum torque on a

**tube**of semicircular cross section that wasmade from a circular

**tube**as in Example 9-15? 9-41. Redo Prob. 9-40 for a**tube**with an equilateral triangular cross section. 9-42. Redo Prob. 9-40 for a

**tube**witha ...

Page 288

From Eq. 11-4, with kL = L, EXAMPLE 11-3 A thin-walled steel

has an outside diameter of 1 .4 in., and has a wall thickness of 0.04 in. Both ends

are pinned. What are the buckling loads in the first and second mode of buckling

...

From Eq. 11-4, with kL = L, EXAMPLE 11-3 A thin-walled steel

**tube**is 20 ft long,has an outside diameter of 1 .4 in., and has a wall thickness of 0.04 in. Both ends

are pinned. What are the buckling loads in the first and second mode of buckling

...

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