The assembly consists of two sections of galvanized steel pipe connected together using a reducing coupling at B. The smaller pipe has an outer diameter of 0.75 in. and an inner diameter of 0.68 in., whereas the larger pipe has an outer diameter of 1 in. and an inner diameter of 0.86 in. If the pipe is tightly secured to the wall at C, determine the maximum shear stress in each section of the pipe when the couple is applied to the handles of the wrench.

The image of the assembly is missing, so i have attached it.

T = 210 Lb.in

τ_ab = 7818.705 lb/in²

τ_bc = 2361.036 lb/in²

Explanation:

We know that formula for shear stress due to the twisting couple is given by;

τ = Tr/J

Where;

T is torque

r is radius of shaft

J is polar moment of inertia.

Let's first find J and T

J = πd⁴/32

For section AB,

J_ab = π(0.75⁴ - 0.68⁴)/32

J_ab = 0.010072 in⁴

Similarly;

J_bc = π(1⁴ - 0.86⁴)/32

J_bc = 0.044472 in⁴

Now, let's find the torque.

From the image attached, we can see that the force acting at point 1 is 15 lb and also the force acting at point 2 is 15 lb. Distance from Point 1 to A; L1 = 6 in

Distance from point 2 to A; L2 = 8 in

Thus, taking moments about point A, we have;

T = PL1 + PL2

T = 15(6) +15(8)

T = 210 Lb.in

r_ab = 0.75/2

r_ab = 0.375 in

r_bc = 1/2

r_bc = 0.5 in

Thus;

τ_ab = 210 × 0.375/0.010072

τ_ab = 7818.705 lb/in²

τ_bc = 210 × 0.5/0.044472

τ_bc = 2361.036 lb/in²

answer:

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