The image of the assembly is missing, so i have attached it.
Answer:
T = 210 Lb.in
Ļ_ab = 7818.705 lb/inĀ²
Ļ_bc = 2361.036 lb/inĀ²
Explanation:
We know that formula for shear stress due to the twisting couple is given by;
Ļ = 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 = Ļdā“/32
For section AB,
J_ab = Ļ(0.75ā“ - 0.68ā“)/32
J_ab = 0.010072 inā“
Similarly;
J_bc = Ļ(1ā“ - 0.86ā“)/32
J_bc = 0.044472 inā“
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;
Ļ_ab = 210 Ć 0.375/0.010072
Ļ_ab = 7818.705 lb/inĀ²
Ļ_bc = 210 Ć 0.5/0.044472
Ļ_bc = 2361.036 lb/inĀ²
