- #1

phyang01

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Does anyone help me to solve this problem in the attached file?

When I solved this problem, I got wrong answers.

Attached file is the image file with the problem and answer. (Question No.3)

I think I have to find allowable shear stress of Pin A to find the safety factor of this pin,

since its ultimate shear strenght is already given. (vertical height is 500mm)

However, My allowable stress for Pin A is 431.5MPa, which is far more than its ultimate

shear stress.

I took moment at B and C to find the shear force of Pin A. (and Pin A is in double shear)

so I used the formula " shear stress = shear force / (area of shear plane) * 2 " to find the

shear force, and I got the answer above(431.5Mpa) while its ultimate shear stress is 350MPa.

I still don't get what is the problem.. (I think there are logical errors on my thought).

Please help!

When I solved this problem, I got wrong answers.

Attached file is the image file with the problem and answer. (Question No.3)

I think I have to find allowable shear stress of Pin A to find the safety factor of this pin,

since its ultimate shear strenght is already given. (vertical height is 500mm)

However, My allowable stress for Pin A is 431.5MPa, which is far more than its ultimate

shear stress.

I took moment at B and C to find the shear force of Pin A. (and Pin A is in double shear)

so I used the formula " shear stress = shear force / (area of shear plane) * 2 " to find the

shear force, and I got the answer above(431.5Mpa) while its ultimate shear stress is 350MPa.

I still don't get what is the problem.. (I think there are logical errors on my thought).

Please help!

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