Question

Consider the figure below. Each of the four pins in this mechanism are the same. If the ultimate shearing stress of the pin material is 180 MPa and the factor of safety is 3, then determine the pin diameter.

          Consider the figure below. Each of the four pins in this mechanism are the same. If the ultimate shearing stress of the pin material is 180 MPa and the factor of safety is 3, then determine the pin diameter.
        
Consider the figure below. Each of the four pins in this mechanism are the same. If the ultimate shearing stress of the pin material is 180 MPa and the factor of safety is 3, then determine the pin diameter.

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University Physics with Modern Physics
University Physics with Modern Physics
Hugh D. Young 14th Edition
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Consider the figure below. Each of the four pins in this mechanism is the same. If the ultimate shearing stress of the pin material is 180 MPa and the factor of safety is 3, then determine the pin diameter. 0.4 m C 0.3 m 0.2 m B D 710 kN
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Transcript

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00:02 So here we need to determine the normal stress acting on each bar, shearing and bearing stress at c.
00:08 So now applying the condition of equation we have sigma fv equal to zero that is fbc times sin of 53 .13 minus 40 equal to zero so we get fbc equal to 50 kilo newton and we have summation fh equal to zero so we have fab minus fbc times cos of 53 .13 equal to zero so we get fab equal to 50 times cos of 53 .13 and this is equal to 30 kilo newton.
00:52 Now let us calculate the normal stress in each bar so we have sigma ab equal to fab over ab and this is equal to 30 times 10 to the 3 over 0 .004 equal to 7 .5 times 10 to the 6 pascals so we have sigma ab equal to 7 .5 mega pascals...
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