Question

A mechanically fastened joint subjected to tension has a total member stiffness of km = 378 * 10^6 lbf/in, bolt stiffness of kb = 3.69 * 10^6 lbf/in. If the joint is held together with a ½ in – 20 UNF x 1 ½ SAE 5 bolt and subjected to an external load of P = 5678.9 lbf, determine the required preload to maintain a factor of safety against joint separation of n0 = 3.

          A mechanically fastened joint subjected to tension has a total member stiffness of km = 378 * 10^6 lbf/in, bolt stiffness of kb = 3.69 * 10^6 lbf/in. If the joint is held together with a ½ in – 20 UNF x 1 ½ SAE 5 bolt and subjected to an external load of P = 5678.9 lbf, determine the required preload to maintain a factor of safety against joint separation of n0 = 3.
        
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University Physics with Modern Physics
University Physics with Modern Physics
Hugh D. Young 14th Edition
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A mechanically fastened joint subjected to tension has a total member stiffness of km = 378 * 10^6 lbf/in, bolt stiffness of kb = 3.69 * 10^6 lbf/in. If the joint is held together with a ½ in – 20 UNF x 1 ½ SAE 5 bolt and subjected to an external load of P = 5678.9 lbf, determine the required preload to maintain a factor of safety against joint separation of n0 = 3.
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Transcript

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00:01 Hello students, it is given that the stiffness of the each bolt is l0 is equal to 1 .0 and the stiffness of the membrane that is em is equal to 2 .6 and the proof strength is equal to 75 .1.
00:26 So the fluctuating external load is applied to the entire joint with maximum power is equal to 6 kilo newton and the minimum power is equal to 20 kilo newton.
00:45 Now let's solve the question.
00:47 So here f1 is equal to 0 .75 fp which is equal to 0 .75 into a into sp which is equal to 0 .75 into a is 20 .1 into 10 raised to minus 6 into 380 into 10 raised to 6 hence it will be equal to 5 .728.
01:14 Now to find out the value of c the equation is kb divided by kd plus km which is equal to 1 by 1 plus 25...
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