Problem 01026 mechanism driven by hydraulic cylinder the...

Problem 01.026 - Mechanism driven by hydraulic cylinder The hydraulic cylinder CF, which partially controls the position of rod DE, has been locked in the position shown. Member BD is 15 mm thick and is connected at Cto the vertical rod by a 9-mm-diameter bolt. Know that P= 2.2 KN and = 75° 100 mm 175 mm C 200 mm 45 mm Problem 01.026.a - Shearing stress in a bolt Determine the average shearing stress in the bolt. The average shearing stress in the bolt is MPa. Problem 01.026.b - Bearing stress at a bolted joint Determine the bearing stress at Cin member BD. The bearing stress at Cin member BD is MPa.

Transcript

00:01 Hi in this question we have the normal strain along bd so in the rod ac and cylinder bd is equal to eac which is equal to e dash ac plus so e dash ac plus esac which is equal to alpha into delta t plus r divided by ae of ac.

00:30 So here this is zero because initial temperature is zero and we have ebd is equal to e dash bd plus e dash of bd so es of bd which is equal to alpha delta t plus r divided by ae into so ae of bd.

00:49 Now we have deformation so deformation in rod ac and bd is given by delta eac is equal to so rl by ae of ac and we have delta ebd is equal to alpha into delta tl plus rl divided by bd so ae of bd.

01:23 So then we have the compatibility relation is given by so compatibility relation is given by it is delta eac so delta eac divided by 0 .7 is equal to delta ebd divided by 0 .3.

01:45 So from this we have the value of delta eac is equal to 2 .5 into delta ebd.

01:52 Now we have the expression for change in the value is 50 minus 20 that is equal to 30 degree celsius and taking the moment along c is equal to so moment along e is equal to zero we have so 0 .75 into ra minus 0 .3 into rb is equal to zero.

02:12 So from this we have ra is equal to 0 .4 into rb.

02:16 Now we have delta eac is equal to ra into 0 .9 divided by area is pi by 4 into 0 .022 square into 200 gigapascal so which is equal to it is 11 .84 into 10 power minus 9 ra.

02:37 So then we have then we have the deformation along bd is equal to 18 .8 into 10 power minus 6 into 30 into 0 .3 minus rb into 0 .3 divided by pi by 4 into 0 .03 whole square into 105 into 10 power 9.

03:04 So from this calculation we have it is 169 .2 into 10 power minus 6 minus 4 .84 into 10 power minus 9 rb...

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