ME211solution_10 - (a) 1 = pr 1.35(106 )(0.75) = = 127 MPa...

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(a) 6 1 1.35(10 )(0.75) 127 0.008 pr MPa t σ == = (b) There are two cover plates inside and outside of the boiler plate. From the force balance, 6 1 127(10 )(0.008) (2) '(0.008) L L = 1 '6 3 . 3 MPa = (c) The gap between the rivet is 0.01 0.05 2 0.04 2 ⎛⎞ −= ⎜⎟ ⎝⎠ The force of the cover plate is 6 63.3(10 )(0.008)(0.04) 20.25 kN = The force of the boiler plate is 6 127(10 )(0.008)(0.04) 40.5 kN = From the shear diagram of the rivet, shear force through the rivet is . 20.25 kN 2 20.25 258 (0.01) 4 avg V MPa A τ π =
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At the section a-a, a normal force N and a moment are balancing against the compressive force of 500lb (FBD) Section property: 2 0.75(0.5) 0.375 A in == 24 1 (0.5)(0.75 ) 0.017578
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ME211solution_10 - (a) 1 = pr 1.35(106 )(0.75) = = 127 MPa...

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