PROBLEM 3: A flat bar of width b = 60 mm and thickness t =...

PROBLEM 3: A flat bar of width b = 60 mm and thickness t = 10 mm is loaded in tension by a force P. The bar is attached to a support by a pin of diameter d that passes through a hole of the same size in the bar. The allowable tensile stress on the net cross section of the bar is $sigma_T$ = 140 MPa, the allowable shear stress in the pin is $ au_S$ = 80 MPa, and the allowable bearing stress between the pin and the bar is $sigma_B$ = 200 MPa. (a) Determine the pin diameter $d_m$ for which the load P will be a maximum. Hints: 1. See Example 11 of the lecture slides on Stress, 2. Plot P versus d for all three failure modes. (b) Determine the corresponding value $P_{max}$ of the load. Ans.: $P_{max}$ = 49.4 kN extless{}

Transcript

00:01 Hello, so the tensile stress in net section of the bar.

00:03 The pin reduces the cross -section area.

00:06 We're left with a net is equal to b minus d times t.

00:13 And so we know that pt, or the sum of t times a net, is equal to the sum of t times b minus d, t.

00:27 The sheer stress in pin, and these are just the equations we're going to be writing out.

00:33 A shear stress right is equal to pi d squared over four and so we're left with p s is equal to the torque of s times a shear and that is equal to the torque of test of s times pi d squared over four and now the bearing stress between pin and bar so we have a bearing is equal to d times t and so pb is equal to the sum of b times a bearing, which is equal to the sum of b times d times t.

01:15 So the first thing after we have this is the expressions with the given value.

01:22 So insert b, which is equal to 60 millimeters and t which is equal to 10 millimeters.

01:27 So we have pt is equal to 140 times 60 minus d times 10.

01:33 You're left with 14 ,000 times 60 minus d .n.

01:43 Ps is equal to 80 times pi d squared over 4, which is going to equal 20 pi d squared newtons.

01:51 And then pb is equal to 200 times d times 10, which is equal to 2 ,000 newtons.

01:58 Now, to find the critical d for maximum load, we know that pt is equal to ps, which is equal to pb so solving numerically you're going to we equate pt is equal to pb so we have 14 000 60 minus d is equal to 2 ,000 d so now we're solving for d and d is going to equal 52 .5 millimeters and now do the same thing equate pt to p s 14 ,060 minus d is equal to 20 pi d squared which leaves you with zero.

02:49 Hello, so then excuse me, we equate this to zero.

02:53 When you expand this, you get 20 pi d squared plus 14 ,000, d minus 840 ,000 is equal to zero.

03:05 Now you can use the quadratic formula, but we get d is equal to 21 .6 millimeters...

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