I believe the solution for problem 56 in SMS MD is incorrect.

The problem is a 500 lbf load applied eccentrically to a plate with 4 bolts.

In the solution the horizontal and vertical shear stresses are reversed. Horizontal should be 2140 psi, and vertical 2570 psi.

The stress in the critical fastener comes out to be 3856 psi. Right between options C 3800 and D 3900.


Anyone confirm?

I believe the final number is correct...I got shear = 3790 psi, however, I see the mistake you are referring to. The solution uses the sine of theta to calculate the horizontal component. Obviously it should be cosine of theta. I agree that this is a mistake. Here is the kicker though...punch in the numbers that 6MS solution uses and you will not get the same answer as the book. They set-up the equation wrong, but give the correct final number!

I believe the shear stress x-component of 2570 psi is correct....3348 * cos (40 deg) = 2565 psi.

I am about ready to ask for my money back for this book!

The author set the equations up correctly. Since the acute, upper, angle was used 3340*sin(39.8) is correct to find the horizontal but the vertical shear was published instead. Same with vertical: 3340*cos(39.8) is correct but the horizontal shear is shown as the answer. I set it up using the normal right triangle orientation with 50.2 deg so 3340*cos(50.2) is shear in x and visa versa. Being that 2570 and 240 are represented on the wrong axis the direct vertical shear is added to wrong component and plugged into the critical fastener stress equation producing and incorrect answer. I'll have to look at it again.

Either I'm going nuts or in the very next problem the area for sheet shear and the sheet tensile areas are both wrong. Sheet shear should be 0.0206 in2 not 0.02953 in2, and Sheet tensile should be 0.0306 in2 not 0.03953 in2.

Confirm?

I'll take a look at the critical fastener problem again, but...Yes, the very next problem calculates the areas wrong.

After looking it over several times I getting the same resultant shear of 3788 psi. It may be a case where I have looked at this problem for so long the mistake is staring me in the face, but I just can't see it. None the less, here are my numbers...

Torsional shear x-component = 3348*cos(39.8) = 3348*sin(50.2) = 2572 psi. Sine vs cosine just depends how the triangle is set-up.
Torsional shear y-component = 3348*sin(39.8) = 3348*cos(50.2) = 2143 psi

Direct shear y-component = 638 psi

Resultant shear = [(2572^2 + (2143 + 638)^2]^1/2 = 3788 psi.

Does the difference come down to how we are setting up the trig?

You're right Brian, I had my triangle flopped. Thanks for posting!

You're right...see the errata for SMS MD.

p. 56, Solution 70: For the horizontal shear stress calculation and the vertical shear stress calculation, the formulae are correct, but the numerical answers were swapped. That is, [tau]x should be “2570 lbf/in2” and [tau]y should be “2140 lbf/in2”. For the direct vertical shear calculation, the formula is correct, but the numerical answer was wrong. So [tau]v should be “638 lbf/in2”. These changes impact the final stress calculation for [tau]. Substituting in the correct values, the final answer is “3784 lbf/in2 (3800 psi)”. The correct answer is ©. In the Why Other Options Are Wrong section, option © becomes (D) and should now read, “This incorrect solution results when the horizontal and vertical stresses are reversed.” (Bert Hartman) 9/26/2005

http://ppi2pass.com/ppi/PPIErrata_pg_Errata-SXMMe1p1.html