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Addendum to last post: I did a quick review of statics, and I see that M, N and V are internal forces. The torsion and shear due to the pipe wrench are external forces. So of course the internal and external reactions would have to oppose each other, just as they show in the answer key's...

OK, here's a summary of the forces at points A & B: I = 1/12(base)(height)3 = 1/12(3)(6)3 = 54 Q = y-bar*area = 1.5 * 9 = 13.5 where area is 3 high and 3 deep For point A: Normal force = P/A = -597/18 this is negative since it is in compression Bending moment= Mc/I =...

Oh, I see. It was just a matter of geometry. Thank you, thank you! :^)

The book's answer key shows, for point A, sigma = P/A = 597.49/18 in compression. The 597 is the sum of [cos(45 degrees)(350LB)] plus 350LB. cos(45 degrees)(350LB) is 247LB, which is bending the boom clockwise, therefore putting point A in compression. The additional 350 must come from the...

The book's answer key shows a normal stress P/A as a result of adding the 350LB force from the cable to the cos(45 degrees)(350LB). It shows both of these as compressive forces. I can see how the cos(45 degrees)(350LB) is compressive; the force of the 350LB load is bending the boom clockwise...

Oh, thank you!!! That was a subtlety that I had missed. And it obviously had a profound effect on the calculations. Thank you again.

OK, for anyone else who is stuck here, I found the answer in another youtube video: The equation of our line is y=mx+b. b=0. m is the slope, which they call 4KN/3m when you examine the left-most 3 meters of the load (back to that in a, ahem, moment. Sorry, I just couldn't resist making that...

OK, so the equation for the load is simply y=4x. To obtain shear, integrate y=4x to yield y=2x2. At a value of x=1.5, the shear is 4.5KN, which agrees with the book's answer key. Is this correct so far? Integrate again and you get y=2x3/3. At a value of 1.5, this yields a moment of 2.25...

TL;DR Summary: A frame with a triangular distributed load is pin-connected to a 2-force member. Find the combined stress at point E on the frame. I am stuck at determining the value of M at the cut. The book shows the value at 8.25KN-meter, but I cannot see how they arrived at that number...

Now that was an embarrassing error.:oops: OK, I now have the correct answer for IB. Next, we look at the top of the frame, examining DHC; If we sum the moments from point D, the moment caused by GH balances the moment caused by the load at point C, and we get the correct answer for the value...

IB would have to be in tension. So at point I, Ix is to the left, and Iy is negative. Oh, I see your point. Let's amend the equation: ΣFx: Ax - (cos60)IB = 0 However, for the Fy equation, which is where we determine the value of Ay, the signs and logic and values are correct, right? I assume...

Homework Statement Given the frame shown, determine the internal loadings at D Homework EquationsThe Attempt at a Solution This should be a simple problem...but I cannot see where I am making a mistake. If I approach it by examining the entire structure, I get a wrong answer: ΣFx: Ax +...

OK, here comes the math. -P -294.3 + Ntop + 0.25Nright=0 We know that Ntop=4Nright, so we can immediately substitute, giving us: -P -294.3 + 4.25Nright Let's call this eqn 1 The torque eqn is: -0.075P-22.07+0.056Nright+.325Nright=0 -0.075P-22.07+0.381Nright=0 Let's call this eqn 2 Let's...

OK, that makes sense. I am embarrassed to admit I overlooked torques. So here we go: Sum of forces in Y: -P -196.2 - 98.1 + Ntop + .25Nright = 0 notation: Ntop is normal force of top block Sum of forces in X: -.25Ntop + Nright =0 Cool! We immediately know a relationship between the...

OK, here's my FBD about point B. From B's point of view, the 10KG is pulling it to the right and down slightly. The downward pull is offset by mu times its N. From B's point of view, the 20KG block is pulling it up and to the left slightly. The leftward pull is offset by mu times its N. And we...