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## Mechanics Problem HELP!

 Quote by D44 So the ground reaction force is the hypotenuse of the triangle used to calculate the other forces? I'd previously written it as the adjacent. Just for future reference, how would you know which it was?
When you first look at the frame in its entirety, always find the reactions first, using sum of y forces =0, sum of x forces = 0, and sum of moments about any point = 0. When you look in the x direction, there are no applied forces in that direction, hence, no horizontal reaction at the support. There is only a vertical reaction P1 +P2, and a moment, P1x1 + P2x2. The vertical reaction is the resultant hypotenuse if you wish to break it up into comonents parallel and perpendicular to the sloped member.
 The shear force, 64.5sin15, does this not act at the rhs (ground)?
yes...
 If not, so far at the lhs I have a moment of 68.1Nm, a shear force of 64.5sin15N - at the rhs I have a horizontal force of 64.5cos15N and an moment of 128.4Nm? Am I also trying to find a vertical force at the rhs too?
yes, the shear force at the right hand side as you noted. Direction of force please at the left and right hand sides?

 Quote by D44 Using the equillibrium equations, the vertical up force at the rhs is equal to the 64.5sin15N force?
yes....
 What happens with the horizontal force at the rhs? Because the sum of Fx=0, but where does the other horizontal force come from to balance the rhs force? 64.5cos15 from lhs from splitting the 64.5N force up?
yes...the horizontal (axial) force in the member is 64.5 cos 15. You are doing very well so far...now just draw the shear and moment diagrams for this slanted member and you've got it nailed...
 Shear forces at both ends then, lhs acting downwards at 64.5sin15 and rhs acting upwards at 64.5sin15? So the axial force of 64.5cos15 is what equals the ground reaction force to make the sum of Fx=0? When drawing the shear and moment diagrams, how is the axial force incorporated? What effect does it have?

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 Quote by D44 Shear forces at both ends then, lhs acting downwards at 64.5sin15 and rhs acting upwards at 64.5sin15?
Yes! Since the sum of all forces perp to the member must be 0, if you have the shear force down at the left end, then it must act up at the right end,equal in mgnitude, because there are no other external forces applied on the beam in that direction.
 So the axial force of 64.5cos15 is what equals the ground reaction force to make the sum of Fx=0?
The axial compressive force of 64.5cos15 at the lhs must be balanced by an equal and opposite axial component of the reaction force at the ground.
 When drawing the shear and moment diagrams, how is the axial force incorporated? What effect does it have?
It has no effect when drawing the shear and moment diagrams. You could draw an axial force digram separately, which would be a constant force throughout. When determining the axial and bending memberstresses later, then you combine those stresses for the final results, per P/A +/- Mc/I.
 Brill, thank you!! My shear force diagram is looking like a vertical line down to -16.69N, horizontal line straight along to the rhs then back up to 0? As for the BM diagram, I'm not so sure, but it has to go vertically staight down to -68.1Nm and then up to 128.4Nm at the rhs? Is this a curve crossing at the centre point of the 0 line, not a linear line? The induced stresses are a combination of which, sorry? The shear forces and the axial force? Although I know what the P/A +/- Mc/I means, how am I to apply this? How do you mean per?

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 Quote by D44 Brill, thank you!! My shear force diagram is looking like a vertical line down to -16.69N, horizontal line straight along to the rhs then back up to 0?
Yes, you are quite correct.
 As for the BM diagram, I'm not so sure, but it has to go vertically staight down to -68.1Nm
yes
 and then up to 128.4Nm at the rhs? Is this a curve crossing at the centre point of the 0 line, not a linear line?
It doesn't cross the line. The moment at the lhs and the shear at the lhs, both produce ccw moments about the support, so the straight line (not a curved line) from left to right slants down, not up. You must remember that the slope of the moment diagram at a given point is equal to the shear at that point. Thus, its slope is -16.69, a negative slope, and the moment at the right end is thus -68.1 - 16.69(4) = about -134 ( there is a round off error somwhere, as it should agee with the 128.4 you calculated earlier).
 The induced stresses are a combination of which, sorry? The shear forces and the axial force? Although I know what the P/A +/- Mc/I means, how am I to apply this? How do you mean per?
It looks like the problem as given on page 1 did not ask for stresses..that's probably the 2nd part of the question.. If you haven't got to stresses yet, don't go any further until you study that topic.

i am confused about the moment at the slanting bit... i have worked it out as 134 .. is that right

 Quote by PhanthomJay Yes, you are quite correct. yes It doesn't cross the line. The moment at the lhs and the shear at the lhs, both produce ccw moments about the support, so the straight line (not a curved line) from left to right slants down, not up. You must remember that the slope of the moment diagram at a given point is equal to the shear at that point. Thus, its slope is -16.69, a negative slope, and the moment at the right end is thus -68.1 - 16.69(4) = about -134 ( there is a round off error somwhere, as it should agee with the 128.4 you calculated earlier). It looks like the problem as given on page 1 did not ask for stresses..that's probably the 2nd part of the question.. If you haven't got to stresses yet, don't go any further until you study that topic.
 Recognitions: Gold Member Homework Help Science Advisor Ok let's crank it out. P1 is (20)(2.15) = 43 and P2 is (20/2)(2.15) = 21.5, so the vert load is 64.5 N. The moment about the support is 43[(2.15/2) +.1 + 4 sin15] + 21.5[(2.15/3) + .1 + 4 sin15] = 134.9 N-m. Or, if you look at the moment at the support in the free body of the slanted member, it's 64.5(4)(sin15) + 68.1 = 134.9 N-m.....checks out OK.
 Using the bending equations, I have a value of approx 88Pa for the induced stresses. Am I right in using the the biggest moment in the equation, which is at the ground?
 Sorry, 44.24MPa

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 Quote by D44 Sorry, 44.24MPa
It's somewhere around there, I didn't calc out the numbers, but yes, max bending stress is Mc/I where M is the max moment (139 Nm) occurring at the support, which controls the overall frame structural design. The I term comes from the properties of the hollow circle, and c is its outside radius.
 Hi, i am having the same trouble with this assignment, mainly drawing the free body diagram of each section. Not sure exactly whether to draw it as a full frame or two seperate parts. thanks

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