# Mechanics Problem HELP!

by mikex24
Tags: mechanics
 P: 87 yes i understand it.
 PF Patron HW Helper Sci Advisor P: 5,801 You can't pass the module with a 1 hour lecture. Unless you studied it on your own for many many hours after that. I don't know what you were expecting.
 P: 87 it is not just bending on beams. There are many others such as second moment of area, kinetics, kinematics...
PF Patron
HW Helper
P: 5,801
 Quote by mikex24 it is not just bending on beams. There are many others such as second moment of area, kinetics, kinematics... but because this is an assignment and the lecturer thinks that the students are experts with one hour lecture and one hour seminar on bending beams.. :( .
This problem as written has nothing to do with bending of beams. It asks for the shear and moment diagrams. The bending stress is simple, it 's just Mc/I. It sounds to me like you enrolled in a course without having the necessary prerequisite courses on the basics.
 P: 87 Hello. I did some work and i need some tips for the next stage:
 P: 87 Hello. I did some work and i need some tips for the next stage: I found the P1, P2, Mx, Qx P1=20*2.15=43 N and it acts (2.15/2)+0.1=1.175 from the right end P2=0.5*2.15*20=21.5 N and it acts (2.15/3)+0.1=0.816 from the right end therefore P1+P2=Qx therefore Qx=64.5 N so Mx=(43*1.175)+(21.5*0.816)= 68 Nm therefore R=Qx=64.5 so Rcos15=62.3 and Rsin15=16.69 is that right? Cheers
 PF Patron HW Helper Sci Advisor P: 5,801 I'm glad you did not give up. Those values are correct, but i'm not sure why you call it Qx and Mx; the value you calculated for Qx is the vertical upward internal reaction at the right end of the horizontal beam (call it Q), and the value for Mx (call it M_right) is the internal clockwise moment at the right end of the horozontal beam. You still need to calculate the moment at the support. Now you need to draw the shear and moment doagrams as requested iin the problem..
 P: 87 yes... here is the trouble. how can i found the moment on the fixed support(i think that is the same as the upward but in different direction)?
 PF Patron HW Helper Sci Advisor P: 5,801 l already gave you the equation for the moment at the fixed base support, in a sketch many posts ago. That is one way of doing it. But you need to calculate the values of x1 and x2, using basic geometry and trig.
 P: 87 I did it. I found 64.5. I have to use the Rcos15 to find the x1 and x2?
 P: 87 I use trigonometry and i found the x1 and x2 and i found the Mx at the fixed end of the framework which is : 134.805. ??
 P: 87 What is the total plan and the next stages to solve this question?
PF Patron
HW Helper
P: 5,801
 Quote by mikex24 What is the total plan and the next stages to solve this question?
Please tell me how you would go about drawing shear and moment diagrams for the more simple case of a simple beam supported at each end with a uniformly distributed load. You have to get back to the basics............as i tried to note before, if you don't understand the basic steps, you will find drawing the shear and moment diagrams for this problem quite difficult.........and I'm not about to do it for you...........we are here to assist, but you must do the work..............
 P: 87 i know that we have to cut the beam in x values to find in each point of x the value of moment and force. ? for A-B the Mb and Qb will zero then if i start from the left of the beam then for B-C part i will cut the just after the first force which is the force of triangular the problem i found there is that is that i cant find many unknows x to have quadtrative equation because always the unknows x is just one like 8x-3 and this is a straight linear equation. ??
 P: 1 We've got a similar problem(the exact same) to this one for our assignment and we were wundering if the strength of the tubeing by workin out the external area then taking the internal area away from it to get the area of the piping and then divide the values of the forces we got from the wieght of the billboard. we did this but the values we got were extremly small compared to the yeild stress of the low carbon steel. the value we got was 0.2MNm^-2 but the yeild stress is 300MNm^-2 thanks for any help you can give
PF Patron
HW Helper
P: 5,801
 Quote by Joshsamuel117 We've got a similar problem(the exact same) to this one for our assignment and we were wundering if the strength of the tubeing by workin out the external area then taking the internal area away from it to get the area of the piping and then divide the values of the forces we got from the wieght of the billboard. we did this but the values we got were extremly small compared to the yeild stress of the low carbon steel. the value we got was 0.2MNm^-2 but the yeild stress is 300MNm^-2 thanks for any help you can give
This is the 3rd separate post on this same problem, and I need to know up front where this assignment is coming from, and for what purpose it is being assigned, before we can provide additional assistance.
 P: 87 This is an assignment of a UK university. As you can there are many person with difficulties on this assignment as we teached only 2 hours on this module for the beams..
PF Patron
HW Helper