# Pls help with my shear and moment diagram problem

1. Nov 24, 2008

### lunaticpimp12

kindly help me with this problem.

i can't solve for the reactions and draw the shear and moment diagrams. can you solve it for me?

pls help me. thanks in advance.

2. Nov 24, 2008

### PhanthomJay

Hi there, we don't solve problems here, we just try to help you solve them. Solve for the reactions using newton's laws of equilbrium for forces in x direction, forces in y direction, and torques about any point, each of which must add to zero. The shear and moment diagrams , which then follow, are a bit time consuming. Please show some attempt, thanks.

3. Nov 24, 2008

### lunaticpimp12

i solve for the reactions and i got:

Reactions for E = 52.44 kN

Reactions for A = 39.56 kN

now my question is what is the shear diagram for the 5 kN/m. i got stuck when i draw it i got for the shear diagram of the reaction = 23.74 and i don't know how to draw for the 5kN/m. thanks.

4. Nov 24, 2008

### PhanthomJay

Your reactions look good, but member AB is the hypotenuse of a 3:4:5 right triangle, which is not scaled on the drawing correctly, probably accounting for your error; the axial load is .6(R_A) = 23.74kN just up from A, and the shear load is .8(R_A) = 31.65kN at that point.

Drawing the shear diagram on AB, it starts at +31.65 at A, continues with that constant value up to the distributed 5kN/m load, at (2,1.5), that is, at 2.5m along the member. At this point, you need to change the 5kN/m load into its axial and shear components (4kN/m shear distributed load), and note that dV/dx = -q, that is, the slope of the shear diagram now becomes 4KN/m within that distributed 4 m long load segment, so at the end of it, the shear is now 15.65kN. Now painstakingly continue, cutting FBD's every now and then to check your numbers and your sanity.

5. Nov 25, 2008

### lunaticpimp12

i got it now it should be 3.2 kN/m and i got it all correctly. anyways thanks for the help.