Question

Hi

I have been stuck on this moments question for a while so please do help, how do i get the shear forces at the points A B and C for this beam.

I know the the sum of upward forces are equal to downward forces so Fa+Fb=225

My problem is how do i take moments about B and C to get the shear forces?

Image of question posted below:

http://imageshack.us/photo/my-images/197/questin.jpg/

Answer 1

From points a to b the load is distributed, so 25x9=225 N total udl load , what confuses me is the point load A.

Answer 2

Is it mentioned that the body is in translational and rotational equilibrium?

Answer 3

no it does not mention neither it just says calculate the maximum loads and the points a , b and c

Answer 4

How do i take moments abouts about support b or a?

Answer 5

i guess the shear force at c is 30kn right?

Answer 6

I've not understood the situation depicted completely...Anyway I don't think u can work out anything if the body is not at equilibrium. I'd say make two equations. One by equating the force upwards = force downwards
Second, Torque clockwise (about CM) = torque anticlockwise (about CM)
Find out the variables, then proceed for watever is asked in the question...

Answer 7

Yeah it is 30 KN

Answer 8

Oh no actually 30 KN is the normal force there!

Answer 9

how do i take get the shear force and support b?

Answer 10

Let me check the pic again and try understand...Hold..

Answer 11

if the value of ax=0 ?

Answer 12

I guess the udl can be divided since it is evenly distributed so 112.5kn, from a to b... im i on the right lines here?

Answer 13

Skull patrol any ideas how to proceed?

Answer 14

iknow how to get the shear force of a if i have b , what i dont know is the method how to solve for support b?

Answer 15

The upwards forces are in equalibrium, i know that.... but how can i get the force at b if ax=0?

Answer 16

i know how the shear force diagram looks like .... i just dont know get the force at b

Answer 17

Can anyone help? im new to this topic ...