Homework checkover, Beams

1. May 25, 2012

Harrison01

Just wondered if someone was able to give the homework the once over as im not feeling confident i have this right.

1. I have a simply supported beam 10metres in length, supported at 2 metres in from both ends. there is a 5Kn U.D.L along the entire length of the beam and downward forces of 10Kn 3 metres in from the LHS and 20Kn at the far RHS of the beam

Determine the vertical reactions at the supports.

Ok, first of all i worked out the U.D.L and its position. Total U.D.L= 10(length of beam) x 5 (Load)= 50Kn. This force is then added to the beam as a downward force of 50Kn in the middle of the distance between the 2 supports.

(force x distance from support Ra) = (Support Rb x distance between the 2 supports)
(10x1)+(50x3)+(20x8) = (Rbx6)
10+150+160 = 6Rb
320 = 6Rb
Rb = 320/6
Total upward force at support Rb = 53.3Kn

(force x distance from support Rb) = (support Ra x distance between the 2 supports)
(20x2)+(Rax6)=(10x5)+(50x3)
40+6Ra=50+150
6Ra=50+150-40
6Ra=160
Ra=160/6
Total upward force at support Ra = 26.7Kn

upward forces = 53.3Kn +26.7Kn=80Kn
downward forces = 10Kn+50Kn+20Kn=80Kn

I'm pretty happy i have this right as my upward forces are the same as my downward forces putting the beam in equilibrium.

I have other Q's but ill do 1 at a time.

2. May 25, 2012

Harrison01

Sketch the shear force diagram for the beam.

Ok, downward forces are Negative and upward forces Positive. Because there is a UDL involved the diagram will have a slant to it.
Working from the far LHS:
@ LHE Q (let the shear force @ Y0 be Q0 and so on) = 0
@ 1M Q1= -1x5 udl (distance from LHE x force) = -5
@ 2M Q2 = -2x5 udl = -10
Include Ra = -2x5 udl+26.7 = +16.7
@ 3M Q3 = -3x5 udl+26.7 = +11.7
Include concentrated load of 10Kn = -3x5 udl+26.7-10 = -1.7
@ 4M Q4 = -4x5 udl+26.7-10 = -3.3
@ 5M Q5 = -5x5 udl+26.7-10 = -8.3
@ 6M Q6 = -6x5 udl+26.7-10 = -13.3
@ 7M Q7 = -7x5 udl+26.7-10 = -18.3
@ 8M Q8 = -8x5 udl+26.7-10 = -23.3
Include Rb = -8x5 udl+26.7-10+53.3 = +30
@ 9M Q9 = -9x5 udl+26.7-10+53.3 = +25
@ 10M Q10 = -10x5 udl+26.7-10+53.3 = +20
Include concentrated load of 20Kn = -10x5 udl+26.7-10+53.3-20 = 0

I'm happy i have this right as the end result is 0.

sorry for the essays..

3. May 25, 2012

Harrison01

Calculate the bending moments along the beam @ 1M intervals.

working from left to right

Bending moment @ far LHE = Ma = 0
@ 1M from far LHE M1 = -M due to udl
@ 1M = -(5x1x0.5) = -2.5
@ 2M = -(5x2x1) = -10
@ 3M = -(5x3x1.5) + (26.7x1) = +4.2
@ 4M = -(5x4x2) + (26.7x2)-(10x1) = +3.4
@ 5M = -(5x5x2.5) + (26.7x3)-(10x2) = -2.4
@ 6M = -(5x6x3) + (26.7x4)-(10x3) = -13.2
@ 7M = -(5x7x3.5) + (26.7x5)-(10x4) = -29
@ 8M = -(5x8x4) + (26.7x6)-(10x5) = -49.8
@ 9M = -(5x9x4.5) + (26.7x7)-(10x6)+(53.3x1) = -22.3

State the position and magnitude of the maximum bending moment in the beam.
-49 @ 2M from the far RHE