Have i done this right? Simply Supported Beams

  • Thread starter Thread starter Vancurt
  • Start date Start date
  • Tags Tags
    Beams
AI Thread Summary
The discussion centers around calculating the reactions Ra and Rb for a simply supported beam under specific loads. The user struggles with finding these values using the sum of moments and has provided examples from their homework and notes. They initially calculated Ra and Rb incorrectly but later confirmed their original calculations were accurate. The user seeks clarification on how to plot the shear force diagram (SFD) and receives feedback that their rough version of the SFD appears acceptable. The conversation emphasizes the importance of checking calculations by taking moments about different points.
Vancurt
Messages
9
Reaction score
0
ƩƩ

Homework Statement



I am having issues finding Ra and Rb using the Sum of the momemnts, it's ƩMCW = MACW.

I need to find the position of the maximum and minimum bending moments.

The Supported beams in question is 3.5m long. Starting from the left and moving towards the right: 1m in and then a 2KN load, then 1.5m further in and then another load of 3KN hits down and then an extra 1m when it meets the end.

Like this: 1m, 2KN, 1.5m, 3KN, 1m.
Ra Rb

Looking at my notes for some example i did (didn't finish the notes, must of forgot and now paying the price), example i did:

3m, 4KN, 4m, 3KN.
Ra Rb

And here are the notes:

ƩMCW = MACW

(4x3) + (3X7) = Rb x 12

12 + 21 =Rb x 12

33/12 = Rb =2.75KN

ƩFup = ƩFdown

Ra + Rb = 4+3

Ra + 2.75 = 7

Ra = 7 - 2.75

Ra = 4.25KN

In the Engineering book i brought: Engineering Science by Mike Tooley and Lloyd Dingle, there is a beam example:

2m, 20KN, 3m, 5KN, 2m, 30KN, 2m, 15KN.
Ra Rb

He did this: 2x20 + 5x5 + 7x30 + 9x15 = 8xRb

Using this example and my own, i attempted to answer the question to find Ra and Rb. The very first and poorly drawn supported beam at the top.

Homework Equations



Direct Stress σ = F/A = Force/Area n/m squared

Sheer stress ζ = F/A = Force/Area n/m squared

Strain ε = Δ x L/L = Change in length/Original length.

Modulus of elasticity E = σ/ε = stress/strain N/M squared.

The Attempt at a Solution



Here is my attempt at the very first diagram (the 3.5 over all length)

(2x1) + (2.5x3) = 9.5

Rb x 3.5 = 9.5 x 3.5 = 33.25

2+7 = Rb x 3.5

9.5/3.5 = 2.71KN

Rb = 2.71KN

ƩFup = ƩFdown

Ra + Rb = 2 + 3 = 5

Ra + 2.71 = 5 - 2.71 = 2.5 KN

Ra = 2.5KN

So Rb = 2.71 KN and Ra = 2.5KN

What i don't undertand is on my previous attempt, Ra = 2.286 KN and Rb = 2.714KN

Where have i gone wrong? And what do i need to do after finding Ra and Rb?
 
Physics news on Phys.org
Found Ra. How did i miss that.

My first notes were correct. Rb = 2.714kn and Ra = 2.286kn

I did the calculations wrong.
 
Last edited:
Ok, i need help, how do i plot the sheer force diagram?

I did a rough example:

+

+2.28 for 1m then with the 2KN i come down to .28 for 2.5m and with the 3KN i come down to -2.72 for 3.5m and then back up to 0 inbetween the plus and minus.
-

Is this rough version ok?
 
You can always check your own reactions by taking moments about ANY point you haven't used to get the initial result. Your SFD seems ok to me, But people use various sign conventions.
 
Back
Top