# Determine forces in beam with hinges

1. Nov 28, 2016

### fonseh

1. The problem statement, all variables and given/known data

Is it possible to find the VA , VB and VC without ' breaking ' the beam into 2 parts ?

2. Relevant equations

3. The attempt at a solution
Here's my working , i gt

Moment about A = 10(1) -VB(2) -VC(4) +5(3)(4 + (3/2) ) = 0 Hence , 92.5-2VB -4VC= 0

moment about B = VA(2) -10-VC(2) +(5)(3)(2 + (3/2) ) = 0 Hence , 42.5 + 2VA -2VC = 0

Moment about C = VA(4) -10(3) +VB(2) + (5)(3)(3/2) =0, Hence , -7.5 +4VA +2VB = 0

and sum of vertical force = VA +VB +VC = 25

Then , i solve the 4 equations using online calculator , i found that some of the values couldnt be found .

Or there's something wrong with my working ?

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2. Nov 29, 2016

### fonseh

Bump

3. Nov 29, 2016

### PhanthomJay

you are making a few errors here, particularly when including the hinge force at B in your equations when looking at the whole beam, because it is not an external support. But anyway, looking at the entire beam, you can sum moments = 0 about countless points and get an infinite number of equations , but you won't get anywhere, because you only have --?-- equilibrium equations to work with. Break the beam apart.

4. Nov 29, 2016

### fonseh

Do you mean for Moment about A = 10(1) -VC(4) +5(3)(4 + (3/2) ) = 0 Hence , 92.5 -4VC= 0 , it should be like this ?

For Moment about C = VA(4) -10(3) + (5)(3)(3/2) =0, Hence , -7.5 +4VA = 0 , is should be like this ?

For moment about B , = VA(2) -10-VC(2) +(5)(3)(2 + (3/2) ) = 0 Hence , 42.5 + 2VA -2VC = 0 ?

sollving 3 equations , i have VA = 1.875 , VC = 23.125 , VB =0 ?
bUT , my ans is still different from the author ans

5. Nov 29, 2016

### PhanthomJay

You are not paying attention , you forgot to include the unknown moment , M, at the fixed end. The problem cannot be solved unless you break up the beam at B. Follow the book solution.

6. Nov 30, 2016

### fonseh

after making correction ,

Moment about A = 10(1) -VC(4)+M +5(3)(4 + (3/2) ) = 0 Hence , 92.5 -4VC +M= 0

For moment about B , = VA(2) -10-VC(2) +(5)(3)(2 + (3/2) )+M = 0 Hence , 42.5 + 2VA -2VC +M= 0 ?

For Moment about C = VA(4) -10(3) +M+ (5)(3)(3/2) =0, Hence , -7.5 +4VA+M = 0 ,

here's my answer , M = 0 , VA = 1.875 , VC = 23.125 , which is still not same as the author's working

Last edited: Nov 30, 2016
7. Nov 30, 2016

### PhanthomJay

well, you didn't do it right. You do not get any more useful equations when summing moments about multiple different points. Suppose you had 2 equations, x + y = 0 and 2x +2y = 0. 2 equations, 2 unknowns, but I bet you can't solve for x and y.
If you do not wish to believe that you must break up the beam to solve, then I can't help you any further.

8. Nov 30, 2016

### fonseh

do you mean my equation is correct ? just my concept is incorrect ?

9. Nov 30, 2016

### PhanthomJay

Yes, your equations are ok but if you try to solve them correctly you get 0 = 0. The concept is wrong. When you look at the entire beam, you have 4 unknowns....the moment at A, VA, HA, and VC. Four unknowns, with only 3 useful equilibrium equations: sum of forces in y direction = 0, sum of forces in x direction =0, and sum of moments about a point = 0. So you need to resort to another method to solve, which is the one described in the solution.