Determine forces in beam with hinges

[fonseh]

Homework Statement

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

Homework Equations

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 couldn't be found .

Or there's something wrong with my working ?

 

[fonseh]

Bump

 

[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.

 

[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

 

[PhanthomJay]

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

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.

 

[fonseh]

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.

after making correction ,

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

 

[PhanthomJay]

after making correction ,

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

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.

 

[fonseh]

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.

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

 

[PhanthomJay]

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

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.