Determine forces in beam with hinges

In summary: Yes, your equations are ok but if you try to solve them correctly you get 0 = 0. The concept is wrong.
  • #1
fonseh
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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 ?
 

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  • #3
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.
 
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  • #4
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
fonseh said:
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.
 
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  • #6
PhanthomJay said:
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
 
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  • #7
fonseh said:
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.
 
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  • #8
PhanthomJay said:
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 ?
 
  • #9
fonseh said:
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.
 
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FAQ: Determine forces in beam with hinges

What is the purpose of determining forces in a beam with hinges?

Determining forces in a beam with hinges is important for structural analysis and design. It allows engineers to ensure that the beam can withstand the expected loads and forces without failure.

How do hinges affect the forces in a beam?

Hinges act as points of rotation or support in a beam, which can change the distribution of forces and moments in the beam. This is why it is important to determine the forces in a beam with hinges.

What factors influence the forces in a beam with hinges?

The forces in a beam with hinges can be influenced by the type and location of the hinges, the magnitude and direction of applied loads, and the properties of the beam material.

What are the methods for determining forces in a beam with hinges?

There are several methods for determining forces in a beam with hinges, including the method of sections, the method of joints, and the method of virtual work. These methods use principles of statics and mechanics to calculate the forces and moments in the beam.

Can computer software be used to determine forces in a beam with hinges?

Yes, there are many computer software programs available for structural analysis that can determine the forces in a beam with hinges. These programs use numerical methods and algorithms to calculate the forces and provide visual representations of the beam's behavior.

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