Statics 2D Virtual Work problem regarding calculation of Normal force

In summary: NIn summary, For problem 1, a frame supported by a hinge and roller is loaded with a couple of 14 kN*m, a force of 12 kN, and a distributed force of 4 kN/m. We can use the principle of virtual work to calculate the normal force in C, which is found to be -12.2666 kN. However, according to equilibrium equations, the answer should be -17.6 kN. For problem 2, a similar frame is loaded with a couple of 4 kN*m, a force of 6 kN, and a distributed force of 1 kN/m. Using the principle of virtual work, the shear force in C is found
  • #1
manan1
7
0
PROBLEM 1


Homework Statement



VIRT_LR_009.jpg


The frame in the figure is supported by a hinge in A and a roller in G. It is loaded by a couple = 14 kN*m in B, a force = 12 kN in D and a distributed force = 4 kN/m on section EG. = 1.4 m.
Calculate the normal force in C. Use the correct signs for tension and compression. Hint: Solve using the principle of virtual work and use previously mastered methods to check your answer.


This is a problem from mastering engineering, the statics book. I can solve the problem using equilibrium equations, but i want to know what i am doing wrong with when I am doing using the virtual work theorem.

Homework Equations



[itex]\delta[/itex]W = 0
[itex]\delta\theta small → tan ( \delta\theta) ≈ \delta\theta ; [/itex]


The Attempt at a Solution



DSC00011.jpg



[itex]\delta\theta_{1} = \delta\theta_{2} = \delta\theta[/itex]


[itex]\delta\theta_{1} = \delta u_{1}/a[/itex]
[itex]\delta\theta_{2} = \delta u_{2}/2a[/itex]


LET
[itex]\delta u_{2} = \delta u[/itex]

[itex]\delta u_{2}/2a = \delta u_{1}/a → \delta u_{1} = \delta u/2[/itex]
[itex]\delta u_{3}/(a/2) = \delta u/2a → \delta u_{3} = \delta u/4[/itex]
[itex]\delta u_{5} = \delta u_{1} = \delta u/2[/itex]
[itex]\delta u_{4} = \delta u_{2} = \delta u[/itex]

[itex]\delta W = 0 [/itex]

Therefore
[itex] -M*\delta\theta-N*\delta u_{5}-N*\delta u_{4}-F*\delta u_{2}-qa*\delta u_{3} = 0 [/itex]
[itex]\delta u ≠ 0 [/itex]
[itex]-M*\delta u/2a-N*\delta u/2-N*\delta u-F*\delta u-qa*\delta u/4 = 0 [/itex]
[itex]-\frac{3}{2}N = M/2a + F + qa/4 [/itex]
N = -12.2666... kN


But the answer should be -17.6 kN according to equilibrium equations.



PROBLEM 2

Homework Statement



attachment.php?attachmentid=40380&stc=1&d=1319724126.jpg


The frame in the figure is supported by a hinge in A and a roller in G. It is loaded by a couple = 4 kN*m in D, a force = 6 kN in B and a distributed force = 1 kN/m on section EG. = 1.5 m
Calculate the shear force in C with the sign convention as shown in the figure. Hint: Solve using the principle of virtual work and use previously mastered methods to check your answer.

Homework Equations



[itex]\delta[/itex]W = 0
[itex]\delta\theta small → tan ( \delta\theta) ≈ \delta\theta ; [/itex]


The Attempt at a Solution



attachment.php?attachmentid=40381&stc=1&d=1319724126.jpg


I have no idea how to calculate the virtual displacement for the force F at B.
 

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Last edited:
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  • #2
For problem 1, why are you sure that δθ1=δθ2 ?
 
  • #3
i induce the same amount of rotation in both, and than follow to measure the change in the distances
 
  • #4
I would try the other way around : assuming a deplacement δu2=δu1 and then calculate the rotation
 
  • #5
tried it, doesn't work.

the new virtual work equation becomes...

-M/a - F - qa/4 = 2N
=> N = -11.7
 
Last edited:

Related to Statics 2D Virtual Work problem regarding calculation of Normal force

What is the definition of Normal force in a 2D virtual work problem?

The Normal force in a 2D virtual work problem refers to the force that is perpendicular to the contact surface between two objects. It is the force that prevents two objects from passing through each other.

How do you calculate the Normal force in a 2D virtual work problem?

The Normal force can be calculated by using the equation N = mgcosθ, where N is the Normal force, m is the mass of the object, g is the acceleration due to gravity, and θ is the angle between the object and the horizontal surface.

What is the significance of the Normal force in a 2D virtual work problem?

The Normal force plays a crucial role in determining the stability and equilibrium of an object. It is also essential in calculating the forces acting on an object in a 2D virtual work problem.

How is the Normal force related to other forces in a 2D virtual work problem?

The Normal force is often related to the weight of an object and the other forces acting on it, such as friction and applied forces. It is essential to consider the Normal force when determining the net force on an object in a 2D virtual work problem.

Can the Normal force be negative in a 2D virtual work problem?

No, the Normal force cannot be negative in a 2D virtual work problem. It is always directed perpendicular to the contact surface and can only have a positive value. If the calculated value of the Normal force is negative, it means that the direction of the force is in the opposite direction as assumed.

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