Statics problem (Simple first year): Calculate the internal forces for this structure

[YNH]

Homework Statement

Static (Simple first year): Me and my teacher gets to different answers

Relevant Equations

The equilibrium equation

I need to calculate the internal forces M3, V3, and N3 for a specific section of a structural system. The known reaction forces are:

• R_AH = -5.7 kN
• R_AV = 0.7425 kN
• R_CV = 5.02 kN

Two different sections have been analyzed: one on the left side of the cut (my choice) and one on the right side (analyzed by my teacher). I have attached both my calculations and my teacher’s calculations, as well as an image of the static system. My image is labeled with "Student" and shows my approach.

Could someone please review the attached calculations and images and help verify the correct values for M3, V3, and N3? Any insights or corrections would be greatly appreciated. Thank you!

 

[Lnewqban]

Welcome, @YNH !

First things first: A clear and precise free body diagram should be the very first step when trying to solve problems of this type.
Otherwise, trying to solve even the simplest problem may lead to errors.

Please, go over the signs of your equation of moment at cut section 3.

As the values of the reactions are given, you could verify the accuracy of each solution (right and left approaches to cut section 3) simply by substituting x by the 4.8 meters value.

 

[YNH]

I have looked through the results and signs but can't see the error still getting the same result.

 

[Lnewqban]

Without that horizontal force (5.7 kN), that cross section 3 would feel only the amount of internal shear and moment stresses induced by the distributed load, and the vertical reactions at both supports, A and C, would be symmetrical and each would be equal to half the distributed load.

The existence of that horizontal load, combined with the incapacity of the support C to offer any horizontal resistance, creates an additional internal moment that is transferred along the beam until our cut section 3.

Therefore, in order to keep the previous balance after we complete the cut, we need to provide that section with all the imaginary external forces and moments needed to compensate.

Based on the above, just estimate mentally, approximately how much compensating moment that section 3 needs.

 

[YNH]

I'm not sure what you mean my first language is not English. Shouldn't I just use the equilibrium equations that I have done? There are no more forces that I can plus or minus with. So I don't really know what I'm doing wrong.

 

[YNH]

Second off all, the two vertical reaction forces are not the same, is that the problem?

 

[YNH]

I think the problem lies with the vertical reaction force is not the same because that is the only difference. But I calculated the reaction forces correctly and got the same result as my teacher.

 

[Lnewqban]

I'm not sure what you mean my first language is not English. Shouldn't I just use the equilibrium equations that I have done? There are no more forces that I can plus or minus with. So I don't really know what I'm doing wrong.

Please, excuse me the complicated previous explanation then.
I only wanted you to understand the need for a concentrated moment at the cut section.

The difficulty that I see:
Not all the signs in your equation of moments balance are correct.

Note how your teacher shows the positive directions of forces and moments in his solution, and follows that convention with his equations.

My calculated numbers for the reactive forces are as follows:
Horizontal reaction in support A = 5.7000 kN to the left
Vertical reaction in support A = 0.7425 kN pointing up
Vertical reaction in support C = 5.0175 kN pointing up

 

[YNH]

I didn't mean to be disrespectful in my previous answer so I apologize but shouldn't the horizontal reaction force in A be -5.7.

 

[YNH]

I have now checked my signs for the equation of moments balance and have now the equation
[M][/3]+1.2*x*(x/2)+5.7*1.8+(-5.7)*3.6-0.7425*x

can you point out what is wrong and also our equation for the shear force is also wrong what is wrong there?

And i like to thank you for your help

 

[Lnewqban]

I didn't mean to be disrespectful in my previous answer so I apologize but shouldn't the horizontal reaction force in A be -5.7.

You were not disrespectful.
The sign of the force depends on what your sign convention is.

I only know that the horizontal reaction force at A must be pointing in opposite direction to the only external horizontal force (5.7 kN pointing to the right).
The horizontal reaction force at A is pointing to the left.

 

[Lnewqban]

I have now checked my signs for the equation of moments balance and have now the equation
[M][/3]+1.2*x*(x/2)+5.7*1.8+(-5.7)*3.6-0.7425*x

That is not an equation (it is missing =0).

Summation of all the clockwise moments (about the cut section) = Summation of all the counterclockwise moments (about the cut section)