logistic_guy
Full Member
- Joined
- Apr 17, 2024
- Messages
- 591
Determine the internal normal force, shear force, and bending moment at point \(\displaystyle C\) in the beam.
stress on a beam
- Thread starter logistic_guy
- Start date
show us your effort/s to solve this problem.
Draw a Freebody diagram of the beam
Calculate the support loads at A and B - using equations of static equilibrium.
You claim to be an engineering student - and you did not know these initial steps ?!
Continue....
Any progress?
logistic_guy
Full Member
- Joined
- Apr 17, 2024
- Messages
- 591
Very valuable information. Thanks for the share professor khan.
I was. I was not only an engineering student, but also the best of all of them, at least in the graduation period.
I did not say that. I said I have never solved this problem before. But I have an accumulating knowledge that could help me crack this problem.
Of course!
Here is my strategy on how to crack this problem. The first thing is I want to understand how the forces mentioned in the question will act on the point \(\displaystyle C\). It seems that they are just a horizontal force (internal normal force) and vertical force (shear force). And the bending moment is the easiest thing to visualize as it is nothing more than a torque acting on that interesting point.
The main idea is to cut the beam into two parts at the point \(\displaystyle C\), and then try to find these two forces and the moment.
Let us assume that I cut the beam, and I am working with left half of the beam. In the horizontal direction, there is only one force, the internal normal force, \(\displaystyle N_C\). Since nothing will be happening to the beam in the horizontal direction, this force is equal to zero.
\(\displaystyle N_C = 0\)
Next, I will calculate the vertical forces. I have only two, one acts on the point \(\displaystyle A, F_A\) and one acts on the point \(\displaystyle C, F_C\), which is the one we are interested to find.
I will assume that the shear force, \(\displaystyle F_C\) is acting upward, and I will take upward as positive.
Then
\(\displaystyle F_A + F_C = 0\)
\(\displaystyle F_C = -F_A\)
Next, I will calculate the bending moment, \(\displaystyle M_C\) taking it as a counterclockwise torque and taking the counterclockwise direction as positive.
Then
\(\displaystyle 60000 + M_C - F_A(1) = 0\)
\(\displaystyle M_C = F_A - 60000\)
Now I have the needed equations in hand. All remained is to calculate the value of the force \(\displaystyle F_A\). That will be the idea of working on the full beam.
When working on the full beam, we don't care about the point \(\displaystyle C\) anymore. Let us find the vertical forces acting on the beam.
\(\displaystyle F_A + F_B - 10000 = 0\)
\(\displaystyle F_A = 10000 - F_B\)
Next, let us work on moments (torques) acting on the beam. Before that we have to choose a turning around point. I think that the point \(\displaystyle B\) is a perfect point for a turning around point as seen in the figure. It is a matter of taste, you can choose anywhere else.
As usual, I will take counterclockwise as positive.
\(\displaystyle 60000 - 2F_A - 2(10000) = 0\)
\(\displaystyle F_A = \frac{60000 - 20000}{2} = \frac{40000}{2} = 20000\) N
This means that I will not need this equation \(\displaystyle F_A = 10000 - F_B\), but I can find \(\displaystyle F_B\) if I want.
Finally, we can find everything we were asked for.
The internal normal force:
\(\displaystyle N_C = 0\)
The shear force:
\(\displaystyle F_C = -F_A = -20000 = -20\) kN
The bending moment:
\(\displaystyle M_C = 20000 - 60000 = -40000 = -40\) kN
The negative sign is an indication that I have chosen the opposite direction of the actual direction. I chose the bending moment to be counterclockwise when it is in fact clockwise. I chose the shear force to act upward when it is in fact acting downward.
In simple words, the shear force, \(\displaystyle F_C\), is an internal force in the beam. If an external force that is larger than the calculated shear force acts on the point \(\displaystyle C\), it will cause the internal force to increase which in turn will cause the internal bending moment, \(\displaystyle M_C\), to increase which in turn will cause a deformation of the shape of the beam, usually bending it, and sometimes even cutting it.
Right idea - but instead of describing with words draw free body diagrams.
Follow thee steps for a "concise" answer.