# Forces and stress

## Homework Statement

1.Explain how one with the help of the free body diagram can calculate forces and moments on the vertical cut line through the beam in a distance 1 m to the right from the point C.

2. Explain how one calculate size and distribution normal stress of the cut line. The beam is a I profile with moment of inerti I= 22,2*106 mm4 and height = 162 mm. ## Homework Equations

Part 1.

∑Fx = 0 ∑Fy = 0 ∑Fz = 0

∑Mx = 0 ∑My = 0 ∑Mz = 0.

Part 2.

The elastic flexure formula:

σx = - (Mr y / I)

## The Attempt at a Solution

Part 1.

I have applied the equations of equilibrium to the full beam to determine the unknown forces acting on it.

+ ∑Ma = 0: -20 kN(2 m) + Cy(5.5 m) = 0 → 5.5Cy = 40 + 112.5 → Cy = 27,7 kN

↑ + ∑Fy = 0: -Ay – 20 kN + 27,7 kN = 0 → Ay = -20 + 27,7 → Ay = 7.7 kN

→ + ∑Fx = 0: Ax = 0 kN

This might be overdoing it however, since we are only asked about the forces and moments on the vertical cut line through the beam in a distance 1 m to the right from the point C. Here I have attempted

+ ∑Ma = 0: -20 kN(2 m) + Cy(4 m) = 0 → 4Cy = 40 → Cy = 10 kN

↑ + ∑Fy = 0: -Ay – 20 kN + 10 kN = 0 → Ay = -20 + 10 → Ay = -10 kN

→ + ∑Fx = 0: Ax = 0 kN

Is this correct.

Part 2.

We are given:

I = 22,2*106 mm4

H = 162 mm.

So do we simply put the numbers in

σx = - (Mr y / I)

= - 27,7 kN x 0,162 m / 22,2*106 mm4 .

This part confuses me. Help would be appriciated.

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PhanthomJay
Homework Helper
Gold Member
No this is not correct. It is almost always beneficial to find the support reactions first, before you start taking free body diagrams at and around a cut section. To find the support reactions, look at the entire beam and apply your equilibrium equations. It is best to sum moments about 2 different points to find the very reactions. Then apply sum of Fy = 0 as a check. If you sum forces before applying the second moment equation, a math error in the first step will give you errors in the rest.
What should you get for reactions?

No this is not correct. It is almost always beneficial to find the support reactions first, before you start taking free body diagrams at and around a cut section. To find the support reactions, look at the entire beam and apply your equilibrium equations. It is best to sum moments about 2 different points to find the very reactions. Then apply sum of Fy = 0 as a check. If you sum forces before applying the second moment equation, a math error in the first step will give you errors in the rest.
What should you get for reactions?
Hi PhantomJay!

Thanks for your suggestion. I have followed your approach and made the following (I can currently not scan my hand drawings of the beam, but hope it make sense anyway):

+ ∑Ma = 0:

-20(2) + RCy (5) – 37,5(5 + 2.5) = 0 →

RCy (5) = 40 + 37,5(7,5) = 0 →

RCy (5) = 40 + 281,25 = 0 →

RCy = 64,25 kN

∑Fy = 0:

RBy = 0

Ay + Cy – 20 – 37,5 = 0 →

Ay + 64,25 – 20 – 37,5 = 0 →

Ay = 6,75 kN

∑Fx = 0

Is this correct. It is of the whole beam, but I am not certain how to adjust it to the specific question in the given exercise.

Chestermiller
Mentor
Have you drawn a free body diagram of the portion of the beam between the cut and either one of the ends? What shear force and moment would you need to apply at the cut to hold that portion of the beam in equilibrium?

Have you drawn a free body diagram of the portion of the beam between the cut and either one of the ends? What shear force and moment would you need to apply at the cut to hold that portion of the beam in equilibrium?
Hi Chestermiller!

∑Fy = 0:

-Fa + Vx + Dy = 0

-80 + Vx + 64,25 = 0

Vx = 15,75 kN

∑Fx = 0

Nx = 0.

+ ∑Ma = 0:

Mx + Fb(2) – F (5) = 0

Mx + 80(2) – 37,5(5) = 0

Mc = 27,5 kN.

Chestermiller
Mentor
I don't confirm your reaction forces. I get a reaction force of 30.5kN at C and a reaction force of 4.5kN at A. These add up to 35 kN, and balance the moments about points C and A. You seem to be using 37.5 kN at D, while the figure shows 15 kN.

I don't confirm your reaction forces. I get a reaction force of 30.5kN at C and a reaction force of 4.5kN at A. These add up to 35 kN, and balance the moments about points C and A. You seem to be using 37.5 kN at D, while the figure shows 15 kN.
Hi!

Thanks for the reply. I thought I needed to multiply the force 15 kN with the distance 2,5 m.

Chestermiller
Mentor
Hi!

Thanks for the reply. I thought I needed to multiply the force 15 kN with the distance 2,5 m.
What made you think that? Then the units of the 37.5 would be kN-m, and then, in the moment balance, you would be multiplying it again by 7.5 m, to get something with units of kN-m^2, which is not a moment or the same units as the other terms in the equation.

I don't confirm your reaction forces. I get a reaction force of 30.5kN at C and a reaction force of 4.5kN at A. These add up to 35 kN, and balance the moments about points C and A. You seem to be using 37.5 kN at D, while the figure shows 15 kN.
Hi!

∑Fy = 0:

Ay + Cy – Dy = 0 →

-20(2) + Cy(2 + 3) – 15(2 + 3 + 2.5) = 0 →

Cy(5) = 40 + 112,5 = 0 →

Cy = 30,5 kN

∑Fy = 0:

Ay + Cy – By – Dy = 0 →

Ay + 30,5 – 20 – 15 = 0 →

Ay = - 30,5 + 20 + 15 →

Ay = 4,5 kN

∑Fx = 0.

1 m to the right of point C.

∑Mx = 0:

Mx + Cy(1 - 3) – Ay(2 + 3) = 0

Mx + 30,5(1 - 3) – 4,5(5) = 0

Mx = - 61 + 22,5

Mx = -38,5 kN.

Is this the complete (and correct answer) to part 1 of the exercise.

Chestermiller
Mentor
I don't get what you did. For the shear force 1 m to the right of point C, I get a downward shear force on the right-hand frame of the left-hand section of 30.5-15 = 15.5 kN. If M is the counterclockwise moment on the right-hand face of the left-hand section, taking moments on the left-hand section about point C, I get $$15(2.5)+M-15.5(1)=0$$So M = -22.0 kN.m

Or taking moments about the point 1 m to the right of point C, I get $$15(3.5)+M-30.5 (1)=0$$
So, M = -22.0 kN.m.

Or, taking moments about point D, I get $$30.5(2.5)+M-15.5 (3.5) = 0$$
So, M = -22.0 kN.m

PhanthomJay
Homework Helper
Gold Member
I don't get what you did. For the shear force 1 m to the right of point C, I get a downward shear force on the right-hand frame of the left-hand section of 30.5-15 = 15.5 kN. If M is the counterclockwise moment on the right-hand face of the left-hand section, taking moments on the left-hand section about point C, I get $$15(2.5)+M-15.5(1)=0$$So M = -22.0 kN.m

Or taking moments about the point 1 m to the right of point C, I get $$15(3.5)+M-30.5 (1)=0$$
So, M = -22.0 kN.m.

Or, taking moments about point D, I get $$30.5(2.5)+M-15.5 (3.5) = 0$$
So, M = -22.0 kN.m
I don't get what you did. For the shear force 1 m to the right of point C, I get a downward shear force on the right-hand frame of the left-hand section of 30.5-15 = 15.5 kN. If M is the counterclockwise moment on the right-hand face of the left-hand section, taking moments on the left-hand section about point C, I get $$15(2.5)+M-15.5(1)=0$$So M = -22.0 kN.m

Or taking moments about the point 1 m to the right of point C, I get $$15(3.5)+M-30.5 (1)=0$$
So, M = -22.0 kN.m.

Or, taking moments about point D, I get $$30.5(2.5)+M-15.5 (3.5) = 0$$
So, M = -22.0 kN.m
You are not drawing your free body diagrams correctly. As Chestermiller has noted , when you want to find the internal forces and moments at a certain point along the beam, you cut the beam at that point away from the whole beam, then analyze that cut portion separately using the equilibrium equations. There is an unknown shear force and moment at that cut. Solve for them as instructed. You can use either the left cut part of the beam, or the right part of the beam. Not both!

• Chestermiller
Chestermiller
Mentor #### Attachments

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Hi!

Thanks for your reply. That explained a lot, and I think that that I got it now. Then only part 2 remain.

2. Explain how one calculate size and distribution normal stress of the cut line. The beam is a I profile with moment of inerti I= 22,2*106 mm4 and height = 162 mm.

We are given the information:

I = 22,2*106 mm4
h = 162 mm,

and have found:

M = -22.0 kN.m

So the way to answer that should be:

Vx = 0

and

σtop = - (Mr y / I)

= - (-22.0 kN.m) (81 mm) / 22,2 x 106 mm4

= 80,3 N/mm2 .

and

σdown = - (Mr y / I)

= - (-22.0 kN.m) (81 mm) / 22,2 x 106 mm4

= -
80,3 N/mm2 .

PhanthomJay
Homework Helper
Gold Member
Hi!

Thanks for your reply. That explained a lot, and I think that that I got it now. Then only part 2 remain.

2. Explain how one calculate size and distribution normal stress of the cut line. The beam is a I profile with moment of inerti I= 22,2*106 mm4 and height = 162 mm.

We are given the information:

I = 22,2*106 mm4
h = 162 mm,

and have found:

M = -22.0 kN.m

So the way to answer that should be:

Vx = 0

and

σtop = - (Mr y / I)

= - (-22.0 kN.m) (81 mm) / 22,2 x 106 mm4

= 80,3 N/mm2 .

and

σdown = - (Mr y / I)

= - (-22.0 kN.m) (81 mm) / 22,2 x 106 mm4

= -
80,3 N/mm2 .
now what is the direction of the stress at the top of the cross section? And at the bottom? What is the stress in the middle? You are asked for the normal stress distribution in the cross section. Requires some thought.

now what is the direction of the stress at the top of the cross section? And at the bottom? What is the stress in the middle? You are asked for the normal stress distribution in the cross section. Requires some thought.
As far as I can see, this is the solution to the question:

I = 22,2*106 mm4

D = 162 mm.

M = -22.0 kN.m

σ = - (Mr y / I)

y = D/2.

σneutral = 0.

σtop = - (Mr y / I)

= - (-22.0 kN.m) (81 mm) / 22,2 x 106 mm4

= 80,3 N/mm2 .

σdown = - (Mr y / I)

= - (-22.0 kN.m) (81 mm) / 22,2 x 106 mm4

= -80,3 N/mm2 .

PhanthomJay
Homework Helper
Gold Member
Yes ok, but one part of the beam above or below the center of the cross section has tensile stresses pointing outward, while the other part above or below the center of the cross section has compressive stresses pointing inward. Which is which??

Chestermiller
Mentor
Yes ok, but one part of the beam above or below the center of the cross section has tensile stresses pointing outward, while the other part above or below the center of the cross section has compressive stresses pointing inward. Which is which??
Are you saying that he doesn't have the signs correct?

PhanthomJay
Homework Helper
Gold Member
Are you saying that he doesn't have the signs correct?
Plus and minus signs have all sorts of meanings and conventions. They need to be explained.

Chestermiller
Mentor
Plus and minus signs have all sorts of meanings and conventions. They need to be explained.
I realize that there are two sign conventions for stress, but in strength of materials , the overwhelmingly more common one is tensile is positive and compressive is negative.

PhanthomJay
Homework Helper
Gold Member
The op uses the same equation for top and bottom stress but assigns one as plus and the other as minus. Good guess???

Chestermiller
Mentor
The op uses the same equation for top and bottom stress but assigns one as plus and the other as minus. Good guess???
Yikes, I didn't notice that.

So we have:

I = 22,2*106 mm4

h = 162 mm.

M = -22.0 kN.m

σ = - (Mr y / I)

y = D/2. Distance from the centroid.

Stress at the beams neutral axis is (the beam is symmetrical):

σneutral = 0.

We are asked for the normal stress distribution in the cross section, that is, what is the direction of the stress at the top of the cross section; at the bottom; the stress in the middle?

Bending moment is counterclockwise; meaning that the I beam section is subjected to a negative bending moment of -22.0 kN.m. So we know the bottom is in tension, which is positive stress, and the top is in compression, which is negative stress.

So for the bottom (tension), we have

σdown = - (Mr y / I)

= - (-22.0 kN.m) x 81 m) / 22,2 x 106 mm4

= 80,3 N/mm2 .

Is this correct? Should one also calculate σdown, or is the above a sufficient answer to the problem.

PhanthomJay