# Yes, that is correct. Good job!

• Connorm1
In summary, the conversation discusses the calculation of the maximum bending moment and stress in a simply supported beam with point loads. It also asks for a sketch of the stress distribution at the point of maximum stress and for the dimensions of a cross section that will minimize the maximum stress value. The calculations for the maximum bending moment and stress are correct, and to sketch the stress distribution, one needs to draw a linear line connecting 0 to +12.5 MPa and -12.5 MPa at the top and bottom fibers respectively. The proposed solution for part (d) is to increase the depth of the beam by 20%, resulting in a 47.4% reduction in the maximum stress value.
Connorm1

## Homework Statement

A beam of rectangular cross section 200 mm deep and 100 mm wide. If the beam is 3m long, simply supported at either end. And carries point loads R1 to far left, R2 to far right, 1m to left 5kN load, 2m to left 10kN load.
(a) Calculate the maximum bending moment
(b) Calculate the maximum stress in the beam
(c) At the point of maximum stress sketch a graph of the stress distribution through the thickness of the beam, indicating which are tensile and compressive stresses.
(d) Determine the dimensions of the cross section which will minimise the maximum stress value if: • the cross sectional area of the beam can be increased by 20% • the beam section is to remain a solid rectangle • neither the breadth or depth of the beam section can be reduced below their original dimensions. Show the dimensions of the proposed beam cross section with the aid of a sketch.
(e) Determine the percentage reduction of the maximum stress value when the new cross section is used.

M/I=σ/y
Ixx=bd^3/12

## The Attempt at a Solution

a) Firstly i started by finding the bending moment. so taking moments at Point A (R1) i have 3R2=(5x1)+(2x10). R2=25/3kN. As upward forces must equal downwards forces. 10+5=R1+25/3 so R1=20/3. I then moved on to find the max bending moment. so at M0 (0meters from left) =0kNm. At M1 at 1m from left = (20/3*1) = 20/3kNm . At M2 at 2m from left = (20/3*2)-(5*1)= (25/3) = 8.333 kNm. At M3 at 3m from left = (20/3*3)-(5*2)-(10*1)=0kNm. Therefore maximum bending moment is 8.333kNm.
b) using the formula i rearrange to make maximum stress the subject. So σ=My/I. Where M=8.333kNm (8.333*10^3) y= d/2 = 0.1m (100mm) and I we need to find using bd^3/12. so I(NA)=0.1*0.2^3/12 = 6.667*10^-5 m^4. Plugging our values for M, y & I we get σ=(8.333*10^3*0.1) / (6.667*10^-5) = 12.5 *10^6 or 12.5 MPa (or Nm^-2 unsure on SI).
c) I have to admit i don't really know what diagram this is referring too and have no idea of what to draw nor figures to use.
d) So my first assumption was that to minimise the stress value in a beam, we would need to increase the depth (again i don't know 100% why but i could have trial and error both width and depth). As we can only increase it by 20% for the cross sectional area. I'd make this 240mm*100mm. The maximum stress now would be using the same formula. except now I=0.1*0.24^3/12 so I(NA)= 1.152*10^-4 and new y=d/2 0.24/2=0.12m (120mm). So σmax = (8.333*10^3*0.12) / (1.152*10^-4) = 6.58 * 10^6 or 6.58MPa (Nm^-2).
e) I'm assuming to show the percentage reduction of maximum stress across the new cross section used i just take my previously calculated max stresses and do 6.58 *10^6 /12.5 *10^6 =52.6%-100% which is a 47.4% reduction.

Am i heading in the right directions. Bare in mind it's hard to type it out and work it on here so if I've made a mistake on figures i apologise!

Last edited:
Hi Connorm, Welcome to PF.

a) You got it correct.
b) I agree as well, just note that the maximum stress will be 12.5 MPa for both tension and compression ( at each extremity of fiber)

c) To solve part (b) the question assumed that the beam is behaving linearly, and with no present cracks. So, both the stress and strain profiles are linear. Simply, draw the rectangular cross section, locate the centroid (which is at y = 100 mm). Here both strain and stress will be 0. Then draw a linear line connecting 0 to +12.5 MPa (top fiber of the beam) and the same line, from 0 to -12.5 MPa (at the bottom fiber of the beam). Do the same for strain, and to find strain you would need the concrete modulus of elasticity.

d) When you increase the depth, you increase the "bending" capacity of the beam (Ixx). I agree.
e) I agree.

@CivilSigma that's brilliant this question baffled me for the best part. D I’m happy with but E just to confirm my thought of a 47.4 % reduction is correct? That seems quite a lot for only a 20mm increas in depth, but if you agree I’m happy! Lastly, sorry I’ve been busy but if I tag you and upload a picture for c) when I can could you just see if if what I upload is correct?

Connorm1 said:
@CivilSigma that's brilliant this question baffled me for the best part. D I’m happy with but E just to confirm my thought of a 47.4 % reduction is correct? That seems quite a lot for only a 20mm increas in depth, but if you agree I’m happy! Lastly, sorry I’ve been busy but if I tag you and upload a picture for c) when I can could you just see if if what I upload is correct?

no problem, let me know when you have sketched something. It should look like something in this lower left quadrant of this picture .

CivilSigma said:
no problem, let me know when you have sketched something. It should look like something in this lower left quadrant of this picture .
Okay so below is my picture. After we’ve gone through c) can I just get a double confirm everything is good :) thanks again!

#### Attachments

• 1CC6A1D0-A306-4E98-93AE-159D4763222E.jpeg
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You have the correct shape.

Tensile force should be -12.5 MPa ( just make sure compressive and tensile forces are of different signs, follow consistent sign convention)
At the N.A , stress is 0 ( you said a stress of 0 is at the same location of the maximum tensile force). So when your line crosses the N.A, at d/2 = 100 mm , stress is 0.

So, you should have the values : 12.5 , 0 , -12.5 in this order from top to bottom ( for the triangle you drew)

CivilSigma said:
You have the correct shape.

Tensile force should be -12.5 MPa ( just make sure compressive and tensile forces are of different signs, follow consistent sign convention)
At the N.A , stress is 0 ( you said a stress of 0 is at the same location of the maximum tensile force). So when your line crosses the N.A, at d/2 = 100 mm , stress is 0.

So, you should have the values : 12.5 , 0 , -12.5 in this order from top to bottom ( for the triangle you drew)

So more like this picture? I’ve placed the 0 next to the neutral axis (just to show it’s at that point), and added a negative sign to the tensile stress. Does this now satisfy the question 2c)? If so is everything overall looking like a correct solution to the whole question?

Ignore the information below the diagram that was some insurance quote figures! Haha

#### Attachments

• image.jpg
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Connorm1 said:
So more like this picture? I’ve placed the 0 next to the neutral axis (just to show it’s at that point), and added a negative sign to the tensile stress. Does this now satisfy the question 2c)? If so is everything overall looking like a correct solution to the whole question?

Ignore the information below the diagram that was some insurance quote figures! Haha

Yea, it looks good !

CivilSigma said:
Yea, it looks good !
phew! that question has killed me! it was a bit brutal. I have another one which my instructor has said will probably test myself (to do with column failure types and length of which buckling will happen etc. in a hollow cylindrical column)... i'll attempt it first and if not i guess i'll be posting it up with my attempt haha! Thanks so much for your help, hopefully all is correct! But if you're saying what I've done is correct and i think it's correct i can't go far wrong!

@CivilSigma Hey buddy, i realized i messed up on figures for e) i figured out the new maximum stress was 8.68MPa (somehow my calculated messed up) but (8.333*10^3*0.12)/(1.152*10^-4) wasn't 6.58*10^6 (6.58MPa) but 8.68MPa, fancy checking my math xD this gives me a new percentage reduction of 30.56% from (8.68*10^6) / (12.5*10^6) = 0.6944*100 for a percentage = 69.44%. 100%-69.44% to find the reduction gives me 30.56%

new stress of beam in e) will be 8.68 MPa. Then

$$\%Diff = \frac{12.5 - 8.68}{12.5} = 0.305 = 30.5 \%$$

CivilSigma said:
new stress of beam in e) will be 8.68 MPa. Then

$$\%Diff = \frac{12.5 - 8.68}{12.5} = 0.305 = 30.5 \%$$
Good! So in agreeance now that the new stress should be 8.68MPa and the percentage reduction is correct?

## 1. What is a simply supported loaded beam?

A simply supported loaded beam is a type of structural element used in construction and engineering. It consists of a horizontal beam supported at both ends by fixed supports, such as walls or columns, and is subject to a vertical load or weight.

## 2. How does a simply supported loaded beam differ from other types of beams?

The main difference between a simply supported loaded beam and other types of beams, such as cantilever or continuous beams, is that it is only supported at two points. This means that the load is evenly distributed between the two supports, rather than being concentrated at one end.

## 3. What are the advantages of using a simply supported loaded beam?

One of the main advantages of a simply supported loaded beam is its simplicity and ease of construction. It also allows for a larger span compared to other types of beams, making it a cost-effective option for certain types of structures.

## 4. How do you calculate the maximum load a simply supported loaded beam can withstand?

The maximum load a simply supported loaded beam can withstand is determined by its material properties, such as its strength and stiffness, as well as its dimensions and the distance between the supports. This can be calculated using various equations and formulas, such as Euler's buckling equation or the bending moment equation.

## 5. What are some common applications of a simply supported loaded beam?

Simply supported loaded beams are commonly used in various construction and engineering projects, such as bridges, roofs, and floors. They are also used in smaller structures, such as shelves and furniture. Additionally, they are often incorporated into larger structural systems, such as trusses and frames, to provide support and distribute loads.

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