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Calculating the maximum stress of the in the beam? Am i correct?

by charger9198
Tags: beam, correct, maximum, stress
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charger9198
#1
Dec11-11, 07:53 AM
P: 60
Illustration attached --> Attachment 42037

I have a question which requires me to calculate the maximum stress in a simple supported beam.

The beam is 3 m long and rectangular and has two forces acting down vertically totalling 20 kN.

The breadth of the cross sectional plane is 100mm and depth is 200mm

I am a bit rusty on this but below shows my working;

- I Calculated the second moment of area about the neutral axis, = (b*d^3)/12
where b = 100mm and d = 20mm
(100*200^3)/12 mm4

thus (100*200^3)/12 *10^-12 m4

Moment area based on neutral axis = 6.666*10^-5 m4

y= 100 mm = 100*10^-3 m

I used the complete bend theory equation to transpose to ;

σ=M*y/I

σ= (30*100*10^-3)/(6.666*10^-5

= 2.25*10^6 Nm^-2

So maximum stress = 2.25 Mn

both edges are equal distances from the neutral axis

Can someone tell me if im correct? or on the right lines?
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pongo38
#2
Dec11-11, 04:49 PM
P: 699
You haven't explained why M = 30
You have minor errors with units.
The end result is probably not right because it is an extremely high stress. What are the units of stress here?
charger9198
#3
Dec11-11, 07:19 PM
P: 60
Sorry m=15, typo. Can you give me the formula for working out maximum stress?

pongo38
#4
Dec12-11, 04:51 PM
P: 699
Calculating the maximum stress of the in the beam? Am i correct?

Why don't you answer my questions, which are designed to help you?
charger9198
#5
Dec13-11, 12:20 AM
P: 60
apologies, milli-newtons are the units of stress,
I got M from the total applied force on the beam which is 15 kN (although i originally put 20 in my first post)
Then i followed which i thought might be somewhat along the right lines but as you say, looking at it the stress seems excessive
charger9198
#6
Dec18-11, 10:03 AM
P: 60
http://www.physicsforums.com/attachm...7&d=1324149378

here is the diagram i was given to calculate the maximum stress
charger9198
#7
Dec19-11, 04:34 AM
P: 60
Here is my new working out;

From left to right, let's label the points where point loads presented as A, B, C, and D.
Use equlibrium equation, we can find the reactions R1 and R2.
ΣMA=0: R2(3) -5(1)-10(2) = 0. R2=25/3 kN
ΣFy = 0: R1 + R2 -5-10 =0 R1=10/3 kN.
The maximum bending moment occurs at C, which is
Mmax = R2(1) = 25/3 kN.m
Maximum shear force occurs in CD, which is
Vmax = 25/3 kN
Let the cross section width be b = 100 mm = 0.1m and the height be h = 200 mm = 0.2m. Therefore, the maximum bending stress occurs at point C, which is
σmax = Mc/I = M(h/2)/[bh3/12 ] = 6M/(bh2) = 6(25/3*1000)/[0.1*0.23/12] = 750 x 106 Pa = 750 MPa
FOr the rectangular cross section, maximum shear stress occurs at the center of the cross section, and is 3/2 of the average shear stress, i.e.,
τmax = (3/2) (Vmax/A ) = 1.5 (25/3*1000)/(0.1*0.2) = 625000 Pa = 625 kPa.
nvn
#8
Dec19-11, 10:35 AM
Sci Advisor
HW Helper
P: 2,124
charger9198: Nice work. However, always leave a space between a numeric value and its following unit symbol. E.g., 0.2 m, not 0.2m. Also, for exponentiation, you must show a caret (^) symbol. E.g., h^3, not h3.

Your shear stress tau_max is correct, but your bending stress sigma_max is currently incorrect. Although you have the correct formula for sigma (if you insert caret symbols), it appears you did not check your calculations, and your current answer for sigma is wrong. Try again.
charger9198
#9
Dec19-11, 02:55 PM
P: 60
Nvn thanks for all your help. I must be going wrong somewhere or something I'm missing right in front of me, everything I do seems to lead me to my original answer,

σmax = Mc/I =
M(h/2)/[b*(h^3)/12 ] = 6M/(b*(h^2)) = 6((25/3)*1000)/
[0.1*(0.2^3)/12] =
750 x 10^6 Pa =
750 MPa
nvn
#10
Dec19-11, 03:25 PM
Sci Advisor
HW Helper
P: 2,124
charger9198: You are simply not reading what you typed, and you are currently not computing the formula you listed. You will need to carefully read what you typed, and enter the numeric values correctly.
charger9198
#11
Dec19-11, 03:55 PM
P: 60
Ok after chopping and changing I'm getting 9 kPa, 12.5 kPa or 12.5 MPa, think I'm gonna look again tomorrow when I've slept on it, all calculated out today I think. See if I can come somewhere near then
charger9198
#12
Dec22-11, 06:09 AM
P: 60
Got it! Realise were I went wrong...
6*M/(b*h^2)

(6*8333.333333)/(.1*.2^2)
=120 x 10^6
=120 MPa
nvn
#13
Dec22-11, 09:51 AM
Sci Advisor
HW Helper
P: 2,124
charger9198: Your formula in post 12 is correct, and your input values are now correct. But it appears you did not check your calculations yet, and your answer is wrong. Try it again on your calculator.

By the way, numbers less than 1 must always have a zero before the decimal point. E.g., 0.1, not .1.
charger9198
#14
Dec22-11, 05:53 PM
P: 60
12.5 kPa.. My input was incorrect on calculator, think I was being a bit slack with my brackets... Got there...eventually (I think)
nvn
#15
Dec22-11, 09:08 PM
Sci Advisor
HW Helper
P: 2,124
charger9198: You are getting closer. Your units are currently incorrect. Try again.
charger9198
#16
Dec23-11, 08:44 AM
P: 60
Ahhh, I've been missin out the (x1000),
I've been calculating
(6*(25/3)/(0.1*(0.2^2)) giving me 12.5 kPa

When I should have used
(6*(25/3)*1000)/(0.1*(0.2^2)) now giving 12.5 MPa which I believe now to be correct.

Nvn I have a question about finding r1 and r2 from the original drawing. Using the equilibrium equation I got 25/3 for r2 and 10/3 for r1. However when you work this out the sum does not equal the total force acting down (15 kN)

I believe r1 to be 20/3, this now equals the 15 kN
nvn
#17
Dec23-11, 10:10 AM
Sci Advisor
HW Helper
P: 2,124
charger9198: Nice work. Your answer for sigma_max is now correct.

And good catch on R1. Your answer for R1 was wrong in post 7, but luckily you did not use it thereafter. You now have the correct answer for R1, which is 20/3.
charger9198
#18
Dec23-11, 12:59 PM
P: 60
Nvn thanks for all your help once again. Life saver


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