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Bending equation, can someone check my answer?

  • Thread starter Mike.
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  • #1
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Homework Statement



The cross-section of an idealised I section beam has overall dimensions 12mm x 240mm deep. If the web and Flange are both 25mm Thick, determine the second moment of area of the section.
If the maximum bending stress is limited to 100MN/m2, determine the maximum bending moment the beam can withstand


Homework Equations



∂ = Stress Applied
Y = Distance form Neutral Axis
M = Applied moment
I = Second moment of area
E = Youngs modulus
R = Radius (measured to neutral axis)

(∂/y) = (M/I) = (E/R)

I = b . (d^3)
----------
12

Y = depth/2


The Attempt at a Solution



I = b . (d^3)
----------
12

b= 125-25 = 95
d= 240-50 = 190

I = [0.095 . (0.19^3)] / 12 = 5.4x10^-5

(∂/y) = (M/I)

100x10^6 = M
____________ _____________
{(.24/2)=0.12} {5.43x10^-5}

100x10^6
---------- x 5.43x10^-5 = M
0.12

M= 45000nm
45KNM

Is this correct.

Mike

Also i've just tryed the next (and last) one

Which i have no clue about.

A steel bar of rectangular cross-section is bent into a circular arc of 15m. The with of the cross-section is twice its depth.
If the maximum stress due to bending is 60MN/m2, determine the dimensions of the section and the magnitude of the applied moment.
Take young's modulus as 200GN/m2

Dont even no where to start on that one

Mike
 
Last edited:

Answers and Replies

  • #2
PhanthomJay
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502
The formula I=bd^3/12 applies to a rectangular shape only. You have to compute the I of each of the 3 rectangles that comprise the section, and use the parallel axis theorem to get the I of the beam. Are you familiar with it?
 
  • #3
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I though something was wrong.

I think i've seen the equation but carnt find it in my notes?
Do you have it?

Mike
 
  • #4
PhanthomJay
Science Advisor
Homework Helper
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502
Basically, the moment of inertia of the section about its neutral axis is equal to sum of the moments of inertia of each rectangle, abouts its centroid, plus the sums of the (area of each rectangle times the square of the distance from its centroid to the neutral axis). From symmetry, the neutral axis is thru the middle of the beam. [tex] I_x = \Sigma I_n + A_nd^2[/tex]. Please show your work in calculating this expression.
 

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