Bending equation, can someone check my answer?

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The discussion focuses on calculating the second moment of area for an I-section beam and determining the maximum bending moment it can withstand under a specified stress limit. The initial calculations provided by the user, Mike, used the incorrect formula for the moment of inertia, which only applies to rectangular shapes. A correction was suggested, emphasizing the need to compute the moment of inertia for each rectangle that makes up the I-section and apply the parallel axis theorem. Additionally, Mike expressed uncertainty about a subsequent problem involving a steel bar bent into a circular arc, indicating a lack of familiarity with the necessary equations. The conversation highlights the importance of correctly applying structural engineering principles to solve bending problems.
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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:
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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?
 
I though something was wrong.

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

Mike
 
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. I_x = \Sigma I_n + A_nd^2. Please show your work in calculating this expression.
 
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