Finding the width of a bar

1. Oct 4, 2016

foo9008

1. The problem statement, all variables and given/known data
There are 4 axial forces that are applied to 25mm thick structural steel bar with 40mm diameter . If the maximum allowable tensile stress in the bar is 135MPa and the maximum allowable deformation (extension or contraction) of bar is 1.25mm , determine the minimum width , w of the bar , E = 200GPa

2. Relevant equations

3. The attempt at a solution

i have divided the bar into 3 sections , namely AB , BC and also CD. For part AB, the force acting on it 90kN tensile force , for BC , the force acting on it is 50kN compressive force , for CD, the force acting is 220kN tensile force. )

We know that δ(defomation ) = PL / AE , where P - force , L = length of bar , A = area

so i let 1.25x10^-3 = ((90x10^3)(250x10^-3) - (50x10^3)(500x10^-3) + (220x10^3)(750x10^-3) / (25x10^-3 x w x 200x10^9) , so w = 26mm ,

but the ans given is w = 65 mm , which part of my working is wrong ?

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Last edited by a moderator: Oct 4, 2016
2. Oct 4, 2016

PhanthomJay

Deformation analysis looks good, but did you check this wording from the problem statement "... the maximum allowable tensile stress in the bar is 135MPa..."?

Last edited by a moderator: Oct 4, 2016
3. Oct 4, 2016

foo9008

So how should I proceed?

4. Oct 4, 2016

foo9008

ok , when using force = 220kPa , i found that w = 65.2mm, but when i use P = 50kN , w = 14.8mm , when using P = 90kPa , my t = 24mm , why is it so ?

if so , then it's maximum t , right ?

Last edited: Oct 4, 2016
5. Oct 4, 2016

PhanthomJay

yes, since the bar is uniform in dimensions, the higher number controls for stress.

6. Oct 4, 2016

foo9008

Just to double check, the 65.2mm is the maximum t, not minimum t, am I right?

7. Oct 4, 2016

PhanthomJay

I may not have responded clearly. The 65 mm thickness is the minimum t required such that no point in the bar is stressed beyond the max allowed stress of 135 MPa. The stress in the left and mid sections will be less, as the right section with the higher load, and hence higher stress, controls the design.