Traction/ Compression stress and strain exercise

[Wolfrider]

Hello everyone, could you help me with this exercise? I am stuck, and I can't find anything on the internet that solves this. Your help is very much appreciated :)

1. Homework Statement
Circular steel bar, clamped at extremities.
Two parts: A(ab)=800mm² A(bc)=400mm²
L(ab)= 400mm L(bc)=300mm
Axial Forces= 100kN
Young modulus E= 210GPa

Need to find: normal stress in each bar, and the horizontal displacement at point B.

Homework Equations

Sigma (Stress) = Axial Force / Area
Youngs modulus = Simga (stress) / Epsilon (strain)
Epsilon (strain)= Delta L / Delta L0

The Attempt at a Solution

I do know how to use the formulas to calculate the normal stress, but only when I have to Axial Force of AB and BC. I don't know how to calulate it, but I know that in this instance it is F AB= 60KN and F BC= -40KN

With this I calculate sigma ab = 60kN / 800mm² = 75MPa and sigma bc = -40kN / 400mm² = -100 MPa
To get Delta AB I use the Young's modulus: (75MPa*0.4m)/210Gpa= 1,42857*10^(-4)m, for Delta BC I switch out the numbers and get the same amount in the opposite direction.

Could you help me understand why in this case the force of AB is 60kN and for BC 40kN?

Regards,
Wolfrider

 

[SteamKing]

Hello everyone, could you help me with this exercise? I am stuck, and I can't find anything on the internet that solves this. Your help is very much appreciated :)

1. Homework Statement
Circular steel bar, clamped at extremities.
Two parts: A(ab)=800mm² A(bc)=400mm²
L(ab)= 400mm L(bc)=300mm
Axial Forces= 100kN
Young modulus E= 210GPa

Need to find: normal stress in each bar, and the horizontal displacement at point B.

Homework Equations

Sigma (Stress) = Axial Force / Area
Youngs modulus = Simga (stress) / Epsilon (strain)
Epsilon (strain)= Delta L / Delta L0

The Attempt at a Solution

I do know how to use the formulas to calculate the normal stress, but only when I have to Axial Force of AB and BC. I don't know how to calulate it, but I know that in this instance it is F AB= 60KN and F BC= -40KN

With this I calculate sigma ab = 60kN / 800mm² = 75MPa and sigma bc = -40kN / 400mm² = -100 MPa
To get Delta AB I use the Young's modulus: (75MPa*0.4m)/210Gpa= 1,42857*10^(-4)m, for Delta BC I switch out the numbers and get the same amount in the opposite direction.

Could you help me understand why in this case the force of AB is 60kN and for BC 40kN?

Regards,
Wolfrider

Is there a diagram or picture which shows this bar?

The bar itself is statically indeterminate, since it is fixed at both ends. In addition to the normal equations of static equilibrium, you must write additional equations describing the compatibility of displacements internal to the bar.

See http://ocw.nthu.edu.tw/ocw/upload/8/251/Chapter_2.pdf, particularly sections 2.3 and 2.4.

Also, see http://www.assakkaf.com/Courses/ENES220/Lectures/Lecture3 .pdf