Simon33355
Badges: 9
Rep:
?
#1
Report Thread starter 1 year ago
#1
A steel bar of length 40mm and cross-sectional area 4.5x10-4 m^2 is places in a vice and compressed by 0.2mm when the vice is tighted. Calculate the compressive force exerted on the bar.

I know the formula is CS = F / A

I'm not given the Force. So how do I work it out?
0
reply
GgbroTG
Badges: 10
Rep:
?
#2
Report 1 year ago
#2
You should use the formula for Young's Modulus:



E = {\sigma}/{\epsilon}

where \sigma is the stress and \epsilon is the strain. In this situation, the stress might be referred to as compressive stress since stress usually refers to stretching, but since we want the compressive force anyway we don't have to worry about the sign (i.e.  \Delta L would normally be negative in this situation, but we want the force in the same direction so it doesn't matter).

This gives:



E = (F/A)/(\Delta L/L)

Re-arranging, you can get an expression for F:



E = (F\Delta L) / (A L)

\\\

F = (EAL)/\Delta L

Then, substituting values (Young's Modulus of steel is around 200 GPa):



F = ((200 \times \ 10^9)\cdot(4.5 \times 10^{-4})\cdot(4 \times10^{-2})) \ / \ (2 \times 10^{-4})

Which gives a final force of F = 1.8 \times 10^{10} \ {\rm N}, or F = 18 \ {\rm GN}.
1
reply
Simon33355
Badges: 9
Rep:
?
#3
Report Thread starter 1 year ago
#3
(Original post by GgbroTG)
You should use the formula for Young's Modulus:



E = {\sigma}/{\epsilon}

where \sigma is the stress and \epsilon is the strain. In this situation, the stress might be referred to as compressive stress since stress usually refers to stretching, but since we want the compressive force anyway we don't have to worry about the sign (i.e.  \Delta L would normally be negative in this situation, but we want the force in the same direction so it doesn't matter).

This gives:



E = (F/A)/(\Delta L/L)

Re-arranging, you can get an expression for F:



E = (F\Delta L) / (A L)

\\\

F = (EAL)/\Delta L

Then, substituting values (Young's Modulus of steel is around 200 GPa):



F = ((200 \times \ 10^9)\cdot(4.5 \times 10^{-4})\cdot(4 \times10^{-2})) \ / \ (2 \times 10^{-4})

Which gives a final force of F = 1.8 \times 10^{10} \ {\rm N}, or F = 18 \ {\rm GN}.
Thank you so much for you detailed answer, I've worked it out now.
0
reply
X

Quick Reply

Attached files
Write a reply...
Reply
new posts
Back
to top
Latest
My Feed

See more of what you like on
The Student Room

You can personalise what you see on TSR. Tell us a little about yourself to get started.

Personalise

Should the school day be extended to help students catch up?

Yes (24)
30%
No (56)
70%

Watched Threads

View All