mt4101=06-1

Prévia do material em texto

1
Activity Questions
1. Shoes
• Look at your shoes.
• While standing still, what forces are acting on the 
soles of your shoes?
• What sort of deformation would you expect to 
occur?
• When you are walking what forces are acting on the 
soles of your shoes?
• What sort of deformation would you expect to 
occur?
• Draw a diagram showing the forces and resulting 
deformations.
• When standing you exert a compressive force 
on the soles of the shoes. Most materials are 
very strong to compressive forces. When 
walking the shoes are also subject to shearing 
forces, you apply a force at the upper surface 
of the sole, and the ground applies a force at 
the lower surface. The shoes are also subject to 
bending, and some shoes develop cracks in the 
soles. It is these shearing and bending forces 
that wear the shoes out.
mg
N
standing walking
2. Rubber bands
• Which rubber band has the largest spring 
constant?
• How could you estimate the elastic modulus of 
the rubber bands?
• Cut a rubber band to form a strip, and hang a 
weight off it. What would happen if you cut 
the strip in half and hung a weight from it?
• What if you joined two strips together in 
parallel?
• The rubber band that stretches the least for a given 
weight (applied force) has the greatest spring 
constant. If you had rubber bands of the same cross 
sectional area then the one that stretches the least also 
has the greatest elastic modulus. The area is the width 
times the thickness, so if the thickness is similar the 
area can be estimated from the width of the band.
• If you cut a strip of rubber in half it would stretch less 
for a given weight, so its spring constant will have 
decreased, but it will stretch by the same proportion. 
The modulus of elasticity depends on the material, 
and will not have changed.
• If you joined two rubber strips or bands together in 
parallel they would also stretch less, but again the 
modulus of elasticity has not changed. Effectively 
you will have doubled the spring constant by 
doubling the cross sectional area.
Example
• A metal wire is 2.5 mm diameter and 2 m 
long. A force of 12N is applied to it and it 
stretches 0.3 mm. assume the material is 
elastic. Determine the following:
i. The stress in the wire σ.
ii. The strain in the wire ε.
A= π d2/4
A = π (2.5)2/4 = 4.909 mm2
σ = F/A = 12/4.909 = 2.49 N/mm2
ε = ΔL/L = 0.3m/2000m = 0.00015
2
• A steel bar is 10 mm diameter and 
2 m long. It is stretched with a force 
of 20 KN and extends by 0.2 mm. 
calculate the stress and strain.
• A rod is 0.5 m long and 5 mm 
diameter. It is stretched 0.06 mm 
by a force of 3 KN. Calculate the 
stress and strain
• A steel column is 3 m and 0.4 m 
diameter. It carries a load of 50 MN. 
Given that the modulus of elasticity 
is 200GPa, calculate the compressive 
stress and strain and determine how 
much the column is compressed 
A= π d2/4
= π (0.4)2/4 = 0.126 m2
σ = F/A = 50x106/0.126 
= 397.9x106 Pa
E= σ /ε
So ε = σ / E 
= (397.9x106)/(200x109)
= 0.001989
ε = ΔL/L 
So ΔL = ε *L 
= 0.001989* 3000 mm 
= 5.97 mm