Mechanical Properties of Solids — Practice

Question 1

A wire of cross-sectional area 2.0 mm^2 carries a load of 40 N. Find the tensile stress.

Accepted Answer

Convert the area first: 2.0 mm^2 = 2.0 x 10^-6 m^2. Stress = F/A = 40/(2.0 x 10^-6) = 2.0 x 10^7 N/m^2 = 20 MPa. Note the conversion carries a factor of 10^-6, not 10^-3, because area involves the square of a length.

Question 2

Why is Young's modulus not defined for a liquid?

Accepted Answer

Young's modulus is the ratio of tensile stress to longitudinal strain, and longitudinal strain is a change in length divided by an original length. A liquid has no definite length or shape of its own — it takes the shape of its container — so there is no length to change and no such strain to define. Only the bulk modulus, which involves volume, applies to liquids.

Question 3

On a stress-strain curve, which feature gives the stiffness of the material and which gives its strength?

Accepted Answer

Stiffness is the slope of the initial straight portion, since Young's modulus is stress divided by strain in that region. Strength is the height of the peak of the curve, the maximum stress the material withstands before fracture. They are independent properties — glass is stiff but breaks easily, so it is stiff and weak.

Question 4

A steel wire 2.0 m long and of cross-sectional area 1.0 mm^2 is stretched by a 100 N load. Find its elongation (Y = 2.0 x 10^11 Pa).

Accepted Answer

A = 1.0 mm^2 = 1.0 x 10^-6 m^2. Using dL = FL/(AY) = (100 x 2.0)/(1.0 x 10^-6 x 2.0 x 10^11) = 200/(2.0 x 10^5) = 1.0 x 10^-3 m, that is 1.0 mm. A 100 N load moving a steel wire by only a millimetre is what a Young's modulus of 200 GPa means in practice.

Question 5

Find the fractional change in volume of a copper block under a hydraulic pressure of 2.0 x 10^7 Pa (B = 1.4 x 10^11 Pa).

Accepted Answer

dV/V = p/B = (2.0 x 10^7)/(1.4 x 10^11) = 1.43 x 10^-4, a decrease. The answer is dimensionless because it is a volume divided by a volume, and the block's actual size never enters since only a fraction was asked for.

Question 6

A wire of volume 1.0 x 10^-6 m^3 is under a strain of 2.0 x 10^-3. If Y = 2.0 x 10^11 Pa, find the elastic energy stored.

Accepted Answer

Energy per unit volume u = (1/2) Y (strain)^2 = 0.5 x 2.0 x 10^11 x (2.0 x 10^-3)^2 = 0.5 x 2.0 x 10^11 x 4.0 x 10^-6 = 4.0 x 10^5 J/m^3. Total energy = u x volume = 4.0 x 10^5 x 1.0 x 10^-6 = 0.4 J.

Question 7

A hollow steel column has inner radius 10 cm and outer radius 20 cm and carries 3.0 x 10^5 N. Find the compressive stress.

Accepted Answer

The load-bearing area is the annulus: A = pi(R^2 - r^2) = pi(0.20^2 - 0.10^2) = pi(0.04 - 0.01) = pi(0.03) = 0.0942 m^2. Stress = F/A = (3.0 x 10^5)/0.0942 = 3.18 x 10^6 N/m^2. Using pi R^2 instead would understate the stress by a quarter.

Question 8

Distinguish ductile from brittle materials, giving one example of each and stating what the difference looks like on a stress-strain curve.

Accepted Answer

A ductile material has a large plastic region between the elastic limit and fracture, so it deforms substantially and visibly before breaking — copper and mild steel behave this way. A brittle material fractures almost immediately after its elastic limit, with very little plastic region, so it fails suddenly and without warning — glass and cast iron are examples. On the curve, ductility shows as a long flat stretch after yielding, brittleness as a curve that ends soon after the straight part.

Question 9

Two wires of the same material and the same length have radii r and 2r. They are joined end to end and a force F is applied to the free ends. Compare the stress, strain and elongation in the two wires.

Accepted Answer

Joined end to end means the same force F passes through both. Stress = F/A and area goes as the square of the radius, so the thinner wire has four times the area disadvantage: stress ratio thin:thick = 4:1. Strain = stress/Y and the material is the same, so Y is the same, giving a strain ratio of 4:1 as well. Elongation dL = strain x L, and the lengths are equal, so the elongation ratio is also 4:1. The thin wire therefore does four fifths of the total stretching, and it is the one that will break first.

Question 10

A steel wire has a breaking stress of 7.8 x 10^8 N/m^2 and density 7800 kg/m^3. Find the greatest length that can hang vertically without breaking under its own weight (g = 9.8 m/s^2), and explain why thickening the wire does not help.

Accepted Answer

The heaviest-loaded point is the top, which carries the whole weight below it. For a wire of area A and length L, that weight is (density x A x L)g, so the stress at the top is (rho A L g)/A = rho g L — the area cancels. Setting this equal to the breaking stress: L = stress/(rho g) = (7.8 x 10^8)/(7800 x 9.8) = 1.02 x 10^4 m, about 10 km. Thickening the wire does not help because doubling the area also doubles the weight being carried, so the stress at the top is completely unchanged.

Practice — Mechanical Properties of Solids

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