The proportional region within the elastic limit of the stress-strain curve (region OA in Fig. 9.3) is of great importance for structural and manufacturing engineering designs. The ratio of stress and strain, called modulus of elasticity, is found to be a characteristic of the material.
Young’s Modulus
Experimental observation show that for a given material, the magnitude of the strain produced is same whether the stress is tensile or compressive. The ratio of tensile (or compressive) stress (σ) to the longitudinal strain (ε) is defined as Young’s modulus and is denoted by the symbol Y.
$Y=\frac{\sigma}{\epsilon}$
From Eqs. (9.1) and (9.2), we have
Y = (F/A)/(∆L/L) = (F × L)/(A × ∆L)
Since strain is a dimensionless quantity, the unit of Young’s modulus is the same as that of stress i.e., N.m$^{–2}$ or Pascal (Pa). Table 1 gives the values of Young’s moduli and yield strengths of some material.
From the data given in Table 1, it is noticed that for metals Young’s moduli are large. Therefore, these materials require a large force to produce small change in length. To increase the length of a thin steel wire of 0.1 cm2 cross-sectional area by 0.1%, a force of 2000 N is required. The force required to produce the same strain in aluminium, brass and copper wires having the same cross-sectional area are 690 N, 900 N and 1100 N respectively. It means that steel is more elastic than copper, brass and aluminium. It is for this reason that steel is preferred in heavy-duty machines and in structural designs. Wood, bone, concrete and glass have rather small Young’s moduli.
Example 1
A structural steel rod has a radius of 10 mm and a length of 1.0 m. A 100 kN force stretches it along its length. Calculate (a) stress, (b) elongation, and (c) strain on the rod. Young’s modulus, of structural steel is 2.0 × $10^{11}$ N$m^{-2}$.
Answer
We assume that the rod is held by a clamp at one end, and the force F is applied at the other end, parallel to the length of the rod. Then the stress on the rod is given by
Stress = $\frac{F}{A}=\frac{F}{\pi r^2}$
Stress = $\frac{100 \times 10^3}{3.14 \times (10^{-2})^2}$
Stress = $3.18 \times 10^8 \ N/m^2$
The elongation,
$\Delta L = \frac{(F/A)L}{Y}$
$= \frac{(3.18 \times 10^8)(1.0)}{2.0 \times 10^{11}}$
$= 1.59 \times 10^{-3}$ m
$\Delta L = 1.59$ mm
The strain is given by
Strain = ∆L/L
= (1.59 × 10$^{–3}$ m)/(1m)
= 1.59 × 10$^{–3}$
= 0.16 %
Example 1
A copper wire of length 2.2 m and a steel wire of length 1.6 m, both of diameter 3.0 mm, are connected end to end. When stretched by a load, the net elongation is found to be 0.70 mm. Obtain the load applied.
Answer
The copper and steel wires are under a tensile stress because they have the same tension (equal to the load W) and the same area of cross-section A. From Eq. (9.7) we have stress = strain × Young’s modulus. Therefore
$\frac{W}{A}=Y_c\times \frac{\Delta L_c}{L_c}=Y_s\times \frac{\Delta L_s}{L_s}$
where the subscripts c and s refer to copper and stainless steel respectively. Or,
$\frac{\Delta L_c}{L_s}=\frac{Y_s}{Y_c}\times \frac{L_c}{L_s}$
Given Lc = 2.2 m, Ls = 1.6 m,
From Table 9.1 $Y_c = 1.1 \times 10^{11} \ N.m^{–2}$, and $Y_s = 2.0 \times 10^{11} \ N.m^{–2}$.
$\frac{\Delta L_c}{L_s}=\frac{2.0\times 10^{11}}{1.1 \times 10^{11}}\times \frac{2.2}{1.6}=2.5$
The total elongation is given to be
$\Delta L_c+\Delta L_s=7.0 \times 10^{-4}$ m
Solving the above equations,
$\Delta L_c = 5.0 \times 10^{-4} \ m^{–4}$, and $\Delta L_s = 2.0 \times 10^{-4} \ m^{–4}$
Therefore
$W=\frac{A\times Y_c\times \Delta L_c}{L_c}$
$=\pi(1.5 \times 10^{-3})^2\times [(5.0 \times 10^{-4})\times 1.1 \times 10^{11}]/2.2$
= 180 N
Example 1
In a human pyramid in a circus, the entire weight of the balanced group is supported by the legs of a performer who is lying on his back (as shown in Fig. 1). The combined mass of all the persons performing the act, and the tables, plaques etc. involved is 280 kg. The mass of the performer lying on his back at the bottom of the pyramid is 60 kg. Each thighbone (femur) of this performer has a length of 50 cm and an effective radius of 2.0 cm. Determine the amount by which each thighbone gets compressed under the extra load.
Answer
Total mass of all the performers, tables, plaques etc. = 280 kg
Mass of the performer = 60 kg
Mass supported by the legs of the performer at the bottom of the pyramid
= 280 – 60 = 220 kg
Weight of this supported mass
= 220 kg wt. = 220 × 9.8 N = 2156 N.
Weight supported by each thighbone of the performer = ½ (2156) N = 1078 N.
From Table 9.1, the Young’s modulus for bone is given by
Y = 9.4 × $10^9$ N.m$^{–2}$.
Length of each thighbone L = 0.5 m
the radius of thighbone = 2.0 cm
Thus the cross-sectional area of the thighbone
A = π × (2 × 10$^{-2})^2$ m2 = 1.26 × $10^{-3} \ m^2$.
Using Eq. (9.8), the compression in each thighbone (∆L) can be computed as
$\Delta L=\frac{F \times L}{\Delta L\times A}$
$\Delta L=\frac{1078 \times 0.5}{(9.4 \times 10^9)\times 1.26 \times 10^{-3}}$
$=4.33 \times 10^{-3}$ cm
This is a very small change! The fractional decrease in the thighbone is ∆L/L = 0.000091 or 0.0091%.
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