ELASTIC MODULI

YOUNG’S MODULUS

SHEAR MODULUS

BULK MODULUS

YOUNG’S MODULUS

SHEAR MODULUS

BULK MODULUS

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.

The ratio of stress and strain, called modulus of elasticity, is found to be a characteristic of the material.

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.

`color{navy}(Y= sigma/in)`

From Eqs. (9.1) and (9.2), we have

`color{orange}(Y = (F//A)/(ΔL//L))`

`color{navy}(= (F × L) /(A × ΔL))`....(9.8)

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 9.1 gives the values of Young’s moduli and yield strengths of some materials.

From the data given in Table 9.1, 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 crosssectional 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.

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.

`color{navy}(Y= sigma/in)`

From Eqs. (9.1) and (9.2), we have

`color{orange}(Y = (F//A)/(ΔL//L))`

`color{navy}(= (F × L) /(A × ΔL))`....(9.8)

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 9.1 gives the values of Young’s moduli and yield strengths of some materials.

From the data given in Table 9.1, 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 crosssectional 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.

Q 3139491312

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 x× 1011 N m^(-2).`

Class 11 Chapter 9 Example 1

Class 11 Chapter 9 Example 1

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

`color{green} {"Stress" = F/A = F/(pir^2)}`

`(100\ \ 10^3\ \ N)/(3.14 10^2m^2)`

The elongation,

`color{orange} {DeltaL=(F//A)L/Y}`

`= (38. 10^8Nm^(-2)1m)/(2\ \ 10^(11)Nm^(-2)`

`= 1.59xx10^(-3)m`

`= 1.59mm`

The strain is given by

`color{purple} {"Strain" = ΔL//L}`

`= (1.59 × 10–3 m)//(1m)`

`= 1.59 × 10^(–3)`

`= 0.16 %`