Advanced Mechanics
of Materials
Lecture-4
By Dr Nagi
Example 1.
An unknown weight falls through 10 mm on a collar rigidly attached to the lower end of a
vertical bar 3 m long and 600 𝑚𝑚2 in section. If the maximum instantaneous extension is
known to be 2 mm, what is the corresponding stress and the value of unknown weight?
Take E = 200 kN/𝑚𝑚2 .
Resilience
When a body is loaded within elastic limit, it changes its dimensions and on the removal
of the load, it regains its original dimensions. So long as it remains loaded, it has stored
energy in itself. On removing the load, the energy stored is given off as in the case of a spring.
This energy, which is absorbed in a body when strained within elastic limit, is known as strain
energy. The strain energy is always capable of doing some work. The strain energy stored in a
body due to external loading, within elastic limit, is known as resilience and the maximum
energy which can be stored in a body up to the elastic limit is called proof resilience. The
proof resilience per unit volume of a material is known as modulus of resilience. It is an
important property of a material and gives capacity of the material to bear impact or shocks.
Mathematically, strain energy stored in a body due to tensile or compressive load or resilience,
Torsional Shear Stress
When a machine member is subjected to the action of two equal and opposite couples acting in parallel planes (or
torque or twisting moment), then the machine member is said to be subjected to torsion. The stress set up by
torsion is known as torsional shear stress. It is zero at the centroidal axis and maximum at the outer surface.
Consider a shaft fixed at one end and subjected to a torque (T) at the other end as shown in Fig. 2. As a result of
this torque, every cross-section of the shaft is subjected to torsional shear stress. We have discussed above that the
torsional shear stress is zero at the centroidal axis and maximum at the outer surface. The maximum torsional
shear stress at the outer surface of the shaft may be obtained from the following equation
Home Work
1- A shaft is transmitting 97.5 kW at 180 r.p.m. If the allowable shear stress in the material is
60 MPa, find the suitable diameter for the shaft. The shaft is not to twist more that 1° in a
length of 3 meters. Take C = 80 GPa.
2- A hollow shaft is required to transmit 600 kW at 110 r.p.m., the maximum torque being
20% greater than the mean. The shear stress is not to exceed 63 MPa and twist in a length of
3 metres not to exceed 1.4 degrees. Find the external diameter of the shaft, if the internal
diameter to the external diameter is 3/8. Take modulus of rigidity as 84 GPa.