mechanics of materials ,strength of materials, bar ,Tensile stress, compressive stress, shear stress, normal stress,Tensile strain, compressive strain, shear strain

1. Stresses and Strains

2. Depending on the response of the material towards loads materials are classified as
• Elastic
• Plastic
• Rigid
Classification of materials based on deformation:

3. Stress:
L
b
d
𝑅 ≈ 𝑃

4. Stress:
Stress is the internal resistance
offered per unit area of a body
against the deformations due
the external applied load.
Units:
𝑠tress = 𝜎 =
𝑅𝑒𝑠𝑖𝑠𝑡𝑖𝑛𝑔 𝐹𝑜𝑟𝑐𝑒
𝐴𝑟e𝑎

5. TYPES OF STRESSES :
only two basic stresses exists :
• normal stress
• shear stress
Other stresses either are similar to these basic stresses or are a combination of these e.g.
bending stress is a combination tensile, compressive and shear stresses.
Torsional stress, as encountered in twisting of a shaft is a shearing stress.
Normal stresses : We have defined stress as resistive force per unit area. If the stresses are
normal to the areas concerned, then these are termed as normal stresses. The normal stresses are
generally denoted by a Greek letter ( s ).
There are two types of normal stresses namely tensile and compressive stresses

6. Tensile Stress:
L b
d
𝑅 ≈ 𝑃
Tensile Stress=
𝑻𝒆𝒏𝒔𝒊𝒍𝒆 𝒍𝒐𝒂𝒅
𝑪𝒓𝒐𝒔𝒔 𝒔𝒆𝒄𝒕𝒊𝒐𝒏 𝒂𝒓𝒆𝒂

7. Compressive Stress:
𝑅 ≈ 𝑃
Compressive Stress=
𝑪𝒐𝒎𝒑𝒓𝒆𝒔𝒔𝒊𝒗𝒆 𝒍𝒐𝒂𝒅
𝑪𝒓𝒐𝒔𝒔 𝒔𝒆𝒄𝒕𝒊𝒐𝒏 𝒂𝒓𝒆𝒂

8. Shear Stress:
Denoting the shearing stress by the Greek letter 𝜏 (tau), we write
Shear Stress=
𝑺𝒉𝒆𝒂𝒓 𝒍𝒐𝒂𝒅
𝒔𝒉𝒂𝒆𝒓𝒊𝒏𝒈 𝒂𝒓𝒆𝒂

10. Q1. The figure shows a support stand designed to carry downward loads. Compute the stress in the square
bar at the upper part of the stand for a load of 120 kN. The line of action of the applied load is centered on
the axis on the shaft, and the load is applied through a thick plate that distributes the force to the entire cross
section of the stand.
Compressive stress on an arbitrary cross section of support
stand shaft.

11. Stress F = 𝜎 = Force/Area = (compressive)
F = 120 kN
A = 35*35= mm2
𝜎 =
Comment This level of stress would be present at any part of any cross section of the
square shaft between its ends.

12. Q2. The Figure shows two circular rods carrying a casting weighing 11.2 kN. If each
rod is 12.0 mm in diameter and the two rods share the load equally, compute the
stress in the rods.
Given Casting weighs 11.2 kN.
Each rod carries half the load.
Rod diameter = d = 12.0 mm

13. Force acting on each rod=
Area =
Stress= 𝜎= MPa

14. Q3. For the punching operation shown in Figure, compute the shear stress in the material if a
force of 5500 kN is applied through the punch. The thickness of the material is 1.5 mm.

15. Given F = 5500 N;
t = 1.5 mm.
The perimeter, p =
The shear area is A=
Then, the shear stress is =

16. Strain:
Strain is found by dividing the total deformation by the original length of
the bar. The lowercase Greek letter epsilon (ε) is used to denote strain
Strain=
Total deformation
Original length
The strains are classified in to three types:
• Tensile strain
• Compressive strain
• Shear strain

17. Tensile Strain:
Tensile Strain=
𝑰𝒏𝒄𝒓𝒆𝒂𝒔𝒆 𝒊𝒏 𝒍𝒆𝒏𝒈𝒕𝒉
𝒐𝒓𝒊𝒈𝒊𝒏𝒂𝒍 𝒍𝒆𝒏𝒈𝒕𝒉
𝛿𝐿
Tensile Strain is assumed as positive

18. Compressive Strain:
Compressive Strain=
𝑫𝒆𝒄𝒓𝒆𝒂𝒔𝒆 𝒊𝒏 𝒍𝒆𝒏𝒈𝒕𝒉
𝒐𝒓𝒊𝒈𝒊𝒏𝒂𝒍 𝒍𝒆𝒏𝒈𝒕𝒉
𝛿𝐿
Compressive Strain is assumed as negative
𝐿 − 𝛿𝐿

19. Shear Strain:
Shear Strain=
𝑻𝒓𝒂𝒏𝒔𝒗𝒆𝒓𝒔𝒆 𝒅𝒊𝒔𝒑𝒍𝒂𝒄𝒆𝒎𝒆𝒏𝒕
𝑯𝒆𝒊𝒈𝒉𝒕 𝒐𝒇 𝒕𝒉𝒆 𝑩𝒍𝒐𝒄𝒌
Change in angle between planes at right angles is called Shear strain
𝛾 =
𝑤
ℎ