Understanding Shear Stresses in Beams: Crucial Insights into Internal Forces and Structure, Slides of Mechanics of Materials

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• Dharmsinh Desai University

SUB-TOPICS

• Introduction and concepts of Stress and Strain
• Analysis of Bars with uniform section (prismatic bars)
• Bars of varying sections
• Uniformly tapering circular & rectangular bars
• Elongation of bar of uniform section due to self weight
• Composite Sections

There are two primary types of stresses: Normal Stress and Shear Stress Normal stress, σ, is calculated when the force is normal to the surface area; whereas the shear stress, 𝜏 , is calculated when the force is parallel to the surface area.

TYPES OF STRESS

When a force acts perpendicular (or "normal/axial/direct") to the surface of an object, it exerts a normal stress. It is denoted by  It can be either compressive or tensile in direction.

NORMAL STRESS

SHEAR STRESS

Shear stresses, 𝜏 , are commonly found in bolts, rivets, and pins

SHEAR STRESS

If the direction of the force is parallel to the plane of the object. The deformation will be along that plane. The stress experienced by the object here is shear stress or tangential stress. Unit is same as normal stress.

Permissible stress or Allowable stress or Working stress is always less than the Ultimate stress using a factor called factor of safety.

Stress at a point is resolved in Normal and Shear Component. Normal Stress can be read as : x is normal stress acting on plane perpendicular the X direction. Shearing Stress can be read as : 𝜏yz is shear stress in the plane perpendicular to Y axis in the z direction.

DESCRIPTION OF STRESS AT A POINT

TYPES OF STRAIN

Lateral strain / Longitudinal strain = Constant = 1 /m =  = Poisson’s ratio Lateral strain = Poisson’s ratio x Longitudinal strain

TYPES OF STRAIN …

SHEAR STRAIN

Example : 1

A short hollow, cast iron cylinder with wall thickness of 10

mm is to carry compressive load of 100 kN. Compute the

required outside diameter `D’ , if the working stress in

compression is 80 N/mm

2

Solution:  = 80N/mm 2 ; P= 100 kN = 100* 3 N A =(/4) *{D 2

• (D-20) 2 } as  = P/A substituting in above eq. and solving. D = 49. mm

D

d

10 mm

Stress- Strain Curve for Mild Steel (Ductile)

• O-A : Proportionality Limit [Hook’s Law is Valid]
• B : Elastic Limit [When load is removed, deformation is completely recovered]
• C/D : Upper and Lower Yield Stress Points [When load is removed, deformation is not completely recovered = Permanent Set]
• D-E : Strain Hardening
• E : Ultimate Stress Point [Maximum Load which a specimen can withstand]
• E-F : Necking
• F : Fracture / Breaking Stress Point

O

A

B

C

D

E

F

ELASTIC PLASTIC NECKING

Stress- Strain Curve for Mild Steel (Ductile )…