Pure Bending
Assumptions:
- Beam is initially straight and constant cross section.
- Beam material is homogeneous and isotropic.
- The material is elastic, Obeys Hook's law (stress in elastic limit).
- Plane transverse sections remains plane after bending.
- Constant B.M along the length of the beam.
 |
| Pure Bending |
The surface described by the lines which do not extend or contract in the process of bending is called neutral surface and the line of intersection between the neutral surface and the transverse exploratory section is called the neutral axis.
Bending equation is
\[\boxed{\frac{σ}{y} = \frac{M}{I} = \frac{E}{R}}\]
where σ = Bending Stress
M = Bending moment
E = Modulus of elasticity
I = Moment of inertia = \(\frac{\pi D^4}{64}\) for solid circular shaft
R = Radius of curvature
y = Distance from Neutral axis
\[\boxed{σ = \frac{My}{I} = \frac{M}{\frac{I}{y}}}\]
- Section modulus is defined as \(Z = \frac{I}{y}\). Higher the Z lower the stress.
- For same cross section area Z for non-circular cross section is higher than circular sections.
- Maximum Bending stress occurs at where y is maximum (either at the top or bottom of the bar depending on the actual direction of the moment).
Torsion
Torsion means twisting of a structural member when it is loaded by couple that produce rotation about longitudinal axis.
Equation of Torsion
Assumptions:
- The material is homogeneous and isotropic.
- The material is elastic, Obeys Hook's law (stress in elastic limit).
- Cross sections remains plane.
- Cross-sections rotate as if rigid, i.e. every diameter rotates through the same angle
 |
\[\boxed{\frac{T}{J} = \frac{τ}{r} = \frac{Gθ}{L}}\]
where, T = Torsion or applied torque
J = Polar moment of inertia
τ = Shear stress induced due to torsion
R = shaft radius
L = Shaft length
G = modulus of rigidity
θ = Angle of twist
$ϕ = \frac{rθ}{L}$ = Shear strain for any plane from fixed end.
 |
| Shear stress Distribution in torsion |
- Torsional stiffness $ k = \frac{T}{θ} = \frac{GJ}{L}$
.
- Polar section modulus $ Z_p = \frac{J}{r}$
- Shear stress is maximum when r = R.
Polar moment of inertia
- For solid circular shaft
\[J = \frac{\pi d^4}{32}\]
- For hollow shaft (Outer dia D and inter dia d)
\[J = \frac{\pi}{32}(D^4 - d^4)\]
Connections of shafts
- Series connection
Torque (T) will be same in all sections.
Total angle of twist at free end will be equal to the sum of each section's angle of twist.
θ = θ1 + θ2 + θ3
\[θ = T \bigg[ \frac{L1}{G1J1} + \frac{L2}{G2J2} + \frac{L3}{G3J3}\bigg]\]
- Parallel connection
Angle of twist will be same for both shaft.
θ1 = θ2 = θ
Total torque will be equal to sum of torque acting on both shafts.
T = T1 + T2
\[θ = \frac{TL}{G1J1 + G2J2}\]
Combined Bending and Torsion
For a solid shaft (dia D) subjected to bending moment M and torque T
 |
| Combined Bending & Torsion |
Bending stress \(σ = \frac{32 M}{\pi D^3}\)
shear stress due to torsion \( τ = \frac{16 T}{\pi D^3}\)
Maximum principal stress $σ_{max}$ and maximum shear stress $τ_{max}$
\[σ_{max} = \frac{16}{\pi D^3}(M + \sqrt{M^2 + T^2})\]
\[τ_{max} = \frac{16}{\pi D^3}\sqrt{M^2 + T^2}\]
Thus equivalent bending moment (Me) alone would produce the same maximum stresses by combination of M and T:
\[M_e = \frac{1}{2}\Big[M + \sqrt{M^2 + T^2}\Big]\]
And equivalent torque (Te) alone would produce the same maximum shear stress:
\[T_e = \sqrt{M^2 + T^2}\]
NOTE: Equivalent torque and equivalent bending moment should not be used for other purposes like the calculation of power transmission by the shaft, power transmitted depends solely on the torque T carried by the shaft not on Te.