Consider a bar to be rigidly attached at one end and twisted at the other end by a torque or twisting moment T equivalent to F × d, which is applied perpendicular to the axis of the bar, as shown in the figure. Such a bar is said to be in torsion.
Assumptions in Torsional equation
1. The material of the shaft is homogeneous and isotropic throughout length of shaft.
2. The shaft is circular in cross-section remains circular even after loading.
3. A plane section of the shaft normal to its axis before loading remains plane even after the torque has been applied.
4. The twist along the length of the shaft is uniform.
5. The distance between any two normal cross-sections remains same even after application of torque.
6. The maximum shear stress induced in the shaft due to application of torque does not exceed its elastic limit value.
Derivation of Torsion Formula
Hear
R is the outer radius of the shaft.
t is the maximum shear stress at the outer surface.
θ is the angle of twist in radians.
φ is the shear strain
T is the torque (N·m ) being applied to on shaft.
l is the length of the shaft the torque
G is the shear modulus or the modulus of rigidity.
J is the polar moment of inertia circular shaft
Shear stress in Shaft :
A shaft of length l, fixed at one end A, and torque is being applied at the another end B
If a line CD is drawn on the shaft, it will be distorted to CD’ on the application of torque.
If The angle of distortion is φ, from triangle CDD'
DD' = φx l --(1)
but Shear strain = shear stress / shear modulus
φ = t /G
Substitute this relation in equation (1),
Therefore DD' = tl /G --(2)
If R is the outer radius of the shaft. The cross-section will be twisted through an angle θ (OD is twisted to OD')
DD' = Rθ --(3)
the equation (2) is equal to equation (3),
tl /G = Rθ
t/R = Gθ/l
Note: we can also write the above relation in any point,
t/R = q /r = Gθ/l ---(A)
Where q is shear stress at a radius of r form center of shaft.
Note that the highest shear stress is at the point where the radius is maximum, the surface of the shaft.
Torsional resistance of shaft:
When a shaft will be subjected to torsion or twisting moment, there will be developed shear stress and shear strain in the shaft material.
We will consider here a circular shaft which will be subjected to torsion
R = Radius of the circular shaft
D = Diameter of the circular shaft
dr = Thickness of small elementary circular ring
r = Radius of the small elementary of circular ring
q = Shear stress at a radius r from the centre of the circular shaftτ = Shear stress at outer surface of shaft
dA = Area of the small elementary of circular ring
dA = 2П x r x dr
Therefore total torque transmitted by a circular solid shaft could be given by
D = Diameter of the circular shaft
dr = Thickness of small elementary circular ring
r = Radius of the small elementary of circular ring
q = Shear stress at a radius r from the centre of the circular shaftτ = Shear stress at outer surface of shaft
dA = Area of the small elementary of circular ring
dA = 2П x r x dr
Shear stress, at a radius r from the center, could be determined as mentioned here
q/r = τ /R
q = τ x r/R
Turning force due to shear stress at a radius r from the center could be determined as mentioned here
dF = q x dA
dF = τ x r/R x 2П x r x dr
dF = τ/R x 2П r.r.dr
Twisting moment at the circular elementary ring could be determined as mentioned here
dT = Turning force x r
dT = τ/R x 2П r.r.r.dr
dT = τ/R x r.r x (2П x r x dr)
dT = τ/R { r.r x dA}
Total torque could be easily determined by integrating the above equation between limits 0 and RTherefore total torque transmitted by a circular solid shaft could be given by
by substituting the polar moment of Inertia,the total torque transmitted by a circular solid shaft could be given by following equation,
Considering above two equations, we can write here the expression for torsion equation for solid or hollow shafts of uniform circular cross-section and constant wall thickness, the torsion relations are:
The angle of twist can be found by using:
Polar moment of inertia:
Polar moment of inertia J = Ixx + Iyy
The polar moment of inertia for a solid shaft is:
where r is the radius of the object.
The polar moment of inertia for a pipe is:
where the o and i subscripts stand for the outer and inner radius of the pipe.
POWER TRANSMITTED BY THE SHAFT
A shaft rotating with a constant revolutions for minute (r.p.m) is N being acted by a twisting moment T.
The power (P) transmitted by the shaft is
P= 2πNT/60
P = 2πfT
where T is the torque in N·m, f is the number of revolutions per second, and P is the power in watts
thanks
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