Torsion force is a twisting force that causes deformation or failure in materials. It acts when one end of an object is twisted while the other end is held fixed. Examples include twisting a ruler or turning a key in a lock. Torsion is caused by applied torque which promotes rotation. Circular objects subjected to torsion experience shear stresses that increase towards the outer edges, potentially causing yielding or fracture. The maximum shear stress and angle of twist can be calculated using formulas involving the applied torque, polar moment of inertia, and material properties. Ductile materials fail in shear planes while brittle materials fail in tension planes under torsional loads.

1. TORSION FORCE
Presented by
TASFIA AHMED
10.01.03.069

2. INTRODUCTION
• Torsion force is a twisting force that is
applied on an object by twisting one end
when the other is held in position or
twisted in the opposite direction. Different
materials have a different way of
responding to torsion. Some will deform,
crack or even break depending on the
type of material.

3. EXAMPLES
• If a plastic ruler is twisted between both hands.
The ruler is said to be in a state of torsion.
• Whenever we turn a key in a lock the handle
/shank of the key is in torsion.
• Propeller shaft on a ship. The engine turns the
shaft, but the water resists the movement of the
propeller. That induces torsion in the propeller
shaft.

4. IMAGES

5. TORQUE
• In solid mechanics torsion is the twisting of
an object due to applied torque.
• Twisting moments or torques are forces acting
through distance so as to promote rotation.
Simple torque : T = F * l

6. TORSIONAL DEFORMATION OF
CIRCULAR SHAFT
•
•
•
•
Consider a cylindrical bar of
circular cross section twisted by
the torques T at both the ends.
Since every cross section of the
bar is symmetrical and is applied
to the same internal torque T we
can say that the bar is in pure
torsion.
Under action of torque T the right
end of the bar will rotate through
small angle known as angle of
twist.
The angle of twist varies along the
axis of the bar at intermediate
cross section denoted byφ ( x )
φ ( x)

7. TORSIONAL FORCE FOR
CIRCULAR BAR
• Rate of twist
θ=
dφ
dx
• Shear Strain at the outer
surface of the bar
γ max =
bb | rdφ
=
= rθ
ab
dx
• For pure torsion the rate of
twist is constant and equal
to the total angle of twist
divided by the length L of
the bar
rφ
γ max = rθ =
L

8. BEAMS ARE SUBJECTED TO
TORSION
A member subjected to torsion moments would twist
about a longitudinal axis through the shear centre of the
cross section. It was also pointed out that when the
resultant of applied forces passed through the
longitudinal shear centre axis no torsion would occur. In
general, torsion moments would cause twisting and
warping.

9. BAR SUBJECTED TO TORSION
Let us now consider a straight bar
supported at one end and acted upon
by two pairs of equal and opposite
forces.
Then each pair of forces P and P2
1
form a couple that tend to twist the
bar about its longitudinal axis, thus
producing surface tractions and
moments.
Then we can write the moments as
T = 1 d1
P
1
T2 = 2 d 2
P

10. TORSIONAL FORMULA
•
Since the stresses act continuously they have
a resultant in the form of moment.
The Moment of a small element dA located at
radial distance ρ and is given by
dM =
τ max 2
ρ dA
r
The resultant moment ( torque T ) is the
summation over the entire cross sectional
area of all such elemental moments.
T = ∫ dM =
A
τ max
r
∫ρ
2
dA =
A
τ max
Ip
r
Polar moment of inertia of circle with radius r
and diameter d
Ip =
π r4 π d 4
=
2
32
Maximum Shear Stress
τ max =
Tr 16T
=
Ip πd3

11. EXAMPLE
• A solid steel bar of circular cross section has diameter
d=1.5in, l=54in,G=11.5*10^6psi.The bar is subjected to
torques T acting at the ends if the torques have magnitude
T=250 lbft.a) what is the maximum shear stress in the bar
b) what is the angle of twist between the ends?
• From torsion formula
τ max =
16T 16* 250*12
=
= 4530 psi
πd3
π *1.53
Angle of twist
Ip =
π d 4 π *1.54
=
= .4970in 4
32
32
φ=
TL 250*12*54
=
GI p
11.5*106

12. TORSIONAL FAILURE
•
Ductile materials generally fail in shear.
Brittle materials are weaker in tension than shear.
• When subjected to torsion a ductile
specimen breaks along the plane of
maximum shear, i.e. a plane perpendicular
to the shaft axis.
• When subjected to torsion a brittle
material breaks along planes perpendicular
to the direction in which tension is
maximum, i.e. along surfaces at 45 to the
shaft axis.

13. CONCLUSION
• Because of torsion force an object can be
turned and twisted.
• Because if torsion force we can tighten a
nut in a bolt by using a wrench.

14. THANK YOU