2. CE-416
Pre-stressed Concrete Lab Sessional
Presented By:
Md. Zahidul Islam
Id No:10.01.03.142
Course teachers:
Munshi Galib Muktadir
Sabreena Nasrin
Department Of Civil Engineering
Ahsanullah University of Science &
technology
3. STRESS
The internal resistance offered by a body per unit area of
the cross section is known as stress
Types of stresses :
Direct stress
Shear stress or Tangential stress
Bending stress
Torsional stress
Thermal stress
4. TORSIONAL STRESS
Shear stress produced when we apply the twisting moment to the end
of a shaft about its axis is known as Torsional stress.
Torsion is the twisting of an object due to an applied torque.
In sections perpendicular to the torque axis, the
resultant shear stress in this section is perpendicular to the
radius.
Typical Figure of Torsional stress
5. TORSION
Torsion is the twisting of a straight bar when it is loaded by twisting moments or
torques that tend to produce rotation about the longitudinal axes of the bar.
For instance, when we turn a screw driver to produce torsion our hand applies
torque „T‟ to the handle and twists the shank of the screw driver.
6. BARS 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 P1 and P2
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
T1 P d1
1
T2 P2 d 2
7. TORSION
The moments that produce twisting of bar, such as
twisting moments.
and
The moments are indicated by the right hand rule for moment vectors,
which states that if the fingers curl in the direction of the moments then the direction of
the thumb represents the direction of vector.
T1
are called torques or
T2
8. TORSIONAL STRESS ON CIRCULAR SHAFTS
Interested in stresses and strains of
circular shafts subjected to twisting
couples or torques.
Turbine exerts torque T on the
shaft.
Shaft transmits the torque to the
generator.
Generator creates an equal and
opposite torque T.
9. NET TORQUE DUE TO INTERNAL STRESSES
Net of the internal shearing stresses is an
internal torque, equal and opposite to the
applied torque,
T
dF
dA
Although the net torque due to the shearing
stresses is known, the distribution of the
stresses is not.
Distribution of shearing stresses is statically
indeterminate – must consider shaft
deformations.
Unlike the normal stress due to axial loads,
the distribution of shearing stresses due to
torsional loads can not be assumed uniform.
10. AXIAL SHEAR COMPONENTS
Torque applied to shaft produces shearing stresses
on the faces perpendicular to the axis.
Conditions of equilibrium require the existence of
equal stresses on the faces of the two planes
containing the axis of the shaft.
The existence of the axial shear components is
demonstrated by considering a shaft made up of axial
slats.
The slats slide with respect to each other when equal
and opposite torques are applied to the ends of the
shaft.
11. SHAFT DEFORMATIONS
From observation, the angle of twist of the shaft is
proportional to the applied torque and to the shaft
T
length.
L
When subjected to torsion, every cross-section of
a circular shaft remains plane and undistorted
then the bar is said to be under pure torsion.
Cross-sections for hollow and solid circular shafts
remain plain and undistorted because a circular
shaft is axisymmetric.
Cross-sections of noncircular (non-axisymmetric)
shafts are distorted when subjected to torsion.
12. 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 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
( x) .
section denoted by
13. TORSIONAL DEFORMATIONS OF A
CIRCULAR BAR
Rate of twist
d
dx
Shear Strain at the outer
surface of the bar
max
bb |
ab
rd
dx
r
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
r
max
r
L
14. FOR LINEAR ELASTIC MATERIALS
From Hooke‟s Law
G
G is shear modulus of elasticity
is shear strain
From Shear Strain equation :
max
Shear Stress at the outer surface of the
bar :
max
Gr
Torsion Formula : To determine the
relation between shear stresses and
torque, torsional formula is to be
accomplished.
r
r
L
Shear Stresses in cylindrical bar with
circular cross section.
15. TORSIONAL FORMULA
Since the stresses act continously they have a
resultant in the form of moment.
The Moment of a small element dA located at radial
distance and is given by
2
max
dM
dA
r
The resultant moment ( torque T ) is the summation
over the entire cross sectional area of all such
elemental moments.
T
2
max
dM
r
A
dA
A
max
r
Ip
Polar moment of inertia of circle with radius r and
diameter d
r4
d4
Ip
2
Maximum Shear Stress
max
32
Tr
Ip
16T
d3
Distribution of stresses acting on a cross
section.
16. EXAMPLE FROM “MECHANICS OF MATERIALS” BY GERE
& TIMOSHENKO
A solid steel bar of circular cross section has diameter d=1.5in, l=54in,
G=11.5*106 psi.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?
Solution
a)From torsional formula
max
b) Angle of twist
Ip
d4
32
TL
GI p
16T
d3
*1.54
32
16*250*12
*1.53
.4970in 4
250 *12 *54
11.5*10 6
4530 psi
17. STRESS AND STRAIN ANALYSIS
A shaft of circular cross section subjected to torques at opposite ends is considered
.
A point in the cross section at the coordinate x3 will be rotated at an angle
x3
u1
x2 x,3
u2
x1 x3
( ui / uk
uk / xi )
The displacements are given by
and u3 0 .Since
is zero in x3 direction.
Components of strain tensor by using
1
0
we have
ij
0
0
0
x2
2
,x1
2
x2 x1
0
2
2
ii
ik
2
0 : Torsion does not result in the change in volume it implies pure shear
deformation.
18. STRESS AND STRAIN ANALYSIS
From Hooke‟s Law we have
ij
ij
kk
2
where
&
are Lame‟s Constant.
And
=G where G is modulus of rigidity.
Components of stress tensor
0
ij
0
G x2
0
0
G x1
G x2 G x1 0
ij
20. ASSUMPTIONS
The shaft has a uniform cross-section.
Plane sections remain plane after the torque is applied.
The shear strain γ varies linearly in the radial direction.
The material is linearly elastic, so that Hooke's law applies.
21. ANALYSIS OF TORSIONAL STRESS
To find the maximum shear stress τmax which occurs
in a circular shaft of radius c due to the application of
a torque T. Using the assumptions above, we have,
at any point r inside the shaft, the shear stress is
τr = r/c τmax
∫τrdA r = T
∫ r2/c τmax dA = T
τmax/c∫r2 dA = T
22.
J = ∫ r2 dA is the polar moment of inertia of the cross
sectonal area.
Thus, the maximum shear stress:
τmax = Tc/J
For a solid circular shaft, we have:
J = π/2(c)4
For any point at distance r from the center of the shaft, we
have, the shear stress τ is given by
τ = Tr/J
23. TORSION TEST
A torsion test can be conducted on most materials to determine the
torsional properties of the material. These properties include but are not
limited to:
Modulus of elasticity in shear
Yield shear strength
Ultimate shear strength
Modulus of rupture
Ductility