Transcript Chapter 6 - Dr. ZM Nizam
BFC 20903 (Mechanics of Materials) Chapter 6: Torsion
Shahrul Niza Mokhatar [email protected]
Chapter Learning Outcome
1. Defined the relationship between stress and strain 2. Understand torsion theory and their applications 3. Analysed and calculate torsion in solid and hollow circular bar 4. Analysed and calculate the torsion with end restraints 5. Analysed and calculate the torsion with combined bar BFC 20903 (Mechanics of Materials) Shahrul Niza Mokhatar ([email protected]
• • •
Introduction
Torsion commonly found in mechanical engineering.
application, for example machinery structures that has
twisting member
. In civil engineering applications; secondary beam will distributes load to main beam as point load and moment at connection part of the beams.
water retaining in a channel will produce moment that are distributed to the beam as torsion advert post will experience torsion due to wind load that acting on advert planks.
•
Torsion theory
When a bar or shaft of circular section is twisted by moment, its called
pure tension
& the deformed element are in a state of
pure shear
.
Few
assumptions
are taken into account in
torsion analysis
, • The longitudinal axis of the shaft remains straight.
• The shaft does not increase or decrease in length.
• Radial lines remain straight and radial as the cross section rotates.
• Cross sections rotate about the axis of the member.
• • Using Right Hand Law, torsion vector can be determined Planar cross-sectional parallel with member axis will remains planar after subjected to torsion. Shear strain, γ is changing linearly along the bar.
•
Shear stress due to torsion
Shear stress in circular section, τ (tau). From Hooke’s Law,
τ
=
TR J
τ : Shear stress in the shaft if τ max =maximum shear stress occurs at the outer surface @ tegasan ricih. (N/m 2 ) J : Polar moment of inertia of the cross sectional area @ moment sifat tekun kutub (m 4 ) T : Applied torque acting at the cross section (Nm) R : Radius of the shaft (m)
Angles of twist
Twisting angle is angle (in radian) produced when a bar is subjected to torsion. = TL
JG
T = applied torque L = length of member G = shear modulus of material/ modulus of rigidity@modulus ketegaran (N/m 2 ) J = polar moment of inertia
• •
Power transmission
Circular bars or shafts are commonly used for transmission of power. Power is defined as the work performed per unit of time. The work transmitted by a rotating shaft equals to the torque applied times the angle of rotation.
Work = Torque x Angular Displacement Power = d/dt (Work)
• • If torque is not a function of time, then the equation for power simply becomes:
P = T ω
ω is the angular velocity of the shaft (rad/s) = 2π f f : frequency (Hz @ hertz) (1 Hz = 1 cycle/s) T : Applied torque acting at the cross section (Nm) P : Power (W)(1W = 1 Nm/s) Use consistent units for P, T, and ω. Power is commonly specified in horsepower, HP. Angular velocity is usually given in revolutions per minute or RPM. It should then be converted to rad/sec. To do this multiply the value in RPM by 2π and divide by 60.
Example 1
Example 2
• Determine the maximum torque of a hollow circular shaft with inside diameter of 60mm and an outside diameter of 100mm without exceeding the maximum shearing stress of 70MPa.
• Solution; Given; d i = 60mm, d o = 100mm, τ max = 70MPa •
J
32 ( 100 4 60 4 ) 8 .
55
x
10 6
mm
4 max
T
max
r T
max
J
max
J r
( 70
x
10 6 )( 8 .
55
x
10 6 ) 0 .
05 11 .
97
kNm
Remember: τ max =maximum shear stress occurs at the outer surface/radius.
Quiz
A hollow steel shaft has an outside diameter of 150mm and an inside diameter 100mm. The shaft is subjected to a torque of 35kNm. The modulus of rigidity for the steel is 80GPa. Determine; a) the shearing stress at the outside surface of the shaft.
b) the shearing stress at the inside surface of the shaft.
c) the magnitude of the angle of twist in a 2.5m length.
Quiz: Solution
Solution; • •
J
32 ( 150 4 100 4 ) 39 .
89
x
10 6
mm
4 the shearing stress at the outside surface of the shaft.
out
Tr J
35
x
10 3 ( 0 .
075 ) 39 .
89
x
10 6 65 .
81
MPa
the shearing stress at the inside surface of the shaft
in
Tr J
35
x
10 3 ( 0 .
05 ) 39 .
89
x
10 6 43 .
9
MPa
• the magnitude of the angle of twist in a 2.5m length.
TL JG
35
x
10 3 ( 2 .
5 ) 80
x
10 9 ( 39 .
89
x
10 6 ) 0 .
027
rad
•
Composite Bars
Combined bar consists of two or more materials to form a structure. An example is shown in Figure. Superposition principle is used to solve this problem.
• The concept to solve combined bar are: (a) Imposed external torsion is equal to total torsion formed in the bar, i.e., (b) Twisting angle first material is equal to twisting angle of second material at connection part, i.e., (c) Total twisting angle can be calculated from formula,
Group Exercise
• The composite bars with the different material is subjected to the torque is shown in figure. Determine the maximum shear stress and the position. Determine the angle of twist at C.
Solution
J AB
d
4 32 ( 100 2 ) 32 9 .
82
x
10 6
mm
4
J BC
d
4 32 ( 50 2 ) 32 0 .
62
x
10 6
mm
4
• Maximum shear stress; end of C;
AB
T AB r AB J AB
6
x
10 6 ( 50 ) 9 .
82
x
10 6 30 .
55
N
/
mm
2
BC
T BC r BC J BC
4
x
10 6 ( 25 ) 0 .
62
x
10 6 161 .
3
N
/
mm
2 The maximum shear stress occurs in the bar of BC.
Angle of twist at the
TL JG
TL JG AB
TL JG BC
( 6
x
10 6 )( 2000 ) 9 .
82
x
10 6 ( 3
x
10 4 ) ( 4
x
10 6 )( 300 ) 0 .
62
x
10 6 ( 8
x
10 4 ) 0 .
0158
rad
Torsion of non-cylindrical member
• Generally, we deal with axisymmetric bodies and the shear strain is linear through the entire body. However, non-circular cross-sections are not axisymmetric causing complex behaviors, which may cause bulging or warping when the shaft is twisted. Condition of bulging of non-circular shaft
Empirical formulas for various shapes
Example 3
The aluminum shaft shown in figure has a cross sectional area in the shape of an equilateral triangle. Determine the largest torque, T that can be applied to the end of the shaft if the allowable shear stress, is = 56MPa and the angle of twist at its end is restricted to = 0.02 rad. Given G al = 26GPa.
Example 3: Solution
Thin-walled Having Closed Cross Sections
• • Thin walled of noncircular shape are often used to construct lightweight frameworks which is used in aircraft. Due the applied torque, T, shear stress is developed on the front face of the element. Shear flow in a solid body is the gradient of a shear stress through the body. Shear flow is the product of the tube’s thickness and the average shear stress. This value is constant at all points along the tube’s cross section. As a result, the largest average shear stress on the cross section occurs where the tube’s thickness is small.
q
avg t
•
Thin-walled Having Closed Cross Sections
In non-circular thin walled shafts for closed segments. We assume that the stress is uniformly distributed across the thickness and that we can assume an average shear stress. The average shear stress in the body is;
ave
T
2
tA m
dF ds τ ave where,
ave
- average shear stress t - the thickness of the shaft at the point of interest A m - mean area enclosed within the boundary of the centerline of the shaft thickness.
T - the applied torque t T τ ave h
Thin-walled Having Closed Cross Sections
h
(
ave t ds
)
ave t
h
2
ave t
2
ave t d A m ds A m
ave
T
2
tA m
Since, we can determine the shear flow throughout the cross section using the equation;
q
T
2
A m
• • • Angle of twist, This angle can be determined by using the energy method. The angle given in radians, can be expressed as;
TL
4
A m
2
G
ds t
Here the integration must be performed around the entire boundary of the tube’s cross sectional area.
where, t - thickness of the interior segment L - length of the section G - modulus of rigidity of the section/shear modulus
Example 4
A square aluminum tube has the dimensions as shown in figure. a) determine the average shear stress in the tube at point A if it is subjected to a torque of 85Nm. b) compute the angle of twist due to this loading. Given G al = 26GPa.
a) The area, A m ;
m
Example 4: Solution
( 50 )( 50 ) 2500
mm
2 50mm The shaded area = A m 50mm
avg
T
2
tA m
85
x
10 3 2 ( 10 )( 2500 ) 1 .
7
N
/
mm
2 Since t is a constant because of the square tube, the average shear stress is the same at all points on the cross section.
Example 4: Solution
b) Angle of twist;
TL
4
A m
2
G
ds t
( 85
x
10 3 )( 1500 ) 4 ( 2500 ) 2 ( 26
x
10 3 ) 4 50
mm
10
mm
1 .
962
x
10 4 ( 20 ) 3 .
92
x
10 3
rad
Here, the integral represents the length around the centerline boundary of the tube.