-
Combined Loadings
CE2155 Structural Mechanics and Materials
byAssoc Professor T. H. Wee
Department of Civil EngineeringEmail: [email protected]
CE2155 Structural Mechanics and Materials
Torsion refers to the twisting of a straight bar when it is
loaded by moments that tend to produce rotation about the
longitudinal axis of the bar.
A straight bar supported at one end and loaded by two pairs of
equal and opposite forces is an idealized case of torsional
loading. The pair of equal and opposite forces forms a couple that
produces a moment which causes the twisting of the bar. The moments
T1 (= P1d1) and T2 (= P2d2), are called torques or twisting
moments.
The moment can be represented by a vector in the form of a
double-headed arrow drawn perpendicular to the plane containing the
couple. The direction of the moment is indicated by the right-hand
rule for moment vectors. An alternative representation of a moment
is a curved arrow acting in the direction of rotation.
Bar Subjected to Torsion
-
CE2155 Structural Mechanics and Materials
Bar Subjected to Torsion (Example)
If T1 = T2 = 4kNm, twisting moment at the support will be equal
to zero. There will be no torsion and rotation between the support
A and point B. Every cross-section between BC will be subjected to
the same torque 4kNm and there will be rotation in section BC.
A B C
If T1 = 4kNm and T2 = 3kNm, twisting moment at the support will
be equal to 1kNm. Every cross-section between AB will be subjected
to the same internal torque 1kNm. Every cross-section between BC
will be subjected to the same internal torque 3kNm. There will be
rotation between both the sections AB and BC.
If T1 = 3kNm and T2 = 4kNm, twisting moment at the support will
be equal to 1kNm. Every cross-section between AB will be subjected
to the same internal torque 1kNm. Every cross-section between BC
will be subjected to the same internal torque 4kNm. There will be
rotation between both the sections AB and BC.
CE2155 Structural Mechanics and Materials
Example Member subjected to Twisting Moment
The cantilever roof at entrance of Busan Exposition Centre,
connected to the horizontal member, will produce a twisting moment
on the horizontal cylindrical member.
-
CE2155 Structural Mechanics and Materials
Torsional Deformations of a Circular Bar
Assume a prismatic bar of circular cross section twisted by
torques Tacting at the ends. The bar is in pure torsion since every
cross section of the bar is identical and subjected to the same
internal torque T.
In other words, when the prismatic bar is sectioned anywhere
between p and q, the torque at the sectioned plane, known as the
internal torque, will be equal to the external torque, T at p and q
to satisfy conditions of equilibrium.
If point p of the bar is assumed fixed, then under the action of
torque, T, the end q will rotate through a small angle, , known as
the angle of twist.
CE2155 Structural Mechanics and Materials
Torsional Deformations of a Circular Bar
For the prismatic bar of length L shown, is also equal to the
total angle of twist.
The angle of twist changes along the axis of the bar and at
intermediate cross sections it will have a value (x) that is
between zero at p and at q.
If under pure torsion, i.e., every cross section of the bar has
the same radius and is subjected to the same torque, the angle (x)
will vary linearly between the ends (that is d/dx = constant).
-
CE2155 Structural Mechanics and Materials
Now consider an element of the bar between two cross-sections
distance dx apart.
A small element abcd is identified on its outer surface with
sides aband cd.
Torsional Deformations of a Circular Bar
During twisting of the bar, the element twist through a small
angle of twist, d so that points band c move to b and c,
respectively.
The lengths of all the four sides of abcd do not change during
this small rotation.
However, the angles at the corners are no longer equal to
90.
CE2155 Structural Mechanics and Materials
The element is therefore in a state of pure shear and the
magnitude of the shear strain, being the decrease in angle bad is
given by:
where max (maximum at surface) is measured in radians. The
equation can also be expressed as:
abbb'
max =
dxrd=max
This equation relates the shear strain at the outer surface of
the bar of radius r to the angle of twist.
Torsional Deformations of a Circular Bar
-
CE2155 Structural Mechanics and Materials
The quantity d/dx is the rate of change of the angle of twist
with respect to the distance x measured along the axis of the bar.
This rate of change d/dx is denoted by the symbol and is referred
to as the angle of twist per unit length, or the rate of twist:
dxd=
The shear strain at the outer surface can now be written as:
== rdxrd
max
In general, both and vary with the distance x along the axis of
the bar. However, in the special case of pure torsion, the rate of
twist is constant and equal to the total angle of twist divided by
the length L of the bar, that is, = /L. Therefore, for pure torsion
only, we obtain
Lrr ==max
Torsional Deformations of a Circular Bar
CE2155 Structural Mechanics and Materials
The shear strain within the interior of the bar can be found by
the same method used to find the shear strain max at the surface.
Because radii () in the cross sections of a bar remain straight and
undistorted during twisting (for elastic, homogeneous and isotropic
material), the corresponding shear strains in the interior of the
bar may be expressed as
max== rThis equation shows that the shear strains in a circular
bar vary linearly with the radial distance from the center, with
the strain being zero at the center and reaching a maximum value at
the outer surface.
rnote max: =
Torsional Deformations of a Circular Bar
-
CE2155 Structural Mechanics and Materials
The equations for the shear strains also apply to circular
tubes. The variation in the shear strain between the maximum strain
at the outer surface and the minimum strain at the interior surface
is linear.
The minimum strain is related to the maximum strain by the
equation:
max2
1min = r
r
Torsional Deformations of a Circular Bar
CE2155 Structural Mechanics and Materials
Shear Stresses in a Circular Bar in Torsion
Knowing the shear strain in a circular bar in torsion, the
directions and magnitudes of the corresponding shear stresses can
be determined. For the example shown below, it can be observed that
the torque T rotates the right-hand end of the bar
counter-clockwise when viewed from the right, and therefore the
shear stresses act in the directions shown:
The magnitude of the shear stress can be determined from the
stress-strain relation for the material of the bar. If the material
is linearly elastic, we can use Hookes law in shear = GWhere G is
the shear modulus of the material and is the shear strain in
radians. Since therefore,= rmax
= Grmax in which max is the shear stress at the outer surface of
the bar.
-
CE2155 Structural Mechanics and Materials
On the other hand, substituting the equation and
into gives
== G
in which is the shear stress at an interior point at a distance
from the centre. This equations show that the shear stresses vary
linearly with the distance from the center of bar.
The shear stresses acting on a cross-sectional plane are
accompanied by shear stresses of the same magnitude acting on
longitudinal planes as equal shear stresses always exist on
mutually perpendicular planes.
max== rG
Also the state of pure shear at the surface of a bar is
equivalent to equal tensile and compressive stresses acting on an
element oriented at angle of 45o.
= Grmax
Torsional Deformations of a Circular Bar
CE2155 Structural Mechanics and Materials
However, by transformation it can be shown that the maximum
normal stress exist in a plane orientated 45 to pure shear stress (
= 0) plane and also equal to . For brittle material which fails in
tension, failure would result in a plane perpendicular to the
direction of maximum normal stress (tensile stress).
3.When a bar is subjected to only torsion (T), the element abcd
would be in a state of pure shear.
(Slide 17 of Stress and Strain Transformation (Part I)
-
CE2155 Structural Mechanics and Materials
The Torsion Formula
Next, to calculate the stresses and strains in a bar due to any
set of applied torques, the relationship between the shear stresses
and the torque, T have to be determined. When the bar is subjected
to torsion, shear stress will be induced at all points on any
cross-section of the bar. For rotational equilibrium, the shear
stresses must have a resultant in the form of a moment equal to the
torque, T acting on the bar. To determine this resultant, we
consider an element of area dA located at radial distance from the
axis of the bar. The shear force acting on this element is equal to
dA, and the elemental moment about the axis of the bar is given
by
dAr
dAdM 2max ==The resultant moment (equal to the torque T) is the
summation over the entire cross-sectional area of all such
elemental moments:
Jr
dAr
dMTAA
max2max === in which is known as the polar moment
of inertia. = A dAJ 2
CE2155 Structural Mechanics and Materials
Recall that the resultant moment obtained by the summation of
all elemental moments over the entire cross-sectional area is
Jr
dAr
dMTAA
max2max === from which is known at the torsion formula. For a
circle of radius r,the polar moment of inertia is
322
44 drJ ==
JTr=max
= A dArJ 22rA =
rdrdA = 2 = r drrJ 32
since
therefore
and hence
Finally, we obtain where d is the diameter of the circle.
The Torsion Formula
-
CE2155 Structural Mechanics and Materials
Now, combining the torsion formula with obtained earlier, the
shear at
distance from the center of the bar can be deduced as . Once
again it isseen that the shear stresses vary linearly with the
radial distance from the center of the bar.
max= rJ
T=
Next, combining the equation obtained earlier (slide 12), with
the torsion
formula, we obtain which shows that the rate of twist is
directlyproportional to the product GJ, known as the torsional
rigidity of the bar. For a bar in
pure torsion, the total angle of twist is equal to
= GrmaxGJT=
GJTLL ==
JTr=max
Total Angle of Twist
CE2155 Structural Mechanics and Materials
Circular Tube in Torsion
Circular tubes are more efficient than solid bars in resisting
torsion.
+=outerinner AA
dAdAT outerouterinnerinner
This is because, the shear stresses in a solid circular bar are
maximum at the outer boundary of the cross section and zero at the
center. Therefore, most of the material in a solid shaft is
stressed significantly below the maximum shear stress.
Furthermore, the stresses near the center of the cross section
have a smaller moment arm and hence resist lower torque.
To illustrate, consider the total torque in a solid circular bar
as T = T1 + T2 where T1 and T2 are
the torque at the inner (i.e. near the center) and outer portion
of the solid shaft. Then,
-
CE2155 Structural Mechanics and Materials
Since shear stress varies linearly with the maximum at the
surface and minimum at the center of the solid shaft,
innerinner
-
CE2155 Structural Mechanics and Materials
Steel shaft (either bar or tube) to transmit torque of 1200Nm
without exceeding allowable shear stress of 40 MPa, nor allowable
rate of twist of 0.75o/m (Assume G=78 GPa). Determine d0 and d2
Solid Shaft:3 6 316 16(1200N-m) 152.8x10 m
(40MPa)o allow
Td = = = 0.0535m 53.5mmod = =
Based on allowable stress:
Based on allowable rate of twist:9 4
o o
1200Nm 1175x10 m(78GPa)(0.75 /m)( rad /180 )allow
TJG
= = =9 4
4 6 4o
32 32(1175x10 m ) 11.97x10 mJd
= = = 0.0588m 58.8mmod = =This is governing dimension
maxTrJ
=TLGJ
=ddx =
Circular Tube in Torsion - Example Problem
CE2155 Structural Mechanics and Materials
For Circular Tube: 1 2 2 2 22 2(0.1 ) 0.8d d t d d d= = =4 4 4 4
4 42 1 2 2 2 2( ) (0.8 ) (0.5904 ) 0.0579632 32 32
J d d d d d d = = = = 2
4 32 2
( / 2)0.05796 0.1159allow
Tr T d TJ d d
= = =Based on allowable stress:
3 6 32
1200Nm 258.8x10 m0.1159 0.1159(40MPa)allow
Td = = = 2 0.0637m 63.7mmd = =
Based on allowable rate of twist:42(0.05796 )
allowT T
GJ G d = =
Solving the above equation gives: 2 67.1mmd =Ratio of tube/bar
diameters:
0
2 67.1 mm 1.1458.8 mm
dd
= =
Circular Tube in Torsion - Example (Contd)
-
CE2155 Structural Mechanics and Materials
0
2 67.1 mm 1.1458.8 mm
dd
= =
These results show that the hollow shaft uses only 47% as much
material as does the solid shaft, while its outer diameter is only
14% larger
hollow bar are more efficient in the use of materials than are
solid bars.
47.0=solid
hollow
WW
Circular Tube in Torsion - Example (Contd)
CE2155 Structural Mechanics and Materials
Example Member subjected to Twisting Moment
The cantilever roof at entrance of Busan Exposition Centre,
connected to the horizontal member, will produce a twisting moment
on the horizontal cylindrical member.
-
CE2155 Structural Mechanics and Materials
Bending Deformation of a Straight Member
Consider the undeformed bar Fig (a) subjected to a bending
moment, the bar tends to distort as shown in Fig (b). It can be
seen that longitudinal lines becomes curved and vertical transverse
lines remain straight and yet undergo a rotation.
CE2155 Structural Mechanics and Materials
The behaviour of any deformable bar subjected to a bending
moment causes the material within the bottom portion of the bar to
stretch and the material within the top portion to compress.
Consequently, between these two regions there must be a surface,
called the neutral surface, in which longitudinal fibres of the
material will not undergo a change in length.
Note that in the bending of the rubber bar, the top line
stretches, the bottom line compresses and the centre line remains
the same.
Bending Deformation of a Straight Member
-
CE2155 Structural Mechanics and Materials
In bending deformation three assumptions are made:
Bending Deformation of a Straight Member
The longitudinal axis x, which lies within the neutral surface,
does not experience any change in length.
All cross sections of the beam remain plane & perpendicular
to the longitudinal axis during the deformation.
Any deformation of the cross sectionwithin its own plane will be
neglected.
CE2155 Structural Mechanics and Materials
To show how this distortion will strain the material, consider
asegment of the beam that has an undeformed thickness x. The normal
strain along s is determined from
After deformation, x has a radius of curvature with centre of
curvature at point O. The deformed length of s becomes
Therefore, by substitution,
ssslim
s =
0
( ) = ys
( )=
=
yylims 0
Bending Deformation of a Straight Member
-
CE2155 Structural Mechanics and Materials
In other words, for any specific cross section, the longitudinal
normal strain will vary linearly with y from the neutral axis. The
maximum strain occurs at the outermost fibre. Hence the normal
strain can be expressed as a factor of the maximum strain as
maxcy
=A linear variation of normal strain must then be the
consequence of a linear variation in normal stress. Hence will vary
from zero at the members neutral axis to maximum value, max a
distance c farthest from the neutral axis.
maxcy
=
Bending Deformation of a Straight Member
CE2155 Structural Mechanics and Materials
The Flexure Formula
The neutral axis on the cross section satisfies the condition
that the resultant force produced by the stress distribution over
the cross section area must be equal to zero. Noting that the force
dF = dA acts on element dA, and summation of moment about z-axis is
zero,
( ) === A maxAA dAcyydAyydFM
= Amax dAycM 2= A dAyI 2 inertia, ofmoment Since
IMc
max =
-
CE2155 Structural Mechanics and Materials
Since
The normal stress, at any distance y from the neutral axis is
given by
where M = moment
I = moment of inertia of the cross-sectional area computed about
the neutral axis
IMy=
ycmax =
The Flexure Formula
CE2155 Structural Mechanics and Materials
The simply supported beam has the cross-sectional area as shown.
Determine the absolute maximum bending stress in the beam and draw
the stress distribution over the cross section at this
location.
Example Problem:
-
CE2155 Structural Mechanics and Materials
Solution:
The maximum internal moment in the beam kN.m 522865
8
22
.wLM ===
By symmetry, the centroid C and thus the neutral axis pass
through the mid-height of the beam, and the moment of inertia
is
( )( )( ) ( )( )( ) ( )( )( ) 46
323
2
m 103.301
3.002.012116.0002.025.002.025.0
1212
=
+
+=+= AdII
Example Problem:
CE2155 Structural Mechanics and Materials
Applying the flexure formula where c = 170 mm,( )( ) (Ans) MPa
712103301 170522 6 .. ..;IMc maxmax === ( )( ) MPa 211103301 150522
6 .. ..;IMy BBmax ===
Example Problem:
-
CE2155 Structural Mechanics and Materials
Combined Loadings
Most often, the cross section of a member is subjected to
several types of loadings, such as bending and torsion,
simultaneously.
The method of superposition can be used to determine the
resultant stress distribution caused by the loads provided a linear
relationship exists between the stress and loads.
For application, the stress distributions due to each loading is
first determined, and then these distributions are superimposed to
determine the resultant stress distribution.
Also, the geometry of the member should not undergo significant
change when the loads are applied. This is necessary in order to
ensure that the stress produced by one load is not related to the
stress produced by any other load.
CE2155 Structural Mechanics and Materials
Combined Loadings
Superposition of stress components can be used to determine the
normal and shear stress at a point in a member subjected to
combined loadings.
To solve, it is first necessary to determine the resultant axial
and shear, and the resultant torsional and bending moment at the
section where the point is located.
Then the normal and shear-stress resultants are determined by
algebraically adding the normal and shear stress components at the
point.
-
CE2155 Structural Mechanics and Materials
The solid rod shown had a radius of 0.75 cm. If it is subjected
to the loading shown, determine the state of stress at point A.
800 N
500 N
8 cm
10 cm
14 cm
Example Problem:
CE2155 Structural Mechanics and Materials
Solution: The rod is sectioned through point A. Using the
free-body diagram of segment AB, the resultant internal loadings
can be determined from the six equations of equilibrium.The normal
force (500N) and shear force (800N) must act through the centroid
of the cross section and the bending-moment components (8000 N.cm
and 7000 N.cm) are applied about centroidal(principal) axes.
800 N (14 cm) = 11200 N.cm
800 N (10 cm) = 8000 N.cm
500 N (14 cm) = 7000 N.cm
800 N
800 N
500 N
500 N
10 cm
14 cm
Example Problem:
-
CE2155 Structural Mechanics and Materials
Stress Components.
Normal Force: The normal-stress distribution is as shown. For
point A,
Shear Force: The shear-stress distribution is as shown. For
point A, 500 N
2.83 MPa
800 N
6.04 MPa
MPa 832750
5002 .)cm.(
NAP
A ===
( ) 32 2831075021
37504 cm.cm.)cm.(AyQ =
==
)semicircle(for 34=ry
( )[ ] MPa 0467502750 28130800 4413
.)cm.(cm.
)cm.(NtIQV
A ===
Example Problem:
CE2155 Structural Mechanics and Materials
Stress Components.
Bending Moment: For the 8000 N.cm component, point A lies on the
neutral axis, so the normal stress is A = 0. For the 7000 N.cm
moment, y = 0.75cm, so the normal stress at point A is,
( )[ ] MPa 26211750 7507000 441 .cm. )cm.(cm.NIMyA ===
( )[ ] MPa 01169750 75011200 421 .cm. )cm.(cm.NJTcA ===
8000 N.cm
211.26 MPa
7000 N.cm
Torsional Moment: At point A, A=0.75 cm. Thus the shear stress
is
169.01 MPa
11200 N.cm
Example Problem:
-
CE2155 Structural Mechanics and Materials
Superposition. When the individual results are superimposed, it
is seen than anelement of material at A is subjected to both normal
and shear stress components.7000 N.cm
8000 N.cm
11200 N.cm800 N
500 N
(800 N)(500 N) (11200 N.cm)(8000 N.cm) (7000 N.cm)
2.83 MPa 6.04 MPa 211.3 MPa 169 MPa
214 MPa
6.04 MPa + 169 MPa
2.83 MPa + 211.3 MPa
175 MPa
Example Problem: