ce2155 - combined loadings

Combined Loadings

CE2155 Structural Mechanics and Materials

byAssoc Professor T. H. Wee

Department of Civil EngineeringEmail: [email protected]


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


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.


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.


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.



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).


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.


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.


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.



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



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: =



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



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.


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



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)


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


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


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


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,


Since shear stress varies linearly with the maximum at the surface and minimum at the center of the solid shaft, innerinner


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


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)


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)


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.


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.


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.



In bending deformation three assumptions are made:


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.


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



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

=



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 =


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


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:


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:


Applying the flexure formula where c = 170 mm,( )( ) (Ans) MPa 712103301 170522 6 .. ..;IMc maxmax === ( )( ) MPa 211103301 150522 6 .. ..;IMy BBmax ===

Example Problem:


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.


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.


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:


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:


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:


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:


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:

ce2155 - combined loadings

Documents