7 Combined Loadings

8/14/2019 7 Combined Loadings

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DMV 4343JAN ~ JUN `07

DEPARTMENT MANUFACTURING / PRODUCT DESIGN /MOULD / TOOL AND DIE

SEMESTER 4

COURSE MECHANICS OF MATERIALS DURATION 8 hrs

COURSE CODE DMV 4343 / DMV 5343 REF. NO.

VTOS NAME MISS AFZAN BINTI ROZALIMR RIDHWAN BIN RAMELI

PAGE 13

TOPICCOMBINED LOADINGS

SUB TOPIC7.1 Stress Elements for Axially Loaded Bar, Shaft in Torsion, and Beam in Bending7.2 Combined Loadings and Complex Stresses

Chapter 7 COMBINED LOADINGS p1

INFORMATION SHEET

REF NO. :PAGE :


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7.1 Stress Elements for Axially Loaded Bar, Shaft in Torsion, and Beam in Bending

In previous chapters we developed methods for determining the stress distributions in a

member subjected to either an internal axial force, a shear force, a bending moment, or a

torsional moment. Most often, however, the cross section of a member is subjected to

several of these types of loadings simultaneously, and as a result, the method of

superposition, if it applies, can be used to determine the resultantstress distribution caused

by the loads. For application, the stress distribution due to each loading is first determined,

and then these distributions are superimposed to determine the resultant stress distribution.

The principle of superposition can be used for this purpose provided a linear relationship

exists between the stress and the loads. 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.

The discussion will be confined to meet these two criteria.

FIGURE 7.1 This chimney is subjected

to the combined loading of wind and

weight. It is important in investigate the

tensile stress in the chimney since

masonry it weak in tension

PROCEDURE FOR ANALYSIS

The following procedure provides a general means for establishing the normal and shear

stress components at a point in a member when the member is subjected to several different

types of loadings simultaneously. It is assumed that the material is homogeneous and

behaves in a linear-elastic manner. Also, Saint-Venant's principles require that the point

where the stress is to be determined is far removed from any discontinuities in the cross

section or points of applied load.



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Internal Resultants.

Section the member perpendicular to its axis at the point where the stress is to be

determined and obtain the resultant internal normal and shear force components and

the bending and torsional moment components.

The force components should act through the centroidof the cross section, and the

moment components should be computed about centroidal axes, which represent the

principal axes of inertia for the cross section.

This,of course, involves drawing free-body diagrams and writing equilibrium

equations.

Average Normal Stress

1. Compute the stress component associated with eachinternal loading. For each case,represent the effect either as a distribution of stress acting over the entire cross-

sectional area, or show the stress on an element of the material located at a

specified point on the cross section.

TABLE 7.1 FORMULA FOR STRESSES

FORCE, LOADINGS FORMULA

NORMAL FORCEThe internal normal force is developed by a uniform

normal-stress distribution

= P I A.

SHEAR FORCEThe internal shear force in a member that is subjected to

bending is developed by a shear-stress distribution

= VQ/It

BENDING

MOMENT

For straight members the internal bending moment is

developed by a normal-stress distribution that varies

linearly from zero at the neutral axis to a maximum at the

outer boundary of the member

= - My/I

TORSIONAL

MOMENT

For circular shafts and tubes the internal torsional

moment is developed by a shear-stress distribution that

varies linearly from the central axis of the shaft to a

maximum at the shaft's outer boundary

= Tp/J

If the member is a closed thin-walled tube = T/2Amt

THIN-WALLED

PRESSURE

VESSELS

If the vessel is a thin-walled cylinder, the internal

pressure p will cause a biaxial state of stress in the

material

2 = pr/2t

If the vessel is a thin-walled sphere, then the biaxial state

of stress is represented by two equivalent components

2 = pr/2t



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EXAMPLE 7.1

An axial compressive load of 800 kN is applied eccentrically to a short rectangular

compression member, as shown below.

An eccentrically loaded short compression member.

Determine

(a) Individual normal stress for the axially loaded member, and

(b) Sketch the stress distribution of each individual normal stress.

Plan the Solution

The eccentric load Pproduces axial deformation plus bending

about the y and z axes. Therefore, this problem involves asuperposition of the stresses due to F, My, and Mz.Solution

Stress Resultants: Figure 2 shows the stress resultants on

cross section ABCD. Applying equilibrium to a free-body

diagram of the member above section ABCD we get the

following expressions for the stress resultants:

F = - PMy = - P dz

Mz = - P dy



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Individual Normal Stresses: Combining all the above equations for the normal stress due to

the axial force F, for the normal stress due to the bending-moment components, we have:

x =F

+Myz

-Mzy

A Iy Iz

Substitute individual stress into the previous resultant

stress equation, we obtain the following:

(x)F =

- P

=

- 800 kN

= -1000 PaA

(40m)

(20m)

The stress contributions as in Figure (a) (a)

(x)My =(-Pdz)z

=(-800 kN)(5m)z

= -150 z kN/m3Iy (1/12)(40m)(20m)

3

Maximum stress occurs at the farthest point along y-axis

that is at z = 10;

Hence

(x)My = -150 z kN/m3 = 1500 kN/m3

The stress contributions as in Figure (b) (b)

(x)Mz =(-Pdy)y

=(-800 kN)(10m)z

= -75 y kN/m3Iy (1/12)(20m)(40m)

3

Maximum stress occurs at the farthest point along z-axis

that is at y = 20;

Hence

(x)My = -75 y kN/m3 = 1500 kN/m3

The stress contributions as in Figure (c)

(c)



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EXAMPLE 7.2 (Continuation of Example 7.1)

Refer Example 7.1.

Determine

(a) The distribution of normal stress on a cross section, say ABCD, that is far enough

from the point of load application that concentration effects may be neglected.

(b) Sketch the stress distribution, and

(c) Identify the location of the neutral axis in cross sectionABCD

Superposition of Stresses:

(a) Using Figure (a) through (c) from solution of Example 7.1, we can combine, algebraically,

the individual stress contributions at four corners to get

(x)A = - 1000 + 1500 - 1500 = - 1000 N/m2

(x)B = - 1000 - 1500 - 1500 = - 4000 N/m2

(x)C = - 1000 - 1500 + 1500 = - 1000 N/m2

(x)D = - 1000 + 1500 + 1500 = + 2000 N/m2

(b)

+ +

(a) (b) (c)

=



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(c) The location of point R and S can be determined from

x (y,z) = - 1000 - 75y - 150z = 0

To find R (we know y = -20)x (y,z) = - 1000 - 75(-20) - 150z = 0

z =-500

= 3.33-150

To find S (we know z = -10)

x (y,z) = - 1000 - 75y- 150(-

10)= 0

y =-500

= 6.67-75

So;

R (-20m, 3.33m)

S (6.67m, -10m)



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7.2.2 Combined Axial Loading and Torsion.

The next example problem illustrates the superposition of stresses due to combined axial

and torsional loading. The solution involves application of Mohrs circle. Following is the

procedure for constructing and using Mohrs circle of stress

Draw Mohr's Circle

1. Establish a set of (, ) axes, with the same scale on both axes. Remember, the

+ axis points downward. It is good idea to use paper that has a grid, like graph

paper or '"engineering paper." Use a scale that will result in a circle of reasonable

size.

2. Assuming that x, y, and xy are given (or can be determined from a given stress

element), locate point X at (x, xy) and point Y at (y, -xy),

3. Connect points X and Y with a straight line, and locate the center of the circle

where this line crosses the axis at (avg, 0).

4. Draw a circle with center at (avg, 0) and passing through points X and Y. It is best

to use a compass to draw the circle.

Compute the required information

5. Form the triangle with sides xy and (x - y)/2, and compute

R = (x - y)2

+ xy226. If the stresses on a particular face, call it face n, are required, locate point N on

the circle by turning an angle 2 counterclockwise (or clockwise) on the circle,

corresponding to rotating an angle counterclockwise (clockwise) from some

reference face on the stress element. Using trigonometry, calculate n and nt

7. If the principal stresses and the orientation of the principal planes are required,

use to calculate the principal stresses, and use trigonometry to determine some

angle, such as 2xp1 that can be used to locate a principal plane, say p 1, with

respect to some known face, say the x face.

8. Use a procedure similar to Step 7 if the maximum in-plane shear stress and the

planes of maximum shear stress are required.



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In this section we will discussed an example which combined more than two types of

loadings. Following example combined normal stress, bending stress and torsional stress on

a member.

EXAMPLE 7.4

Wind blowing on a sign produces a pressure whose resultant,

P, acts in the -y direction at point C, as shown. The weight of

the sign, Ws, acts vertically through point C, and the thin-wall

pipe that supports the sign has a weight Wp.

Determine the principal stresses at points A and B, where the

pipe column is attached to its base. Use the following

numerical data.

Pipe OD = 3.5 m, A = 2.23 m2, Iy = Iz = 3.02m4, Ip = 6.03m

4

Ws = 125 N, Wp = 160 N, P = 75 N, b = 40 m L = 220 m

Plan the Solution

It will be a good idea to tabulate the stress resultants, stress

formulas, and so forth, so that no stress contribution will be

missed. The weight Wscontributes to the axial force, and it also

produces a moment about the y axis. The wind force P

produces a transverse shear force in the ydirection, and it alsocauses a torque about the xaxis and a moment about the z

axis. A correct free-body diagram is essential.

Figure (a)

Solution

Stress Resultants: All six stress resultants on the cross section at the base of the pipe are

shown in Figure (a). The upper portion of Figure (a) can serve as a free-body diagram for

determining these six stress resultants. Let us tabulate the equilibrium equations and indicate

what stress is produced by each stress resultant and label each individual stress.

Individual Stresses: Using the formulas from Table 7.1, we can compute the numerical value of

each of the nonzero stresses listed in Table 1.

A1 = B1 =F

=

- (125N)

(160N) = -128 Pa (1)

A 2.23 m2



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In summary, the principal stresses at points A and B, rounded to three significant figures are:

(1)A =9510

Pa; (2)A = 0 ; (3)A = - 80 Pa

(1)B = 267 Pa ; (2)B = 0 ; (3)B = -3290 Pa

Review the Solution

By showing all six possible internal resultants at the cross section where stresses are to be

calculated, by writing down and solving all six possible equilibrium equations, and by

carefully considering what stress(es) is (are) produced by each stress resultant, we have

accounted for the effects of all loads on the structure. As noted earlier, we have been careful

to make sure that each stress component acts in the direction that "makes sense." For

example, the force Pbends the pipe in the direction that produces tension at pointA, and soforth.

The maximum flexural stress at the base occurs at neitherA norB.


7 Combined Loadings

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