CHAPTER 8COMBINED LOADING

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CHAPTER OBJECTIVES

Analyze the stress developed in thin-walled pressure vessels

Review the stress analysis developed in previous chapters regarding axial load, torsion, bending and shear

• Discuss the solution of problems where several of these internal loads occur simultaneously on a member’s x-section

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CHAPTER OUTLINE1. Thin-Walled Pressure Vessels2. State of Stress Caused by Combined Loadings

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8.1 THIN-WALLED PRESSURE VESSELS Cylindrical or spherical vessels are commonly

used in industry to serve as boilers or tanks When under pressure, the material which they

are made of, is subjected to a loading from all directions

However, we can simplify the analysis provided it has a thin wall

“Thin wall” refers to a vessel having an inner-radius-thickness ratio of 10 or more (r/t ≥ 10)

When r/t = 10, results of a thin-wall analysis will predict a stress approximately 4% less than actual maximum stress in the vessel

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8.1 THIN-WALLED PRESSURE VESSELS Assumption taken before analysis is that the

thickness of the pressure vessel is uniform or constant throughout

The pressure in the vessel is understood to be the gauge pressure, since it measures the pressure above atmospheric pressure, which is assumed to exist both inside and outside the vessel’s wall

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8.1 THIN-WALLED PRESSURE VESSELSCylindrical vessels A gauge pressure p is

developed within the vessel by a contained gas or fluid, and assumed to have negligible weight

Due to uniformity of loading, an element of the vessel is subjected to normal stresses 1 in the circumferential or hoop direction and 2 in the longitudinal or axial direction

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8.1 THIN-WALLED PRESSURE VESSELSCylindrical vessels We use the method of sections and apply the

equations of force equilibrium to get the magnitudes of the stress components

For equilibrium in the x direction, we requireprt1 =Equation 8-1

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8.1 THIN-WALLED PRESSURE VESSELSCylindrical vessels As shown, 2 acts uniformly throughout the wall,

and p acts on the section of gas or fluid. Thus for equilibrium in the y direction, we require

pr2t2 =

Equation 8-2

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8.1 THIN-WALLED PRESSURE VESSELSCylindrical vessels For Eqns 8-1 and 8-2,

1, 2 = normal stress in the hoop and longitudinal directions, respectively. Each is assumed to be constant throughout the wall of the cylinder, and each subjects the material to tension

p = internal gauge pressure developed by the contained gas or fluidr = inner radius of the cylindert = thickness of the wall (r/t ≥ 10)

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8.1 THIN-WALLED PRESSURE VESSELSCylindrical vessels Comparing Eqns 8-1 and 8-2, note that the hoop

or circumferential stress is twice as large as the longitudinal or axial stress

When engineers fabricate cylindrical pressure vessels from rolled-form plates, the longitudinal joints must be designed to carry twice as much stress as the circumferential joints

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8.1 THIN-WALLED PRESSURE VESSELSSpherical vessels The analysis for a spherical pressure vessel can

be done in a similar manner Like the cylinder, equilibrium in the y direction

requirespr2t2 =

Equation 8-3

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8.1 THIN-WALLED PRESSURE VESSELSSpherical vessels Note that Eqn 8-3 is similar to Eqn 8-2. Thus, this

stress is the same regardless of the orientation of the hemispheric free-body diagram

• An element of material taken from either a cylindrical or spherical pressure vessel is subjected to biaxial stress; normal stress existing in two directions

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8.1 THIN-WALLED PRESSURE VESSELSSpherical vessels The material is also subjected to radial stress, 3.

It has a value equal to pressure p at the interior wall and decreases to zero at exterior surface of the vessel

• However, we ignore the radial stress component for thin-walled vessels, since the limiting assumption of r/t = 10, results in 3 and 2 being 5 and 10 times higher than maximum radial stress (3)max = p

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EXAMPLE 8.1

Cylindrical pressure vessel has an inner diameter of 1.2 m and thickness of 12 mm. Determine the maximum internal pressure it can sustain so that neither its circumferential nor its longitudinal stress component exceeds 140 MPa. Under the same conditions, what is the maximum internal pressure that a similar-size spherical vessel

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EXAMPLE 8.1 (SOLN)

Cylindrical pressure vesselMaximum stress occurs in the circumferential direction. From Eqn 8-1, we have

1 = ;prt 140 N/mm2 =

p(600 mm)12 mm

p = 2.8 N/mm2

Note that when pressure is reached, from Eqn 8-2, stress in the longitudinal direction will be 2 = 0.5(140 MPa) = 70 MPa.

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EXAMPLE 8.1 (SOLN)

Cylindrical pressure vesselFurthermore, maximum stress in the radial direction occurs on the material at the inner wall of the vessel and is (3)max = p = 2.8 MPa. This value is 50 times smaller than the circumferential stress (140 MPa), and as stated earlier, its effects will be neglected.

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EXAMPLE 8.1 (SOLN)

Spherical pressure vesselHere, the maximum stress occurs in any two perpendicular directions on an element of the vessel. From Eqn 8-3, we have

2 = ;pr2t 140 N/mm2 =

p(600 mm)2(12 mm)

p = 5.6 N/mm2

Although it is more difficult to fabricate, the spherical pressure vessel will carry twice as much internal pressure as a cylindrical vessel.

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8.2 STATE OF STRESS CAUSED BY COMBINED LOADINGS In previous chapters, we developed methods for

determining the stress distributions in a member subjected to internal axial forces, shear forces, bending moments, or torsional moments.

Often, the x-section of a member is subjected to several of these type of loadings simultaneously

We can use the method of superposition to determine the resultant stress distribution caused by the loads

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8.2 STATE OF STRESS CAUSED BY COMBINED LOADINGS

Application of the method of superposition1. The stress distribution due to each loading is

determined2. These distributions are superimposed to determine

the resultant stress distributionConditions to satisfy A linear relationship exists between the stress and

the loads Geometry of the member should not undergo

significant change when the loads are applied

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8.2 STATE OF STRESS CAUSED BY COMBINED LOADINGS This is necessary to

ensure that the stress produced by one load is not related to the stress produced by any other load

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8.2 STATE OF STRESS CAUSED BY COMBINED LOADINGS

Procedure for analysisInternal loading Section the member perpendicular to its axis at

the pt where the stress is to be determined Obtain the resultant internal normal and shear

force components and the bending and torsional moment components

Force components should act through the centroid of the x-section, and moment components should be computed about centroidal axes, which represent the principal axes of inertia for x-section

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8.2 STATE OF STRESS CAUSED BY COMBINED LOADINGS

Procedure for analysisAverage normal stress Compute the stress component associated with

each internal loading. For each case, represent the effect either as a

distribution of stress acting over the entire x-sectional area, or show the stress on an element of the material located at a specified pt on the x-section

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8.2 STATE OF STRESS CAUSED BY COMBINED LOADINGS

Procedure for analysisNormal force Internal normal force is developed by a uniform

normal-stress distribution by = P/AShear force Internal shear force in member subjected to

bending is developed from shear-stress distribution determined from the shear formula, = VQ/It. Special care must be exercised as highlighted in section 7.3.

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8.2 STATE OF STRESS CAUSED BY COMBINED LOADINGS

Procedure for analysisBending 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 outer boundary of the member.

Stress distribution obtained from flexure formula, = –My/I.

For curved member, stress distribution is nonlinear and determined from = My/[Ae(R – y)]

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8.2 STATE OF STRESS CAUSED BY COMBINED LOADINGS

Procedure for analysisTorsional moment For circular shafts and tubes, internal torsional

moment is developed by a shear-stress distribution that varies linearly from the central axis of shaft to a maximum at shaft’s outer boundary

Shear-stress distribution is determined from the torsional formula, = T/J.

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

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8.2 STATE OF STRESS CAUSED BY COMBINED LOADINGS

Procedure for analysisThin-walled pressure vessels If vessel is a thin-walled cylinder, internal

pressure p will cause a biaxial state of stress in the material such that the hoop or circumferential stress component is 1 = pr/t and longitudinal 2 = pr/2t.

If vessel is a thin-walled sphere, then biaxial state of stress is represented by two equivalent components, each having a magnitude of 2 = pr/2t

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8.2 STATE OF STRESS CAUSED BY COMBINED LOADINGS

Procedure for analysisSuperposition Once normal and shear stress components for

each loading have been calculated, use the principle of superposition and determine the resultant normal and shear stress components

Represent the results on an element of material located at the pt, or show the results as a distribution of stress acting over the member’s x-sectional area