Combined Loadings Chapter 8

7/29/2019 Combined Loadings Chapter 8

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Chapter Objectives Determine stresses developed in thin-walled pressure

vessels

Determine stresses developed in a members cross

section when axial load, torsion, bending and shear

occur simultaneously.

Copyright 2011 Pearson Education South Asia Pte Ltd


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1. Thin-Walled Pressure Vessels

2. Review of Stress Analyses

In-class Activities



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THIN-WALLED PRESSURE VESSELS


Assumptions:

1. Inner-radius-to-wall-thickness ratio 102. Stress distribution in thin wall is uniform or constant

Cylindrical vessels:

2

:stressalLongitudin

:directionHoop

2

1

t

pr

t

pr


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THIN-WALLED PRESSURE VESSELS (cont)


Spherical vessels:

22

t

pr


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EXAMPLE 1


A cylindrical pressure vessel has an inner diameter of 1.2 m

and a thickness of 12 mm. Determine the maximum internalpressure 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 can sustain?


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EXAMPLE 1 (cont)


The stress in the longitudinal direction will be

The maximum stress in the radial direction occurs on the material at

the inner wall of the vessel and is

The maximum stress occurs in the circumferential direction.

Solutions

(Ans)MPa28N/mm8.212

600140

2

1

p

p

t

pr

MPa701402

12

MPa8.2(max)3 p


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EXAMPLE 1 (cont)


The maximum stress occurs in any two perpendicular directions on an

element of the vessel is

Solutions

(Ans)MPa5.6N/mm6.5

122

600

140

2

2

2

p

p

t

pr


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REVIEW OF STRESS ANALYSES


Normal forceP leads to:

Shear force V leads to:

Bending moment M leads to:

A

Pstressnormaluniform ,

It

VQondistributistressshear ,

beam)curved(foror

beam)straight(for,

yRAe

MyI

Myondistributistressallongitudin


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REVIEW OF STRESS ANALYSES (cont)


Torsional moment T leads to:

Resultant stresses by superposition:

Once the normal and shear stress components for each

loading have been calculated, use the principal of

superposition to determine the resultant normal and shearstress components.

tube)walled-thinclosed(for

2

shaft)circular(for,

tA

T

JTondistributistressshear

m


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EXAMPLE 2


A force of 15 kN is applied to the edge of the member shown in

Fig. 83a. Neglect the weight of the member and determine the

state of stress at points B and C.


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EXAMPLE 2 (cont)


For equilibrium at the section there must be an axial force of 15 000 N

acting through the centroid and a bending moment of 750 000 Nmm

about the centroidal or principal axis.

The maximum stress is

Solutions

MPa75.3

40100

15000

A

P

MPa25.1110040

12

1

5075000

3max I

Mc


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EXAMPLE 2 (cont)


The location of the line of zero stress can be determined by proportional

triangles

Elements of material at B and Care subjected only to normal oruniaxialstress.

Solutions

mm3.33

100

1575

x

xx

(Ans)on)(compressiMPa15

(Ans)(tension)MPa5.7

C

B


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EXAMPLE 3


The tank in Fig. 84a has an inner radius of 600 mm and athickness of 12 mm. It is filled to the top with water having a

specific weight ofw = 10 kNm3. If it is made of steel having aspecific weight ofst = 78 kNm

3, determine the state of stress atpointA. The tank is open at the top.


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EXAMPLE 3 (cont)


The weight of the tank is

The pressure on the tank at levelA is

For circumferential and longitudinal stress, we have

Solutions

kN56.311000

600

1000

61278

22

ststst VW

kPa10110 zp w

(Ans)kPa9.77

56.3

(Ans)kPa50010

2

10006002

1000612

2

100012

1000600

1

st

st

A

W

t

pr


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EXAMPLE 4


The member shown in Fig. 85a has a rectangular cross

section. Determine the state of stress that the loading produces

at point C.


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EXAMPLE 4 (cont)


The resultant internal loadings at the section consist of a normal force, a

shear force, and a bending moment.

Solving,

Solutions

kN89.32kN,21.93kN,45.16 MVN


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EXAMPLE 4 (cont)


The uniform normal-stress distribution acting over the cross section is

produced by the normal force.

At Point C,

In Fig. 85e, the shear stress is zero.

Solutions

MPa32.125.005.0

1045.16 3

A

Pc


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EXAMPLE 4 (cont)


Point Cis located at y = c = 0.125m from the neutral axis, so the normal

stress at C, Fig. 85f, is

Solutions

MPa16.6325.005.0

125.01089.323

21

3

I

Mcc


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EXAMPLE 4 (cont)


The shear stress is zero.

Adding the normal stresses determined above gives a compressive

stress at Chaving a value of

Solutions

MPa5.6416.6332.1 I

Mcc


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EXAMPLE 5


The rectangular block of negligible weight in Fig. 86a is

subjected to a vertical force of 40 kN, which is applied to its

corner. Determine the largest normal stress acting on a section

throughABCD.


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EXAMPLE 5 (cont)


For uniform normal-stress distribution the stress is

For 8 kN, the maximum stress is

For 16 kN, the maximum stress is

Solutions

kPa125

4.08.0

40

A

P

kPa3754.08.0

2.083

121max

x

xx

I

cM

kPa3758.04.0

4.0163

121max

y

xy

I

cM


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EXAMPLE 5 (cont)


By inspection the normal stress at point Cis the largest since each

loading creates a compressive stress there

Solutions

(Ans)kPa875375375125 c


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EXAMPLE 6



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EXAMPLE 5 (cont)



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EXAMPLE 5 (cont)



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EXAMPLE 5 (cont)



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EXAMPLE 5 (cont)



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EXAMPLE 5 (cont)

Combined Loadings Chapter 8

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