36

Chapter4

AXIALLY LOADED MEMBERS:4.1 Introduction4.2 Deformation of Axially Loaded Members4.3 Statically Indeterminate Structures4.4 Method of Superposition4.5 Thermal Deformation and Stress4.6 Stresses on Inclined Planes4.7 Saint-Venant’s Principle4.8 Stress Concentrations

4.1 Introduction:• As we learn in chapter 2, which introduced the concepts of axial stress.

Additionally, in chapter 3 the deformation and strain concept are considered

and so the deflections are assumed to be small.

• Suitability of a structure or machine may depend on the deformations in the

structure as well as the stresses induced under loading. Statics analyses alone

are not sufficient.

• Considering structures as deformable allows determination of member forces

and reactions which are statically indeterminate.

• Determination of the stress distribution within a member also requires

consideration of deformations in the member.

• Chapter 4 is concerned with deformation of axial loaded member (structural

components subjected only to tension or compression, such as trusses,

connecting rods, columns, etc.). Change in length for prismatic bars,

nonuniform bars are determined, it will be used to solve the statically

indeterminate structures, change in length by thermal effect is also considered

stresses on inclined sections will be calculated.

4.2 Deformation of Axially Loaded Members:1)Prismatic bars: As mentioned in Chapter 1, a

prismatic bar is straight structural member having the

same (or constant) cross section throughout its length.

Consider the deformation of a prismatic bar subjected

to load P (tension) shown in Figure 4.1. The load P

produces uniform elongation δ in the bar, which is

said to be in tension. How can we determine δ?

Make use of Hooke’s law and expressions for normal stress and strain as,

i.Geometry of Deformation: normal strain(ε), )a....(L

Figure 4.1: (a) elongationof the prismatic bar, and(b) freebody diagram of apart.

36

Chapter4

AXIALLY LOADED MEMBERS:4.1 Introduction4.2 Deformation of Axially Loaded Members4.3 Statically Indeterminate Structures4.4 Method of Superposition4.5 Thermal Deformation and Stress4.6 Stresses on Inclined Planes4.7 Saint-Venant’s Principle4.8 Stress Concentrations

4.1 Introduction:• As we learn in chapter 2, which introduced the concepts of axial stress.

Additionally, in chapter 3 the deformation and strain concept are considered

and so the deflections are assumed to be small.

• Suitability of a structure or machine may depend on the deformations in the

structure as well as the stresses induced under loading. Statics analyses alone

are not sufficient.

• Considering structures as deformable allows determination of member forces

and reactions which are statically indeterminate.

• Determination of the stress distribution within a member also requires

consideration of deformations in the member.

• Chapter 4 is concerned with deformation of axial loaded member (structural

components subjected only to tension or compression, such as trusses,

connecting rods, columns, etc.). Change in length for prismatic bars,

nonuniform bars are determined, it will be used to solve the statically

indeterminate structures, change in length by thermal effect is also considered

stresses on inclined sections will be calculated.

4.2 Deformation of Axially Loaded Members:1)Prismatic bars: As mentioned in Chapter 1, a

prismatic bar is straight structural member having the

same (or constant) cross section throughout its length.

Consider the deformation of a prismatic bar subjected

to load P (tension) shown in Figure 4.1. The load P

produces uniform elongation δ in the bar, which is

said to be in tension. How can we determine δ?

Make use of Hooke’s law and expressions for normal stress and strain as,

i.Geometry of Deformation: normal strain(ε), )a....(L

Figure 4.1: (a) elongationof the prismatic bar, and(b) freebody diagram of apart.

36

Chapter4

AXIALLY LOADED MEMBERS:4.1 Introduction4.2 Deformation of Axially Loaded Members4.3 Statically Indeterminate Structures4.4 Method of Superposition4.5 Thermal Deformation and Stress4.6 Stresses on Inclined Planes4.7 Saint-Venant’s Principle4.8 Stress Concentrations

4.1 Introduction:• As we learn in chapter 2, which introduced the concepts of axial stress.

Additionally, in chapter 3 the deformation and strain concept are considered

and so the deflections are assumed to be small.

• Suitability of a structure or machine may depend on the deformations in the

structure as well as the stresses induced under loading. Statics analyses alone

are not sufficient.

• Considering structures as deformable allows determination of member forces

and reactions which are statically indeterminate.

• Determination of the stress distribution within a member also requires

consideration of deformations in the member.

• Chapter 4 is concerned with deformation of axial loaded member (structural

components subjected only to tension or compression, such as trusses,

connecting rods, columns, etc.). Change in length for prismatic bars,

nonuniform bars are determined, it will be used to solve the statically

indeterminate structures, change in length by thermal effect is also considered

stresses on inclined sections will be calculated.

4.2 Deformation of Axially Loaded Members:1)Prismatic bars: As mentioned in Chapter 1, a

prismatic bar is straight structural member having the

same (or constant) cross section throughout its length.

Consider the deformation of a prismatic bar subjected

to load P (tension) shown in Figure 4.1. The load P

produces uniform elongation δ in the bar, which is

said to be in tension. How can we determine δ?

Make use of Hooke’s law and expressions for normal stress and strain as,

i.Geometry of Deformation: normal strain(ε), )a....(L

Figure 4.1: (a) elongationof the prismatic bar, and(b) freebody diagram of apart.

37

ii.Condition of Equilibrium: normal stress )b.........(A

P

iii.Material Behaviour: Hooke’s law (linearly elastic) )c.........(E

• Substituting Equs. (a) and (b) into Equ. (c) A

PL

L

A

P

E

)1.4.........(AE

PL , (for both tension and compression).

EA is known as axial rigidity.

The change in length of a member is normally very small when compared to its length.

The stiffness k and flexibility f of a prismatic bar are defined in the same way as for a spring. Since

Stiffness k = force required to produce a unit elongation (k = P/), while

Flexibility f = elongation produced by a unit load (f = / P).

What is the stiffness k and flexibility f for a prismatic bar?

Working:

Stiffness P

k

Now useL

AEP

AE

PL

)2.4(..........L

EAPk

Flexibility P

f

Now usingAE

L

PAE

PL

)3.4..(..........AE

L

Pf

, (for both tension and compression)

Again, stiffness k and flexibility f are reciprocal to each other as,

f

1k , and

k

1f

2) Multiple Prismatic bars: As mentioned earlier, a prismatic bar of linearly elastic material is loaded only at

the ends Its change in length can be obtained from = PL / EA.

Figure 4.2 (a) A prismatic bar is loaded by one or more axial loads acting at intermediate points. Change in

length of this bar can be determined by adding algebraically the elongations and shortenings of the

individual segments. The method to solve this problem:

38

Figure 4.2 : (a) Bar with external loads

acting at intermediate points; (b) (c), and (d)

free-body diagrams showing the internal axial

forces N1, N2, and N3.

part.

Figure 4.3 Stepped bar with multipleloadings.

Figure 4.4: (a) Bar with varyingcross-sectional area and varyingaxial force.

(1) Identify the individual segments.

(2) Cut each segment and consider the free-body diagram.

(3) Determine the internal axial forces Ni, where i = number

of individual segments. Ni can be determined using

equilibrium (Fv = 0). Ni Tension (+ve) and compression

(ve).

(4) Determine the changes in lengths (δi) of each segment as :

iAiEiLiN

i , where : Li = Length of each segment, Ei =

Young’s modulus of each segment, and Ai = Cross section area

of each segment.

(5) Obtain the total changes in lengths () of the entire bar by

adding i of each segment:

n

1i)4.4(..........

iAiEiLiNn

1i i ,

Where n : Total number of segments. = Elongations (+ve)

and shortenings (ve).

This equation is true for linearly elastic materials.

Ni not an external load, but the internal axial force.

This method can be used when a bar consists of several

prismatic segments and having different axial forces with different dimensions and materials, as shown Figure

4.3.

Non Prismatic Bars: Are bars with continuously varying loads

or dimensions, (tapered bar) as shown in Figure 4. 4 Figure 4.

4 (a) The load consists of two parts:

(1) Single force PB acting at the end B.

(2) Distributed forces p(x) acting along the axis.

Since the bar with continuously varying loads and dimensions,

38

©2001 Brooks/Cole, a division of Thomson Learning, Inc. Thomson Learning™ is a trademark used herein under license.©2001 Brooks/Cole, a division of Thomson Learning, Inc. Thomson Learning™ is a trademark used herein under license.

Figure 4.2 : (a) Bar with external loads

acting at intermediate points; (b) (c), and (d)

free-body diagrams showing the internal axial

forces N1, N2, and N3.

part.

Figure 4.3 Stepped bar with multipleloadings.

Figure 4.4: (a) Bar with varyingcross-sectional area and varyingaxial force.

(1) Identify the individual segments.

(2) Cut each segment and consider the free-body diagram.

(3) Determine the internal axial forces Ni, where i = number

of individual segments. Ni can be determined using

equilibrium (Fv = 0). Ni Tension (+ve) and compression

(ve).

(4) Determine the changes in lengths (δi) of each segment as :

iAiEiLiN

i , where : Li = Length of each segment, Ei =

Young’s modulus of each segment, and Ai = Cross section area

of each segment.

(5) Obtain the total changes in lengths () of the entire bar by

adding i of each segment:

n

1i)4.4(..........

iAiEiLiNn

1i i ,

Where n : Total number of segments. = Elongations (+ve)

and shortenings (ve).

This equation is true for linearly elastic materials.

Ni not an external load, but the internal axial force.

This method can be used when a bar consists of several

prismatic segments and having different axial forces with different dimensions and materials, as shown Figure

4.3.

Non Prismatic Bars: Are bars with continuously varying loads

or dimensions, (tapered bar) as shown in Figure 4. 4 Figure 4.

4 (a) The load consists of two parts:

(1) Single force PB acting at the end B.

(2) Distributed forces p(x) acting along the axis.

Since the bar with continuously varying loads and dimensions,

38

©2001 Brooks/Cole, a division of Thomson Learning, Inc. Thomson Learning™ is a trademark used herein under license.©2001 Brooks/Cole, a division of Thomson Learning, Inc. Thomson Learning™ is a trademark used herein under license.

Figure 4.2 : (a) Bar with external loads

acting at intermediate points; (b) (c), and (d)

free-body diagrams showing the internal axial

forces N1, N2, and N3.

part.

Figure 4.3 Stepped bar with multipleloadings.

Figure 4.4: (a) Bar with varyingcross-sectional area and varyingaxial force.

(1) Identify the individual segments.

(2) Cut each segment and consider the free-body diagram.

(3) Determine the internal axial forces Ni, where i = number

of individual segments. Ni can be determined using

equilibrium (Fv = 0). Ni Tension (+ve) and compression

(ve).

(4) Determine the changes in lengths (δi) of each segment as :

iAiEiLiN

i , where : Li = Length of each segment, Ei =

Young’s modulus of each segment, and Ai = Cross section area

of each segment.

(5) Obtain the total changes in lengths () of the entire bar by

adding i of each segment:

n

1i)4.4(..........

iAiEiLiNn

1i i ,

Where n : Total number of segments. = Elongations (+ve)

and shortenings (ve).

This equation is true for linearly elastic materials.

Ni not an external load, but the internal axial force.

This method can be used when a bar consists of several

prismatic segments and having different axial forces with different dimensions and materials, as shown Figure

4.3.

Non Prismatic Bars: Are bars with continuously varying loads

or dimensions, (tapered bar) as shown in Figure 4. 4 Figure 4.

4 (a) The load consists of two parts:

(1) Single force PB acting at the end B.

(2) Distributed forces p(x) acting along the axis.

Since the bar with continuously varying loads and dimensions,

39

Figure 4.4: (b) Bar with varying cross-sectional area and varying axial force,and (c) free body of an element.

therefore, the change in length cannot be obtained using equ.(4.4).

However, we can determine the change in length of a differential element of the bar and then integrate over the

length of the bar.

Figure 4.4 (a), select a differential element at distance x from the left-hand end of the bar.

Figure 4.4 (b) and (c) Internal axial force N(x) acting at this cross section can be determined from

equilibrium using either segment AC or CB as free body diagram.

The internal axial force N(x) and cross section area A(x) Function of distance x.

For differential element (or extremely small element), the

elongation d may be obtained using:

EA

PL

Now substitute N(x) for P, dx for L, and A(x) for A, hence,

dx)x(EA

)x(Nd

EA

PL

The elongation of the entire bar can be obtained using

integration over the entire length:

)5.4.(..........dx)x(EA

)x(Nd

L

0

L

0

This equation is true for linearly elastic materials.

The Procedure for analysis can be summarised as:

1) Internal force:

Use method of sections to determine internal axial force P in the member.

If the force varies along member’s strength, section made at the arbitrary location x from one end of

member and force represented as a function of x, i.e., P(x).

If several constant external forces act on member, internal force in each segment, between two external

forces, must then be determined.

For any segment, internal tensile force is positive and internal compressive force is negative. Results of

loading can be shown graphically by constructing the normal-force diagram.

2) Displacement:

When member’s x-sectional area varies along its axis, the area should be expressed as a function of its

position x, i.e., A(x).

39

Figure 4.4: (b) Bar with varying cross-sectional area and varying axial force,and (c) free body of an element.

therefore, the change in length cannot be obtained using equ.(4.4).

However, we can determine the change in length of a differential element of the bar and then integrate over the

length of the bar.

Figure 4.4 (a), select a differential element at distance x from the left-hand end of the bar.

Figure 4.4 (b) and (c) Internal axial force N(x) acting at this cross section can be determined from

equilibrium using either segment AC or CB as free body diagram.

The internal axial force N(x) and cross section area A(x) Function of distance x.

For differential element (or extremely small element), the

elongation d may be obtained using:

EA

PL

Now substitute N(x) for P, dx for L, and A(x) for A, hence,

dx)x(EA

)x(Nd

EA

PL

The elongation of the entire bar can be obtained using

integration over the entire length:

)5.4.(..........dx)x(EA

)x(Nd

L

0

L

0

This equation is true for linearly elastic materials.

The Procedure for analysis can be summarised as:

1) Internal force:

Use method of sections to determine internal axial force P in the member.

If the force varies along member’s strength, section made at the arbitrary location x from one end of

member and force represented as a function of x, i.e., P(x).

If several constant external forces act on member, internal force in each segment, between two external

forces, must then be determined.

For any segment, internal tensile force is positive and internal compressive force is negative. Results of

loading can be shown graphically by constructing the normal-force diagram.

2) Displacement:

When member’s x-sectional area varies along its axis, the area should be expressed as a function of its

position x, i.e., A(x).

39

Figure 4.4: (b) Bar with varying cross-sectional area and varying axial force,and (c) free body of an element.

therefore, the change in length cannot be obtained using equ.(4.4).

However, we can determine the change in length of a differential element of the bar and then integrate over the

length of the bar.

Figure 4.4 (a), select a differential element at distance x from the left-hand end of the bar.

Figure 4.4 (b) and (c) Internal axial force N(x) acting at this cross section can be determined from

equilibrium using either segment AC or CB as free body diagram.

The internal axial force N(x) and cross section area A(x) Function of distance x.

For differential element (or extremely small element), the

elongation d may be obtained using:

EA

PL

Now substitute N(x) for P, dx for L, and A(x) for A, hence,

dx)x(EA

)x(Nd

EA

PL

The elongation of the entire bar can be obtained using

integration over the entire length:

)5.4.(..........dx)x(EA

)x(Nd

L

0

L

0

This equation is true for linearly elastic materials.

The Procedure for analysis can be summarised as:

1) Internal force:

Use method of sections to determine internal axial force P in the member.

If the force varies along member’s strength, section made at the arbitrary location x from one end of

member and force represented as a function of x, i.e., P(x).

If several constant external forces act on member, internal force in each segment, between two external

forces, must then be determined.

For any segment, internal tensile force is positive and internal compressive force is negative. Results of

loading can be shown graphically by constructing the normal-force diagram.

2) Displacement:

When member’s x-sectional area varies along its axis, the area should be expressed as a function of its

position x, i.e., A(x).

40

If x-sectional area, modulus of elasticity, or internal loading suddenly changes, then Eqn 4.4 should be

applied to each segment for which the quantity is constant.

When substituting data into equations, account for proper sign for P, tensile loadings +ve, compressive

−ve. Use consistent set of units. If result is +ve, elongation occurs, −ve means it’s a contraction.

Example 4.1:Given: Aluminium [E = 70 GPa] member ABC supports a load of 28 kN, as shown in

Figure 4.5.

Find:

(a) The value of load P such that the deflection of joint C is zero.

(b) The corresponding deflection of joint B.

SOLUTION:

41

Example 4.2:

Given: A slightly tapered bar AB of solid circular cross section and length L (Fig.ure 4. 6 a) is supported at end B

and subjected to a tensile load P at the free end A. The

diameters of the bar at ends A and B are dA and dB

respectively.

Find: The elongation of the bar due to the load P.

SOLUTION:

©2001 Brooks/Cole, a division of Thomson Learning, Inc. Thomson Learning™ is a trademark used herein under license.©2001 Brooks/Cole, a division of Thomson Learning, Inc. Thomson Learning™ is a trademark used herein under license.

©2001 Brooks/Cole, a division of Thomson Learning, Inc. Thomson Learning™ is a trademark used herein under license.©2001 Brooks/Cole, a division of Thomson Learning, Inc. Thomson Learning™ is a trademark used herein under license.

42

Example 4.3:Given: Composite A-36 steel bar (Est = 210 GPa) shown in Figure 4.7 (a) is made from two

segments AB and BD. Area AAB = 600 mm2 and ABD = 1200 mm2.

Find:

1) The vertical displacement of end A?

2) Displacement of B relative to C?

SOLUTION: Due to external loadings, internal axial forces in regions AB, BC and CD are

different. Apply method of sections and equation of vertical force equilibrium as shown.

Variation is also plotted.

42

Example 4.3:Given: Composite A-36 steel bar (Est = 210 GPa) shown in Figure 4.7 (a) is made from two

segments AB and BD. Area AAB = 600 mm2 and ABD = 1200 mm2.

Find:

1) The vertical displacement of end A?

2) Displacement of B relative to C?

SOLUTION: Due to external loadings, internal axial forces in regions AB, BC and CD are

different. Apply method of sections and equation of vertical force equilibrium as shown.

Variation is also plotted.

42

Example 4.3:Given: Composite A-36 steel bar (Est = 210 GPa) shown in Figure 4.7 (a) is made from two

segments AB and BD. Area AAB = 600 mm2 and ABD = 1200 mm2.

Find:

1) The vertical displacement of end A?

2) Displacement of B relative to C?

SOLUTION: Due to external loadings, internal axial forces in regions AB, BC and CD are

different. Apply method of sections and equation of vertical force equilibrium as shown.

Variation is also plotted.

43

Example 4.4:Given: A hollow steel [E = 30,000 ksi] tube (1) with an outsidediameter of 2.75 in. and a wall thickness of 0.25 in. is fastened to asolid aluminum [E = 10,000 ksi] rod (2) that has a 2in.-diameter anda solid 1.375-in.-diameter aluminum rod (3). The bar is loaded asshown in Figure 4.8.Find:(a) the change in length of steel tube (1).(b) the deflection of joint D with respect to the fixed support at A.(c) the maximum normal stress in the entire axial assembly.

SOLUTION:

43

Example 4.4:Given: A hollow steel [E = 30,000 ksi] tube (1) with an outsidediameter of 2.75 in. and a wall thickness of 0.25 in. is fastened to asolid aluminum [E = 10,000 ksi] rod (2) that has a 2in.-diameter anda solid 1.375-in.-diameter aluminum rod (3). The bar is loaded asshown in Figure 4.8.Find:(a) the change in length of steel tube (1).(b) the deflection of joint D with respect to the fixed support at A.(c) the maximum normal stress in the entire axial assembly.

SOLUTION:

43

Example 4.4:Given: A hollow steel [E = 30,000 ksi] tube (1) with an outsidediameter of 2.75 in. and a wall thickness of 0.25 in. is fastened to asolid aluminum [E = 10,000 ksi] rod (2) that has a 2in.-diameter anda solid 1.375-in.-diameter aluminum rod (3). The bar is loaded asshown in Figure 4.8.Find:(a) the change in length of steel tube (1).(b) the deflection of joint D with respect to the fixed support at A.(c) the maximum normal stress in the entire axial assembly.

SOLUTION:

44

4.3 Statically Indeterminate Structures:

Figure 4.9 statically indeterminate bars.

Figure 4.10 (a) Analysis of a staticallyindeterminate bar.

Statically determinate Reactions and internal forces (axial,

shear and moment) can be determined only from the 3

equations of equilibrium:

0vF ; 0hF ; 0M

Statically determinate Unknown forces can be determined

without knowing the properties of the materials (such as the

Young’s modulus E).

Statically indeterminate Reactions and internal forces

cannot be found by statics alone (the 3 equations of

equilibrium).

Hence, additional equation is needed (equation of

compatibility).

Figure 4.9 statically indeterminate bars. Consider

equilibrium:

Fv = 0

RA +RB-P=0

RA + RB = P

One useful equation (Fv = 0) cannot solve 2 unknowns

(RA, RB).

Figure 4.10 Analysis of a statically indeterminate bar.

Prismatic bar AB Attached to rigid supports at both ends

and is axially loaded by a force P at point C.

How to determine RA and RB?

As mentioned earlier, one equation (Fv = 0) cannot solve

the 2 unknowns (RA, RB).

We need one more equation. How?

Change in length of a bar must be compatible with the

conditions at the supports.

Supports of the bar are both fixed at the ends Total change

in length is zero (AB = 0).

©20

01 B

rook

s/C

ole,

a di

visi

on o

f T

hom

son

Lea

rnin

g,In

c.T

hom

son

Lea

rnin

g ™is

a tr

adem

ark

used

her

ein

unde

r lic

ense

.©

2001

Bro

oks/

Col

e,a

divi

sion

of

Tho

mso

n L

earn

ing,

Inc.

Tho

mso

nL

earn

ing ™

is a

trad

emar

k us

ed h

erei

n un

der

licen

se.

©20

01 B

rook

s/C

ole,

a di

visi

on o

f T

hom

son

Lea

rnin

g,In

c.T

hom

son

Lea

rnin

g ™is

a tr

adem

ark

used

her

ein

unde

r lic

ense

.©

2001

Bro

oks/

Col

e,a

divi

sion

of

Tho

mso

n L

earn

ing,

Inc.

Tho

mso

nL

earn

ing ™

is a

trad

emar

k us

ed h

erei

n un

der

licen

se.

45

Equation of compatibility In this case, the change in length

of the bar must be compatible with the conditions at the

supports (AB = 0).

For linearly elastic materials = Now determine RA and RB:

Working:

Example 4.5:

Figure 4.11 (a) Analysis of a statically

indeterminate structure.

Figures 4.11 (a-b) show a solid circular steel cylinder S is encased

in a hollow circular copper tube C. The cylinder and tube are

compressed between the rigid plates of a testing machine by

compressive forces P. Both parts have length L.

Steel cylinder

As = Cross-sectional area

Es = Modulus of elasticity (Young’s modulus)

Copper tube

Ac = Cross-sectional area

Ec = Modulus of elasticity (Young’s modulus)

Determine the following:

(a) Compressive forces Ps in the steel cylinder and Pc in the

copper tube.

©2001 Brooks/Cole, a division of Thomson Learning, Inc. Thomson Learning™ is a trademark used herein under license.©2001 Brooks/Cole, a division of Thomson Learning, Inc. Thomson Learning™ is a trademark used herein under license.

46

(b) Compressive stresses s in the steel cylinder and c in the

copper tube.

(c) Shortening of the assembly.

SOLUTION:

©2001 Brooks/Cole, a division of Thomson Learning, Inc. Thomson Learning™ is a trademark used herein under license.©2001 Brooks/Cole, a division of Thomson Learning, Inc. Thomson Learning™ is a trademark used herein under license.

47

4.4 Method of Superposition• After subdividing the load into components, the principle of superposition states that the resultant

stress or displacement at the point can be determined by first finding the stress or displacement causedby each component load acting separately on the member.

• Resultant stress/displacement determined algebraically by adding the contributions of each component.

Conditions:

1. The loading must be linearly related to the stress or displacement that is to be determined.

2. The loading must not significantly change the original geometry or configuration of the member

When to ignore deformations?

3. Most loaded members will produce deformations so small that change in position and direction ofloading will be insignificant and can be neglected

4. Exception to this rule is a column carrying axial load, discussed in Later.

Superposition principle procedure:

47

4.4 Method of Superposition• After subdividing the load into components, the principle of superposition states that the resultant

stress or displacement at the point can be determined by first finding the stress or displacement causedby each component load acting separately on the member.

• Resultant stress/displacement determined algebraically by adding the contributions of each component.

Conditions:

1. The loading must be linearly related to the stress or displacement that is to be determined.

2. The loading must not significantly change the original geometry or configuration of the member

When to ignore deformations?

3. Most loaded members will produce deformations so small that change in position and direction ofloading will be insignificant and can be neglected

4. Exception to this rule is a column carrying axial load, discussed in Later.

Superposition principle procedure:

47

4.4 Method of Superposition• After subdividing the load into components, the principle of superposition states that the resultant

stress or displacement at the point can be determined by first finding the stress or displacement causedby each component load acting separately on the member.

• Resultant stress/displacement determined algebraically by adding the contributions of each component.

Conditions:

1. The loading must be linearly related to the stress or displacement that is to be determined.

2. The loading must not significantly change the original geometry or configuration of the member

When to ignore deformations?

3. Most loaded members will produce deformations so small that change in position and direction ofloading will be insignificant and can be neglected

4. Exception to this rule is a column carrying axial load, discussed in Later.

Superposition principle procedure:

48

3.5 Thermal Stress:Most engineering materials expand when heated and contract when cooled. The strain due to a

temperature change of ΔT is called thermal strain and is obtained by:= ∆ , = ∆ ………… . (3.6)Where : ℎ , : ℎ ,: ℎ, : ℎ .Total Strains: strains caused by temperature changes and strains caused by applied loads are

essentially independent.the total normal strain in a body acted on both temperature changes and applied

load is given by: = + ……………(3.7)Example 4.6:

Given: At a temperature of 40°F, a 0.08-in. gap exists

between the ends of the two bars shown in Figure

4.12. Bar (1) is an aluminum alloy [α = 12.5 ×

10−6/°F] and bar (2) is s ta in l ess s tee l [α = 9.6 ×

10−6/°F]. The supports at A and C are rigid.

Find: The lowest temperature at which the two bars

contact each other?

SOLUTION:

48

3.5 Thermal Stress:Most engineering materials expand when heated and contract when cooled. The strain due to a

temperature change of ΔT is called thermal strain and is obtained by:= ∆ , = ∆ ………… . (3.6)Where : ℎ , : ℎ ,: ℎ, : ℎ .Total Strains: strains caused by temperature changes and strains caused by applied loads are

essentially independent.the total normal strain in a body acted on both temperature changes and applied

load is given by: = + ……………(3.7)Example 4.6:

Given: At a temperature of 40°F, a 0.08-in. gap exists

between the ends of the two bars shown in Figure

4.12. Bar (1) is an aluminum alloy [α = 12.5 ×

10−6/°F] and bar (2) is s ta in l ess s tee l [α = 9.6 ×

10−6/°F]. The supports at A and C are rigid.

Find: The lowest temperature at which the two bars

contact each other?

SOLUTION:

48

3.5 Thermal Stress:Most engineering materials expand when heated and contract when cooled. The strain due to a

temperature change of ΔT is called thermal strain and is obtained by:= ∆ , = ∆ ………… . (3.6)Where : ℎ , : ℎ ,: ℎ, : ℎ .Total Strains: strains caused by temperature changes and strains caused by applied loads are

essentially independent.the total normal strain in a body acted on both temperature changes and applied

load is given by: = + ……………(3.7)Example 4.6:

Given: At a temperature of 40°F, a 0.08-in. gap exists

between the ends of the two bars shown in Figure

4.12. Bar (1) is an aluminum alloy [α = 12.5 ×

10−6/°F] and bar (2) is s ta in l ess s tee l [α = 9.6 ×

10−6/°F]. The supports at A and C are rigid.

Find: The lowest temperature at which the two bars

contact each other?

SOLUTION:

49

Example 4.7:

50

2.5 Stresses on Inclined Sections

• Consider the bar loaded as shown in Figure 4.13.

• Pass a section through the member forming an angle q with

the normal plane.

• From equilibrium conditions, the distributed forces

(stresses) on the plane must be equivalent to the force P.

• Resolve P into components normal and tangential to the

oblique section,

)8.......(..........sinPP,&cosPPyx

• The average normal and shear stresses on the oblique

plane are:

)9......(..........cossinAP

cosA

sinPA

P

2cosAP

cosA

cosPA

P

x

y

x

x

yx

x

• The maximum normal stress occurs when the reference plane is

perpendicular to the member axis,

)10(..........0&A

P

0max

• The maximum shear stress occurs for a plane at + 45o withrespect to the axis,

2A2P

45cos45sinAP

max ………..(11)

50

2.5 Stresses on Inclined Sections

• Consider the bar loaded as shown in Figure 4.13.

• Pass a section through the member forming an angle q with

the normal plane.

• From equilibrium conditions, the distributed forces

(stresses) on the plane must be equivalent to the force P.

• Resolve P into components normal and tangential to the

oblique section,

)8.......(..........sinPP,&cosPPyx

• The average normal and shear stresses on the oblique

plane are:

)9......(..........cossinAP

cosA

sinPA

P

2cosAP

cosA

cosPA

P

x

y

x

x

yx

x

• The maximum normal stress occurs when the reference plane is

perpendicular to the member axis,

)10(..........0&A

P

0max

• The maximum shear stress occurs for a plane at + 45o withrespect to the axis,

2A2P

45cos45sinAP

max ………..(11)

50

2.5 Stresses on Inclined Sections

• Consider the bar loaded as shown in Figure 4.13.

• Pass a section through the member forming an angle q with

the normal plane.

• From equilibrium conditions, the distributed forces

(stresses) on the plane must be equivalent to the force P.

• Resolve P into components normal and tangential to the

oblique section,

)8.......(..........sinPP,&cosPPyx

• The average normal and shear stresses on the oblique

plane are:

)9......(..........cossinAP

cosA

sinPA

P

2cosAP

cosA

cosPA

P

x

y

x

x

yx

x

• The maximum normal stress occurs when the reference plane is

perpendicular to the member axis,

)10(..........0&A

P

0max

• The maximum shear stress occurs for a plane at + 45o withrespect to the axis,

2A2P

45cos45sinAP

max ………..(11)

51

Example 4.8:

Given: An axial load P is applied to the 1.75 in. by 0.75 in. rectangular bar

shown in Fig.ure 4.15. The bar is subjected to an axial loadof P= 18 kips.

Find: The normal stress perpendicular to plane AB and the shear stress

parallel to plane AB?

Assumptions: the bar weight is neglected.

SOLUTION:

4.7 Saint-Venant’s Principle:• Localized deformation occurs at each end, and the deformations

decrease as measurements are taken further away from the ends

as shown Figure 4.15.

• At section c-c, stress reaches almost uniform value as compared

to a-a, b-b.

• c-c is sufficiently far enough away from P so that localized

deformation “vanishes”, i.e., minimum distance.

• General rule: min. distance is at least equal to largest dimension

of loaded x-section. For the bar, the min. distance is equal to

width of bar.

• This behavior discovered by Barré de Saint-Venant in 1855, this

name of the principle.

• Saint-Venant Principle states that localized effects caused by any

load acting on the body, will dissipate/smooth out within regions

that are sufficiently removed from location of load. Saint-

Venant’s Principle: Stress distribution may be assumed

51

Example 4.8:

Given: An axial load P is applied to the 1.75 in. by 0.75 in. rectangular bar

shown in Fig.ure 4.15. The bar is subjected to an axial loadof P= 18 kips.

Find: The normal stress perpendicular to plane AB and the shear stress

parallel to plane AB?

Assumptions: the bar weight is neglected.

SOLUTION:

4.7 Saint-Venant’s Principle:• Localized deformation occurs at each end, and the deformations

decrease as measurements are taken further away from the ends

as shown Figure 4.15.

• At section c-c, stress reaches almost uniform value as compared

to a-a, b-b.

• c-c is sufficiently far enough away from P so that localized

deformation “vanishes”, i.e., minimum distance.

• General rule: min. distance is at least equal to largest dimension

of loaded x-section. For the bar, the min. distance is equal to

width of bar.

• This behavior discovered by Barré de Saint-Venant in 1855, this

name of the principle.

• Saint-Venant Principle states that localized effects caused by any

load acting on the body, will dissipate/smooth out within regions

that are sufficiently removed from location of load. Saint-

Venant’s Principle: Stress distribution may be assumed

51

Example 4.8:

Given: An axial load P is applied to the 1.75 in. by 0.75 in. rectangular bar

shown in Fig.ure 4.15. The bar is subjected to an axial loadof P= 18 kips.

Find: The normal stress perpendicular to plane AB and the shear stress

parallel to plane AB?

Assumptions: the bar weight is neglected.

SOLUTION:

4.7 Saint-Venant’s Principle:• Localized deformation occurs at each end, and the deformations

decrease as measurements are taken further away from the ends

as shown Figure 4.15.

• At section c-c, stress reaches almost uniform value as compared

to a-a, b-b.

• c-c is sufficiently far enough away from P so that localized

deformation “vanishes”, i.e., minimum distance.

• General rule: min. distance is at least equal to largest dimension

of loaded x-section. For the bar, the min. distance is equal to

width of bar.

• This behavior discovered by Barré de Saint-Venant in 1855, this

name of the principle.

• Saint-Venant Principle states that localized effects caused by any

load acting on the body, will dissipate/smooth out within regions

that are sufficiently removed from location of load. Saint-

Venant’s Principle: Stress distribution may be assumed

52

independent of the mode of load application except in the

immediate vicinity of load application points.

• Thus, no need to study stress distributions at that points near

application loads or support reactions.

4.8 Stress Concentrations:Force equilibrium requires magnitude of resultant force

developed by the stress distribution to be equal to P. In

other words, =This integral represents graphically the volume under each

of the stress-distribution diagrams shown.

In engineering practice, actual stress distribution not needed,

only maximum stress at these sections must be known.

Member is designed to resist this stress when axial load P is

applied.

K is defined as a ratio of the maximum stress to the

nominal (average) stress acting at the smallest cross

section: =K is independent of the bar’s geometry and the type of

discontinuity, only on the bar’s geometry and the type of

discontinuity.

As size r of the discontinuity is decreased, stress

concentration is increased.

It is important to use stress-concentration factors in design

when using brittle materials, but not necessary for ductile

materials

52

independent of the mode of load application except in the

immediate vicinity of load application points.

• Thus, no need to study stress distributions at that points near

application loads or support reactions.

4.8 Stress Concentrations:Force equilibrium requires magnitude of resultant force

developed by the stress distribution to be equal to P. In

other words, =This integral represents graphically the volume under each

of the stress-distribution diagrams shown.

In engineering practice, actual stress distribution not needed,

only maximum stress at these sections must be known.

Member is designed to resist this stress when axial load P is

applied.

K is defined as a ratio of the maximum stress to the

nominal (average) stress acting at the smallest cross

section: =K is independent of the bar’s geometry and the type of

discontinuity, only on the bar’s geometry and the type of

discontinuity.

As size r of the discontinuity is decreased, stress

concentration is increased.

It is important to use stress-concentration factors in design

when using brittle materials, but not necessary for ductile

materials

52

independent of the mode of load application except in the

immediate vicinity of load application points.

• Thus, no need to study stress distributions at that points near

application loads or support reactions.

4.8 Stress Concentrations:Force equilibrium requires magnitude of resultant force

developed by the stress distribution to be equal to P. In

other words, =This integral represents graphically the volume under each

of the stress-distribution diagrams shown.

In engineering practice, actual stress distribution not needed,

only maximum stress at these sections must be known.

Member is designed to resist this stress when axial load P is

applied.

K is defined as a ratio of the maximum stress to the

nominal (average) stress acting at the smallest cross

section: =K is independent of the bar’s geometry and the type of

discontinuity, only on the bar’s geometry and the type of

discontinuity.

As size r of the discontinuity is decreased, stress

concentration is increased.

It is important to use stress-concentration factors in design

when using brittle materials, but not necessary for ductile

materials

53

Stress concentrations also cause failure structural members

or mechanical elements subjected to fatigue loadings.

Example 4.9:

Given: Steel bar shown in next (Figure 4.18) has

allowable stress, σallow = 115 MPa.

Find: largest axial force P that the bar can carry.

SOLUTION:

Because there is a shoulder fillet, stress-concentrating

factor determined using the graph next (Figure 4.19).

Calculating the necessary geometric parameters yields= 1020 = 0.5Also,

ℎ = 4020 = 2Thus, from the graph, K = 1.4.

Average normal stress at smallest x-section,

= 20 × 10 = 0.005 /=115 = 1.4(0.005 )= 16.43(10 ) = 16.43

53

Stress concentrations also cause failure structural members

or mechanical elements subjected to fatigue loadings.

Example 4.9:

Given: Steel bar shown in next (Figure 4.18) has

allowable stress, σallow = 115 MPa.

Find: largest axial force P that the bar can carry.

SOLUTION:

Because there is a shoulder fillet, stress-concentrating

factor determined using the graph next (Figure 4.19).

Calculating the necessary geometric parameters yields= 1020 = 0.5Also,

ℎ = 4020 = 2Thus, from the graph, K = 1.4.

Average normal stress at smallest x-section,

= 20 × 10 = 0.005 /=115 = 1.4(0.005 )= 16.43(10 ) = 16.43

53

Stress concentrations also cause failure structural members

or mechanical elements subjected to fatigue loadings.

Example 4.9:

Given: Steel bar shown in next (Figure 4.18) has

allowable stress, σallow = 115 MPa.

Find: largest axial force P that the bar can carry.

SOLUTION:

Because there is a shoulder fillet, stress-concentrating

factor determined using the graph next (Figure 4.19).

Calculating the necessary geometric parameters yields= 1020 = 0.5Also,

ℎ = 4020 = 2Thus, from the graph, K = 1.4.

Average normal stress at smallest x-section,

= 20 × 10 = 0.005 /=115 = 1.4(0.005 )= 16.43(10 ) = 16.43