Chapter 4 Axialy Loaded Members Final

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Chapter

4

AXIALLY LOADED MEMBERS:4.1 Introduction

4.2 Deor!ation o A"ia##$ Loaded Me!%er&

4.' Statica##$ Indeter!inate Structure&

4.4 Method o Superpo&ition

4.( )her!a# Deor!ation and Stre&&

4.* Stre&&e& on Inc#ined +#ane&

4., Saint-enant/& +rincip#e

4.0 Stre&& Concentration&

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 reuires

consideration of deformations in the member.

• Chapter ! is concerned with deformation of axial loaded member "structural

components sub#ected 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 sol%e the statically

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

stresses on inclined sections will be calculated.

4.2 Deor!ation o A"ia##$ Loaded Me!%er&:

+ri&!atic %ar& : As mentioned in Chapter &, a prismatic

bar is straight structural member ha%ing the same "or

constant) cross section throughout its length.

• Consider the deformation of a prismatic bar sub#ected

to load P "tension$ shown in 'igure !.&. (he load P

produces uniform elongation δ in the bar, which is said to be in tension. )ow can we determine δ*

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

'igure !.&- "a$ elongation

of the prismatic bar, and

"b$ freebody diagram of a



37

. eo!etr$ o Deor!ation: normal strain"$, ⇒

Condition o Eui#i%riu!: normal stress ⇒

. Materia# Beha3iour: )ooe/s law "linearly elastic$ ⇒

Substituting 0us. "a$ and "b$ into 0u. "c$

∴ , "for both tension and compression$.

• EA is nown as axial rigidity.

• (he change in length of a member is normally %ery small when compared to its length.

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

• Stiffness k 1 force reuired to produce a unit elongation (k = P/ δ ), while

• 'lexibility f 1 elongation produced by a unit load (f =δ / P).

hat is the stiffness k and flexibility f for a prismatic bar *

oring-

Stiffness⇒

ow use

'lexibility ⇒

ow using



, , . ., , . .

38

∴ , "for both tension and compression$

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

, and

Mu#tip#e +ri&!atic %ar &: As mentioned earlier, a prismatic bar of linearly elastic material is loaded only at the

ends 9ts change in length can be obtained from δ = PL / EA.

• 'igure !.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. (he method to sol%e this problem-

9dentify the indi%idual segments.

Cut each segment and consider the free;body diagram.

Determine the internal axial forces N i, where i 1 number of

indi%idual segments. N i can be determined using euilibrium

"Σ F v 1 5$. N i (ension "<%e$ and compression "−%e$.

Determine the changes in lengths "δ i$ of each segment as -

, where - Li 1 8ength of each segment, E i 1

=oung/s modulus of each segment, and Ai 1 Cross section area

of each segment.

>btain the total changes in lengths ( δ ) of the entire bar by adding δ i of each segment-

,

here n - (otal number of segments. δ 1 0longations "<%e$ and shortenings "−%e$.

• (his euation is true for linearly elastic materials.

• N i → not an external load, but the internal axial force.

• (his method can be used when a bar consists of se%eral prismatic segments and ha%ing different axial forces



..

. . . .

39

with different dimensions and materials, as shown 'igure

!.3.

• 5on +ri&!atic Bar&- Are bars with continuou&#$

3ar$in6 loads or dimensions, "tapered bar$ as shown in

'igure !. ! 'igure !. ! "a$ → (he load consists of two parts-

"&$ Single force P B acting at

the end B.

"2$ Distributed forces p()

acting along the axis.

• Since the bar with

continuously %arying loads and dimensions, therefore, the change

in length δ cannot be obtained using eu."!.!$.

• )owe%er, we can determine the change in length of a differential element of the bar and then integrate o%er the

length of the bar.

• 'igure !.! "a$, select a differential element at distance from the left;hand end of the bar.

• 'igure !.! "b$ and "c$ → 9nternal axial force N() acting at this cross section can be determined from

euilibrium using either segment A! or !B as free body diagram.

• (he internal axial force N() and cross section area A()→ 'unction of distance .

• 'or differential element "or extremely small element$, the

elongation d δ may be obtained using-

• ow substitute N() for P , d for L" and A() for A" hence,

• (he elongation of the entire bar can be obtained using

integrati#n o%er the entire length-

7i6ure 4.4: 8a 6ar with %arying

cross;sectional area and %arying

axial force.

'igure !.3 Stepped bar with multiple

loadin s.

7i6ure 4.2 : "a$ 6ar with external loads

acting at intermediate points? "b$ "c$, and "d$

free;body diagrams showing the internal axial

forces N 1, N 2, and N 3.

7i6ure 4.4: 8% 6ar with %arying cross;

sectional area and %arying axial force,

and "c$ free body of an element.



40

• (his euation is true for linearly elastic materials.

• (he Procedure for analysis can be summarised as-

1) nternal force!

♦ @se method of sections to determine internal axial force P in the member.

♦ 9f the force %aries along member/s strength, section made at the arbitrary location from one end of

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

♦ 9f se%eral c#nstant eternal f#rces act on member, internal force in each segment , between two external

forces, must then be determined.

♦ 'or any segment, internal tensile force is positi%e and internal compressi%e force is negati%e. esults of

loading can be shown graphically by constructing the normal;force diagram.2) "isplacement!

• hen member/s x;sectional area varies along its axis, the area should be expressed as a function of its

position , i.e., A().

• 9f x;sectional area, modulus of elasticity, or internal loading suddenly changes, then 0n !.! should be

applied to each segment for which the uantity is constant.

• hen substituting data into euations, account for proper sign for P , tensile loadings <%e, compressi%e

B%e. @se consistent set of units. 9f result is <%e, elongation occurs, B%e means it/s a contraction.

E"a!p#e 4.1:

i3en: Aluminium 0 1 5 EFaG member A6C supports a load of 2H , as shown in

'igure !.I.

7ind:

"a$ (he %alue of load F such that the deflection of #oint C is Jero.

"b$ (he corresponding deflection of #oint 6.

SOL9)IO5:



41

E"a!p#e 4.2:

i3en: A slightly tapered bar AB of solid circular cross section and length L "'ig.ure !. K a$ is supported at end B

and sub#ected to a tensile load P at the free end A. (he diameters of the bar at ends A and B are d A and d B

respecti%ely.

7ind: (he elongation of the bar due to the load F.

SOL9)IO5:

. . . .



42

E"a!p#e 4.':

i3en: Composite A;3K steel bar " E st 1 2&5 EFa$ shown in 'igure !. "a$ is made from two

segments A6 and 6D. Area AA6 1 K55 mm2 and A6D 1 &255 mm2.

7ind:

&$ (he %ertical displacement of end A*

2$ Displacement of B relati%e to !$

SOL9)IO5: Due to external loadings, internal axial forces in regions A6, 6C and CD are

different. Apply method of sections and euation of %ertical force euilibrium as shown.

Lariation is also plotted.



%

A

1

PL

AE

∑<I G"& m$"&5K$

K55 mm2

"2&5$"&53

$ 7m2

G

1<3I G"5.I m$"&5K$

&255 mm2 "2&5$"&53$ 7m2G<

B!I G"5.I m$"&5K$&255 mm2 "2&5$"&53$ 7m2G

<1 <5.K& mm% A

1

P B!

L B!

A B! E

<3I G"5.I m$"&5K$

&255 mm2 "2&5$"&53$ 7m2G11 <5.&5! mm

43

E"a!p#e 4.4:

i3en: A hollow steel 0 1 35,555 siG tube "&$ with an outside

diameter of 2.I in. and a wall thicness of 5.2I in. is fastened to a

solid aluminum 0 1 &5,555 siG rod "2$ that has a 2in.;diameter and

a solid &.3I;in.;diameter aluminum rod "3$. (he bar is loaded as

shown in 'igure !.H.

7ind:

"a$ the change in length of steel tube "&$.

"b$ the deflection of #oint D with respect to the fixed support at A.

"c$ the maximum normal stress in the entire axial assembly.

SOL9)IO5:



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4.' Statica##$ Indeter!inate Structure&:

7i6ure 4. statically indeterminate bars.

7i6ure 4.1; 8a Analysis of a statically

indeterminate bar.

• Statically determinate → eactions and internal forces "axial,

shear and moment$ can be determined #nl& from the 3

euations of euilibrium-

? ?

• Statically determinate → @nnown forces can be determined

without nowing the properties of the materials "such as the

=oung/s modulus E $.

• Statically indeterminate → eactions and internal forces

cann#t be found by statics alone "the 3 euations of

euilibrium$.

• )ence, additional euation is needed "euation of

c#mpatibilit&$.

• 'igure !.M statically indeterminate bars. Consider

euilibrium-

Σ F v 1 5

' A ' BP 15

∴ ' A < ' B 1 P

• >ne useful euation "Σ F v 1 5$ cannot sol%e 2 unnowns

" ' A, ' B$.

• 'igure !.&5 Analysis of a statically indeterminate bar.

• Frismatic bar AB Attached to rigid supports at both ends

4 2 5 5 & 6 r o o s 7 C o l e , a d i % i s i o n o f ( h o m s o n 8 e a r n i n g , 9 n c . ( h o m s o n 8 e a r n i n g :

i s a t r a d e m a r u s e d h e r e i n u n d e r l i c e n s e .

4 2 5 5 & 6 r o o s 7 C o l e , a d i % i s i o n o f ( h o m s o n 8 e a r n i n g , 9 n c . ( h o m s o n 8 e a r n i n g :

i s a t r a d e m a r u s e d h e r e i n u n d e r l i c e n s e .



45

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

• )ow to determine ' A and ' B*

• As mentioned earlier, one euation "Σ F v 1 5$ cannot sol%e

the 2 unnowns " ' A, ' B$.

• e need one more euation. )ow*

• Change in length of a bar δ must be c#mpatible with the

conditions at the supports.

• Supports of the bar are both fixed at the ends (otal change

in length is Jero "δ AB = *).

• 0uation of compatibility 9n this case, the change in length

of the bar must be compatible with the conditions at the

supports "δ AB = *).

• 'or linearly elastic materials → ¿

PL

AE

• ow determine ' A and ' B-

oring-

E"a!p#e 4.(:



46

7i6ure 4.11 8a Analysis of a statically

indeterminate structure.

'igures !.&& "a;b$ show a solid circular steel c&linder + is encased

in a hollow circular c#pper tube ! . (he cylinder and tube are

compressed between the rigid plates of a testing machine by

compressi%e forces P . 6oth parts ha%e length L.Steel cylinder

A s 1 Cross;sectional area

E s 1 +odulus of elasticity "=oung/s modulus$

Copper tube

Ac 1 Cross;sectional area

E c 1 +odulus of elasticity "=oung/s modulus$

Determine the following-

Compressi%e forces P s in the steel cylinder and P c in the

copper tube.

Compressi%e stresses σ s in the steel cylinder and σ c in the

copper tube.

"c$ Shortening δ of the assembly.

SOL9)IO5:

..

. . . .



47

4.4 Method of

Superposition• After subdi%iding 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 caused

by each component load acting separately on the member.

• esultant stress7displacement determined algebraically by adding the contributions of each component.



48

#onditions,

&. (he loading must be linearly related to the stress or displacement that is to be determined.

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

$%en to ignore deformations&

3. +ost loaded members will produce deformations so small that change in position and direction of loading will be insignificant and can be neglected

!. 0xception to this rule is a column carrying axial load, discussed in 8ater.

'uperposition principle procedure!

'.( )her!a# Stre&&:

+ost engineering materials expand when heated and contract when cooled. (he strain due to a

temperature change of N( is called thermal strain and is obtained by-

εT =α ∆ T ,∧δ T =α ∆ TL ………… . (3.6)

hereεT :T hermal strain, α :coefficient of t hermalexpantion,

L: originallengt h,∧δ T : t hermalelongation.



49

(otal 'trains! 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 gi%en by-

εtotal=εσ +εT … … … … …(3.7)

Example 4.6:

i3en: At a temperature of !5O', a 5.5H;in. gap exists

between the ends of the two bars shown in 'igure

!.&2. 6ar "&$ is an aluminum alloy P 1 &2.I Q

&5BK

7O'G and bar "2$ is s ta in le ss s te el P 1 M. K Q

&5BK7O'G. (he supports at A and ! are rigid.

7ind: (he lowest temperature at which the two bars contact each other ?

SOL9)IO5:

E"a!p#e 4.,:



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2.5 Stresses on Inclined

Sections• Consider the bar loaded as shown in 'igure !.&3.

• Fass a section through the member forming an angle - it

the normal plane.

• 'rom euilibrium conditions, the distributed forces

"stresses$ on the plane must be eui%alent to the force F.

• esol%e F into components normal and tangential to the

obliue section,

• (he a%erage normal and shear stresses on the obliue

plane are-

• (he maximum normal stress occurs when the reference plane is

perpendicular to the member axis,



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• (he maximum shear stress occurs for a plane at < !Io with

respect to the axis,

………..(11)

Example 4.8:Given: An axial load F is applied to the &.I in. by 5.I in. rectangular

bar shown in 'ig.ure !.&I. (he bar is sub#ected to an axial load of F 1 &H

ips.

ind: (he normal stress perpendicular to plane A6 and the shear stress

parallel to plane A6*

!ssumptions: the bar weight is neglected.

S"#$%I"&:



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4., Saint-enant/& +rincip#e:

• 8ocaliJed deformation occurs at each end, and the deformations

decrease as measurements are taen further away from the ends

as shown 'igure !.&I.

• At section cc, stress reaches almost uniform %alue as compared

to aa, bb.

• cc is sufficiently far enough away from P so that localiJed

deformation R%anishes, i.e., minimum distance.

• Eeneral rule- min. distance is at least eual to largest dimensi#n

of loaded x;section. 'or the bar, the min. distance is eual to

width of bar.

• (his beha%ior disco%ered by 6arrT de Saint;Lenant in &HII, this

name of the principle.

• Saint;Lenant Frinciple states that l#cali0ed effects caused by any

load acting on the body, will dissipate/sm##t #ut within regions

that are sufficientl& rem#ved from location of load. Saint;

Lenant/s Frinciple- Stress distribution may be assumed

independent of the mode of load application except in the

immediate %icinity of load application points.

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

application loads or support reactions.

4.0 Stre&& Concentration&:

F#rce e-uilibrium reuires magnitude of resultant force

de%eloped by the stress distribution to be eual to P. 9n other

words,

P=∫ A

1

σ dA

• (his integral represents graphically the v#lume under each

of the stress;distribution diagrams shown.

• 9n engineering practice, actual stress distribution not needed,

only maimum stress at these sections must be nown.

+ember is designed to resist this stress when axial load P is

applied.



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• 1 is defined as a ratio of the maximum stress to the

nominal a3era6e ) stress acting at the smallest cross

section-

K =

σ max

σ nom

• 1 is independent of the bar/s geometry and the type of

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

discontinuity.

• As siJe r of the discontinuity is decreased, stress

concentration is increased.

• 9t is important to use stress;concentration factors in design

when using brittle materials, but not necessary for ductile

materials

• Stress concentrations also cause failure structural members

or mechanical elements sub#ected to fatigue l#adings.

Example

4.':



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i3en: Steel bar shown in next "'igure !.&H$ has

allowable stress, Uallow 1 &&I 2Pa.

7ind: largest axial force P that the bar can carry.

SOL9)IO5:6ecause there is a shoulder fillet, stress;concentrating

factor determined using the graph next "'igure !.&M$.

Calculating the necessary geometric parameters yields

r

n=10mm

20mm=0.5

Also,

h =

40mm

20mm=2

Thus, from the grah, K ! 1.4.

A"erage #ormal stress at smallest $%se&t'o#,

σ nom= P

20!10=0.005 P " /mm

2

σ allo= K σ max

115 "

mm2 =1.4(0.005 p)

P=16.43 (103 ) " =16.43 #"

Chapter 4 Axialy Loaded Members Final

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