4.5 FORCE METHOD OF ANALYSIS (FLEXIBILITY METHOD) It is...

4.5 FORCE METHOD OF ANALYSIS (FLEXIBILITY METHOD)

• It is also possible to solve statically indeterminate problem by writing the compatibility equation using the superposition of the forces acting on the free body diagram.

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0

p B

AC ACBB

BCA B A

PL LF L F PAE AE L

LF F P F PL

EXAMPLE 4.8

The A-36 steel rod shown in Fig. 4–17a has a diameter of 10 mm. It is fixed to the wall at A, and before it is loaded there is a gap between the wall at B’ and the rod of 0.2 mm. Determine the reactions at A and Neglect the size of the collar at C. Take Est = 200GPa.


EXAMPLE 4.8 (cont.)

• Using the principle of superposition,

• From Eq. 4-2,

• Substituting into Eq. (1), we get

Solutions

B

BABBB

ACP

FFAELF

AEPL

992

392

3

103944.7610200005.0

2.1

105093.010200005.04.01020

1 0002.0 BP

(Ans) kN 05.41005.4

103944.76105093.00002.03

93

B

B

F

F


EXAMPLE 4.8 (cont.)

• From the free-body diagram,

Solutions

(Ans) kN 0.16 005.4200

A

A

x

FF

F


APPLICATIONS

Thermal Stress Stress Concentration

Inelastic Axial Deformation


4.6 THERMAL STRESS

• Ordinarily, the expansion or contraction δT is linearly related to the temperature increase or decrease ΔT that occurs.

• If the change in temperature varies throughout the length of the member, i.e. ΔT = ΔT(x), or if α varies along the length, then

TLT

= linear coefficient of thermal expansion, property of the material= algebraic change in temperature of the member= original length of the member= algebraic change in length of the member

TTT

dxTT


T TL

TFLTLAE

F TAETE

EXAMPLE 4.10 The rigid bar is fixed to the top of the three posts made of A-36 steel and 2014-T6 aluminum. The posts each have a length of 250 mm when no load is applied to the bar, and the temperature is T1 = 20°C. Determine the force supported by each post if the bar is subjected to a uniform distributed load of 150 kN/m and the temperature is raised to T2 = 80°C.


EXAMPLE 4.10 (cont.)

• From the free-body diagram we have

• The top of each post is displaced by an equal amount and hence,

• Final position of the top of each post is equal to its displacement caused by the temperature increase and internal axial compressive force.

Solutions

(2) alst

(1) 010902 ;0 3 alsty FFF

FalTalal

FstTstst


EXAMPLE 4.10 (cont.)

• Applying Eq. 2 gives

• With reference from the material properties, we have

• Solving Eqs. (1) and (3) simultaneously yields

Solutions

FalTstFstTst

(3) 109.165216.1

101.7303.025.025.020801023

1020002.025.025.020801012

3

926

926

alst

alst

FF

FF

(Ans) kN 123 and kN 4.16 alst FF


4.7 STRESS CONCENTRATION

• When a member’s cross section area is suddenly changed, the maximum normal stress occurs at the smallest cross section area.

• The stress concentration factor K is a ratio of the maximum stress to the average stress acting at the smallest cross section; i.e.

avg

Kmax


• K is independent of the material properties.• K depends only on the specimen’s geometry and the type

of discontinuity.


max . .

ys uall

orF S

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