CIVL 222 STRENGTH OF

MATERIALS

Chapter 4-b

Axially Loaded Members

AXIAL LOADED MEMBERS

Todays Objectives:Students will be able to:

a) Determine the elastic deformation of axially loaded member

b) Apply the principle of superposition for total effect of different loading cases

c) Deal with compatibility conditions

d) Use force method of analysis.

In-class Activities:

Applications

Saint-Venants principle

Elastic deformation in axially

loaded member

Principle of superposition

Compatibility conditions

Force method of analysis

APPLICATIONS

Most concrete columns are reinforced with steel rods; and

these two materials work together in supporting the applied

load. Are both subjected to axial stress?

St. Venants Principle

The stresses and strains in a body at points that

are sufficiently remote from points of application

of load depend only on the static resultant of the

loads and not on the distribution of the loads.

ELASTIC DEFORMATION OF AN AXIALLY LOADED

MEMBER

and = P (x)

A (x) = d

dx

Provided these quantities do not exceed

the proportional limit, we can relate

them using Hookes Law, i.e. = E

P (x)

A (x)= E ( ) d

dx d =

P (x) dx

A (x) E

= L

0

P (x) dx

A (x) E

Fig. 4-2

Fig. 4-4

EXAMPLE 1

EXAMPLE 1 (continued)

PRINCIPLE OF SUPERPOSITION It can be used to simply problems having complicated loadings.

This is done by dividing the loading into components, then

algebraically adding the results.

It is applicable provided the material obeys Hookes Law and the

deformation is small.

Fig. 4-10

If P = P1 + P2And d d1 d2Then the deflection at location x is sum of two cases, ie

x = x1 + x2

COMPATIBILITY CONDITIONS When the force equilibrium condition alone cannot determine

the solution, the structural member is called statically

indeterminate.

In this case, compatibility conditions at the constraint locationsshall be used to obtain the solution. For example, the stresses

and elongations in the 3 steel wires are different, but their

displacement at the common joint A must be the same.

Fig. 4-51

COMPATIBILITY CONDITIONS

The distributed loading w= 46kN/m is supported by threesuspender bars. AB and EF aremade from aluminum and CD ismade from steel. If each bar has a

cross-sectional area of 450 mm2,determine the forces in each bar

when the distributed loading is

applied. Est =200GPa, Eal= 70Gpa.

EXAMPLE

FORCE METHOD OF ANALYSIS

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.

Fig. 4-16

EXAMPLE

EXAMPLE (continued)

THERMAL STRESS

A temperature change results in a change in length or

thermal strain. There is no stress associated with the thermal strain unless the elongation is restrained by

the supports.

coef.expansion thermal

AE

PLLT PT

Treat the additional support as redundant and apply

the principle of superposition.

0

0

AE

PLLT

PT

The thermal deformation and the deformation from

the redundant support must be compatible.

TEA

P

TAEPPT

0

Static Indeterminacy

Structures for which internal forces and reactions

cannot be determined from statics alone are said

to be statically indeterminate.

0 RL

Deformations due to actual loads and redundant

reactions are determined separately and then added

or superposed.

Redundant reactions are replaced with

unknown loads which along with the other

loads must produce compatible deformations.

A structure will be statically indeterminate

whenever it is held by more supports than are

required to maintain its equilibrium.

Determine the reactions at A and B for the steel bar and loading shown, assuming a close fit at

both supports before the loads are applied.

Solve for the reaction at A due to applied loads and the reaction found at B.

Require that the displacements due to the loads

and due to the redundant reaction be compatible,

i.e., require that their sum be zero.

Solve for the displacement at B due to the redundant reaction at B.

SOLUTION:

Consider the reaction at B as redundant, release the bar from that support, and solve for the

displacement at B due to the applied loads.

Example

SOLUTION:

Solve for the displacement at B due to the applied loads with the redundant constraint released,

EEA

LP

LLLL

AAAA

PPPP

i ii

ii9

L

4321

2643

2621

34

3321

10125.1

m 150.0

m10250m10400

N10900N106000

Solve for the displacement at B due to the redundant constraint,

i

B

ii

iiR

B

E

R

EA

LP

LL

AA

RPP

3

21

262

261

21

1095.1

m 300.0

m10250m10400

Example

Require that the displacements due to the loads and due to

the redundant reaction be compatible,

kN 577N10577

01095.110125.1

0

3

39

B

B

RL

R

E

R

E

Find the reaction at A due to the loads and the reaction at B

kN323

kN577kN600kN 3000

A

Ay

R

RF

kN577

kN323

B

A

R

R

Example

Question # 1: Coaxial tube and core

An aluminum tube (1) encases a

brass core (2). The two

components are bonded together

to comprise an axial member that

is subjected to a downward force

of 30 kN. Tube (1) has an outer

diameter of D= 30 mm and an

inner diameter of d= 22 mm. The

elastic modulus of the aluminum

is 70 GPa. The brass core (2) has

a diameter of D= 22 mm and an

elastic modulus of 105 GPa.

Compute the normal stresses in

tube (1) and core (2).

Question # 2: Coaxial tube and core

Question # 3: Compound bars with temperature change

The compound bar, composed of three segments shown; bronze

segment [A= 2000 mm2, E= 83 GPA: =19 x 10-6

mm/mm/C], aluminum segment [A= 1400 mm2, E= 70 GPA:=23 x 10-6 mm/mm/C], and steel segment [A= 800 mm2, E=200 GPA: =11.7 x 10-6 mm/mm/C], is initially stress-free.Compute the stress in each material if the temperature drops

25C . Assume that the walls do not yield.

Question # 4: Coaxial bars with temperature change

A rectangular bar 30-mm wide

and 24-mm thick made of

aluminum [E= 70 GPA: =23 x10-6 mm/mm/C] and tworectangular copper bars 30-mm

wide and 12-mm thick [E= 120

GPA: =16 x 10-6 mm/mm/C]are connected (in longitudinal

direction) by two smooth 11-mm

diameter pins. When the pins are

initially inserted into the bars,

both the copper and aluminum

bars are stress free. After the

temperature of the assembly has

increased by 65C.

(a) Determine the internal

axial force in the

aluminum bar.

(b) Determine the normal

strain in the copper

bars.

(c) Determine the shear

stress in the 11-mm

diameter pins.

READING QUIZ1) The stress distributions at different cross sections are different

(see figure below).

However, at locations far enough away from the support and the

applied load, the stress distribution becomes uniform. This is due to

A)Principle of superposition C) Poissons effect

B) Inelastic property D) Saint Venants Principle

Fig. 4-1

READING QUIZ

2) The principle of superposition is valid provided that

a) The loading is linearly related to the stress or displacement

b) The loading does not significantly change the original geometry

of the member

c) The Poissons ratio v 0.45

d) Youngs Modulus is small

A)a, b and c C) a and b only

B) a, b and d D) All

CONCEPT QUIZ1) The assembly consists of two posts made from material 1

having modulus of elasticity of E1 and a cross-sectional area

A1 and a material 2 having modulus of elasticity E2 and cross-

sectional area A2. If a central load P is applied to the rigid cap,

determine the force in each post. The support is also rigid.

Let r = E1A1E2A2

A) P1 = ( ) Pr

2r + 1

P2 = ( ) P1

2r + 1

B) P1 = ( ) P1

2r + 1

P2 = ( ) Pr

2r + 1

C) P1 = r P

P2 = (2r-1) P

D) P1 = r (r+1) P

P2 = (r+1) P

Prob. 4-62