ecs 238 - chapter 1 (stress and strain)

BY

AIMI MUNIRAH BINTI JALILLUDDIN

ECS 238

BASIC SOLID MECHANICS

CHAPTER 1:

1D & 2D LINEAR STRESS & STRAIN SYSTEM

Learning Outcomes

CONTENTS

1

CHAPTER 1: 1D & 2D Linear Stress and Strain System

Stressa

Strain

Stress Strain / Shear Stress - Strain Diagram

3

Introduction2

Factor of Safety

Deformation

Deformation of Composite Materials

c

d

e

f

b

LEARNING OUTCOMES

At the end of this topic, students should be able to :

1. Differentiate and compute the normal and shearing

stresses (CO1: PO1, CO1: PO3)

2. Compute the deformation caused by normal and

shearing stresses (CO1: PO1, CO1: PO3)

3. Solve the composite material problem using stress

and strain concept (CO1: PO1, CO1: PO3)

4. Solve theoretically OR graphically the plane stress

problem (CO1: PO3)

1D & 2D LINEAR STRESS &

STRAIN SYSTEM

CHAPTER 1

INTRODUCTION Direct Stress

Symbol: Unit : N/m2 (Pascal) or N/mm2

Formula: = P/A or F/A Shear Stress

Symbol: Unit : N/m2 or N/mm2

Formula: = V/A

Strain Symbol: Unit : Dimensionless Formula: = L/L

Load

Tensile Load (+ve)

Compressive Load (-ve)

Tensile Load (Tension )

Example: towing ropes

or lifting hoist

Compression

Example: column/pillar

INTRODUCTION

Equilibrium of a Particle

A particle is in equilibrium when the resultant of all the forces acting on that particle is zero

REVIEW OF PHY190:

Equilibrium of a Rigid Body

A rigid body is in equilibrium when the external

forces acting on it form a system of forces

equivalent to zero

EQUILIBRIUM OF RIGID BODIES

If the resultant force is zero, and the

resultant moment about one axis is zero, then

the resultant moment about any other axisin the body will be zero also.

VISUALIZATION OF

CONCEPT

A B

Rigid body AB is in statically equilibrium under set of 2 forces.

F1 = 2kN F2 = 2kN

Check whether true the rigid body is in statically equilibrium

Horizontal force = -2 + 2 = 0 kN (correct)

Say, cut the rigid body into 2 equal sections at X, as shown below;

A BF1 = 2kN F2 = 2kNX

AF1 = 2kN B F2 = 2kNX X

But, rigid body AX and XB are also in statically equilibrium under the force.

F1X = ? F2X = ?

Horizontal force = 0

Applying to rigid body AX;-F1 + F1x = 0 -2 + F1x =0; F1x = 2 kN

Applying to rigid body XB;

-F2x + F2x = 0 -F2x + 2 = 0, F2x = 2 kN

Remember sign convention for direct forces,

-ve

-ve

+ve

+ve

DIRECT STRESS

Normal stress results from a uniformly or equally

applied direct force across a cross section.

The average normal stress in the member is obtained

by dividing the magnitude of the resultant internal

force F by the cross sectional area A.

Formula => = P/A

where P = load, A = cross section area

Unit : N/m2 = Pa, N/mm2

DIRECT / NORMAL STRESS

Force perpendicular to area

Cross sectional area

NOTE ON UNITS:

The fundamental unit of stress is 1 N/m2 and this is

called a Pascal (Pa).

This is a small quantity in most fields of

engineering so we use the multiples kPa, Mpa

and GPa.

Areas may be calculated in mm2 and units of

stress in N/mm2 are quite acceptable.

Since 1 N/mm2 converts 1 000 000 N/m2 then it

follows that the N/mm2 is the same as Mpa.

DIRECT / NORMAL STRESS

EXAMPLE 1Determine the stress in a bar of 20 mm

diameter if it is subjected to an axial load of

30 kN in tension

A = x 202/4 = 314.16 mm2

= 30 x 103 N = 95.5 N/mm2

314.16 mm2

EXAMPLE 2Determine the stress in each of the bar in the

Figure below.

BC = 3 x 103 N = 38.2 N/mm2

78.54 mm2

AB = 7 x 103 N = 39.6 N/mm2

176.71 mm2

4 kN

3 kNA

B

C

D = 15 mm

D = 10 mm

EXAMPLE 3

Given: Bar width = 35 mm, thickness = 10 mm

Determine max. average normal stress in bar when

subjected to loading shown.

SOLUTIONInternal loading

Normal force diagram

Average normal stress

BC =PBC

A

30(103) N

(0.035 m)(0.010 m)= = 85.7 MPa

SHEAR STRESS

SHEAR STRESS

Shear stress is the

type of stress that

acts tangential to

plane

Shear stress,

= Shear force, V

Area, A

Unit : N/m2 = Pa,

N/mm2

Single shear

Double shear

SHEAR STRESSShear occurs typically:

1. When a pair of shear

cuts a material

2. When a material is punched

3. When a beam has a

transverse load

4. When a pin carries a

load

SHEAR STRESS

EXAMPLE 1

EXAMPLE 2

DIRECT STRAIN

Description: When a force is applied to an elastic

body, the body deforms. The way in

which the body deforms depends

upon the type of force applied to it.

DIRECT STRAIN

A tensile force makes

the body longer

A compressive force

makes the body shorter

Direct strain is the deformation per unit of the

original length. It has no units since it is a ratio of

length to length

Formula => = L/L

where L = change in length

L = original length

DIRECT STRAIN

L L

P P

EXAMPLE 1A metal wire is 2.5mm diameter and 2m long. A force

of 12N is applied to it and it stretches 0.3mm. Assumethe material is elastic, determine the following:

1. The stress in the wire,

2. The strain in the wire,

A = x 2.52/4 = 4.909mm2

= 12 N = 2.44N/mm2

4.909

= 0.3 = 0.000152000

SHEAR STRAIN

Shear strain is defined as the ratio of distance

deformed to the height.

Formula => = x/L

where x = deformation

L = length

SHEAR STRAIN

EXAMPLE 1The plate is deformed into the dashed shape as

shown in the figure below. Determine the average

normal strain along side AB and average shear strain

in the plate relative to the x and y- axes.

SOLUTION

(a) Line AB, coincident

with y axis, becomes

line AB after deformation. Length

of line AB is

AB = (250 2)2 + (3)2 = 248.018 mm

SOLUTION(a) Therefore, average normal strain for AB is,

= 7.93(103) mm/mm

(AB)avg = ABAB AB 248.018 mm 250 mm

250 mm=

Negative sign means strain

causes a contraction of AB.

SOLUTION(b) Due to displacement of B to B, angle BAC

referenced from x, y axes changes to . Since xy = /2 , thus

xy = tan1 3 mm

250 mm 2 mm= 0.0121 rad

( )

FACTOR OF SAFETY

(FOS)

Every material has a certain capacity to carry

load, but it is unsafe to load a material to the full

capacity it would have no reserve strength.

This is dangerous because:

1. May experience a load greater than anticipated

2. Material may be defective

3. Construction may be faulty (fabrication/erection/

workamanship etc.)

4. Other unforeseen situation (calculation errors etc.)

Remedy: Apply a Factor of Safety (FOS) that provides

a margin for error and uncertainty

FACTOR OF SAFETY

Factor of Safety (FOS) = Maximum Stress

Maximum stress is obtained from experimental testing

of the material

FACTOR OF SAFETY

Allowable Working Stress

MAXIMUM OR

ULTIMATE STRESS

ALLOWABLE

STRESS

FOS for normal stress = max/allow

FOS for shear stress = max/allow

In above equation, FOS will be greater or equal to 1.0 to

avoid potential failure.

FACTOR OF SAFETY

MAXIMUM OR

ULTIMATE STRESS

ALLOWABLE

STRESS

ALLOWABLE STRESS

To determine area of section subjected to a normal force, use

A = P

allow

A = V

allow

To determine area of section subjected to a shear force, use

EXAMPLE 1

SOLUTION

STRESS-STRAIN

DIAGRAM

Used primarily to determine the relationship

between the average normal stress and average

normal strain in common engineering materials

Before testing, 2 small punch marks identified along

specimens length

Measurements are taken of both specimens initial x-sectional area A0 and gauge-length distance L0;

between the two marks

TENSION &

COMPRESSION TEST

Performing the tension or compression test

Set the specimen into a testing machine shown

below

TENSION & COMPRESSION TEST

The machine will stretch specimen at slow

constant rate until breaking point

At frequent intervals during test, data is

recorded of the applied load P.

Elongation = L L0 is measured using either a caliper or an extensometer

is used to calculate the normal strain in

the specimen

L + L

= change in length/original length= L/L

GAUGED

LENGTH

GAUGED

LENGTH

ELONGATED

TENSILE TEST

Using recorded data, we can determine nominal or engineering stress by

Assumption: Stress is constant over the x-section and throughout region between gauge points

Likewise, nominal or engineering strain is found directly from strain gauge reading, or by

Assumption: Strain is constant throughout region between gauge points

By plotting (ordinate) against (abscissa), we get a conventional stress-strain diagram

STRESS-STRAIN DIAGRAM

A

P

L

Figure shows the

characteristic stress-

strain diagram for steel,

a commonly used

material for structural

members and

mechanical elements


Elastic behavior

A straight line

Stress is proportional to strain, i.e., linearly elastic

Upper stress limit, or proportional limit; pl

If load is removed upon reaching elastic

limit, specimen will

return to its original

shape


Strain hardening

Ultimate stress, u While specimen is elongating, its x-sectional area

will decrease

Decrease in area is fairly uniform over

entire gauge length


Necking

At ultimate stress, x-sectional area begins to

decrease in a localized region

As a result, a constriction or neck tends to form in this

region as specimen

elongates further


Necking

Specimen finally breaks at fracture stress, f

51


Ductile materials

Offset method to determine yield strength

1. Normally, a 0.2 % strain is

chosen.

2. From this point on the axis, a line parallel to initial

straight-line portion of

stress-strain diagram is

drawn.

3. The point where this line

intersects the curve defines

the yield strength.


HOOKESLAW

Most engineering materials exhibit a linear

relationship between stress and strain with the elastic region

Discovered by Robert Hooke in 1676 using

springs, known as Hookes law

E represents the constant of proportionality, also

called the modulus of elasticity or Youngs modulus

E has units of stress, i.e., pascals, MPa or GPa.

HOOKES LAW

E

Most grades of steel have same modulus of elasticity,

Est = 200 GPa

Modulus of elasticity is a mechanical property that

indicates the stiffness of a material

Materials that are still have large E values, while spongy materials (vulcanized rubber) have low values

IMPORTANT

Modulus of elasticity E, can be used only if a material has linear-elastic behavior.

Also, if stress in material is greater than the

proportional limit, the stress-strain diagram ceases to

be a straight line and the equation is not valid

HOOKES LAW

MODULUS OF ELASTICITY/

YOUNGS MODULUS,E E, sometimes called as elastic modulus, is a

constant value of a material.

i.e. Steel has E = 200 GPa = 200 x 109 N/m2

We can measure this value by finding

the gradient of the stress-strain graph

over the elastic region

EXAMPLE 1An aluminium rod specimen has an initial gauge length

of 254 mm before the tensile test. After been pulled by a

force of 165 kN, the gauged length increase to 300 mm.

Determine the modulus of elasticity of the specimen if the

rod diameter is 30 mm.

1. The elongation, L = 300-254 = 46 mm

2. Force required to elongate, P = 165 kN

3. Cross sectional area, A = /4 x 302 = 706.86 mm2

4. Direct stress, = P/A = 165 x 103/706.86 = 233.42 N/mm2

5. Original length = 254 mm

6. Strain, = L/L = 46/254 =0.1811

7. From Hookes Law , E = / = 1289 N/mm2 = 1.3 kN/mm2

EXAMPLE 2

SHEAR STRESS-STRAIN

DIAGRAM

Use thin-tube specimens and subject it to

torsional loading

Record measurements of applied torque and

resulting angle of twist

SHEAR STRESS-STRAIN

DIAGRAM

Material will exhibit linear-elastic behavior till its

proportional limit, pl Strain-hardening continues till it reaches ultimate

shear stress, u

SHEAR STRESS-STRAIN

DIAGRAM

Material loses shear strength till it

fractures, at stress of f

Hookes law for shear

G is shear modulus of elasticity or modulus of rigidity

G can be measured as slope of

line on - diagram, G = pl/ pl The three material constants E,

, and G is related by

G

)1(2

EG

SHEAR STRESS-STRAIN

DIAGRAM

POISSONS RATIO

When a body is subjected to axial tensile force, it

elongates and contracts laterally

Similarly, it will contract and its sides expand

laterally when subjected to an axial compressive

force

POISSONS RATIO

Strains of the bar are:

Early 1800s, S.D. Poisson realized that within

elastic range, ration of the two strains is a

constant value, since both are proportional.

rLlatlong

'

long

latRatiosPoisson

,'

POISSONS RATIO

is unique for homogenous and isotropicmaterial

Why negative sign? Longitudinal elongation

cause lateral contraction (-ve strain) and vice

versa

Lateral strain is the same in all lateral (radial)

directions

Poissons ratio is dimensionless, 0 0.5

POISSONS RATIO

EXAMPLE 1

VOLUMETRIC STRAIN

CONCEPT

INTRODUCTIONConsider a volume of a cube under multi

tensile stresses acting on three surfaces.

Under those stresses, the cube deforms into a

rectangular paralelpiped, and its volume, V

X

Y

Z

V = ( 1 + X) ( 1 + Y) (1 + z)

= ( 1 + X ) ( 1 + y) ( 1 + z)

= 1 + x + y + z + xy + xz + xyz

Since xy , xz, yz and xyz are small

quantities, ignore those terms,

V = 1 + x + y + z

Changes in volume, V = V V = x + y + z

Volumetric strain , v = V/V = x + y + z

For a multi-stressed condition, each strain can

be expressed as follows;

x = x/E - yv/E- zv/E = x/E v/E(x + z)

y = y/E v/E(x + z)

z = z/E v/E(x + y)

v = x + y + z

= [ x + y + z ] / E ( 1- 2v)

If x = y = z =

v = 3/E ( 1- 2v)

Bulk Modulus, K = /v

E =(3/v)(1-2v) = 3K(1-2v)

K = E/(3(1-2v))

G = E/(2(1+v))

G = 3K (1-2v)/(2(1+v))

A steel bar 25mm x 15 mm in cross section is 300 mm long

and is subjected to a tensile force of 70 kN. Find the

change in the dimensions of the bar and the change in

volume. (Take E = 200 kPa)

EXAMPLE 1

70 kN15 mm

25 mm300 mm

x

y

z

SOLUTIONLongitudinal strain, = / E

= 70000 N/[(15 x 25)(200000)

= 0.00093

Lateral strain, y = z = vx = 0.00093 x 0.3

= 0.00028

Change in length = 0.00093 x 300 = 0.279 mm

Change in width = 0.00028 x 25 = 0.007 mm

Change in height = 0.00028 x 15 = 0.0042 mm

Volume of bar = 300 x 25 x 15 = 112500 mm3

Volume strain, v = x + y + z = 0.00037

Change in volume = Vv = 341.625 mm3

What is the stresses, , acting on the cube 1m side

shown below should be so that the change of the

volume is 0.05 %? Given that E = 200 Gpa, v = 0.3

EXAMPLE 2

x

y

z

SOLUTION

x = /E (1-2v) = y = z

v = x + y + z = 3/E (1-2v)

Change in volume = 0.05/100 V

= 0.05/100 (1) = v = 3/E (1-2v)

= 0.05/100 (E/3(1-2v))

= 0.05/100 (200000/3(1-2(0.3)) = 83.33 N/m2

DEFORMATION OF AN

AXIALLY LOADED

MEMBER

DEFORMATION UNDER

DIRECT FORCE For constant x-sectional area A, and

homogenous material, E is constant

With constant external force P, applied at each end, then internal force P throughout length of bar is constant

Load-displacement relationship is:

=PL

AE

EXAMPLE 1Composite A-36 steel bar shown made from two

segments AB and BD. Area AAB = 600 mm2 and

ABD = 1200 mm2.

Determine the vertical

displacement of end A and

displacement of B relative to

C.

SOLUTIONInternal force

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.

SOLUTIONDisplacement

From tables, Est = 210(103) MPa.

Use sign convention, vertical displacement of A relative

to fixed support D is

A =PL

AE [+75 kN](1 m)(10

6)

[600 mm2 (210)(103) kN/m2]=

[+35 kN](0.75 m)(106)

[1200 mm2 (210)(103) kN/m2]+

[45 kN](0.5 m)(106)[1200 mm2 (210)(103) kN/m2]+

= +0.61 mm

SOLUTIONDisplacement

Since result is positive, the bar elongates and so

displacement at A is upward

Apply load-displacement equation between B and C,

A =PBC LBCABC E

[+35 kN](0.75 m)(106)

[1200 mm2 (210)(103) kN/m2]=

= +0.104 mm

Here, B moves away from C, since segment elongates

DEFORMATION DUE TO

TEMPERATURE CHANGES

DEFORMATION DUE TO

TEMPERATURE CHANGES

It is a common experience that materials

expand on heating and contract on cooling.

Consider a rod of a material has a length L at

any temperature TO, then it increases to a

length L + L when heated to a temperature T1

or decreases to a length L-L if T1 is less than TO.

The increase/decrease of material due to temperature effect depends on coefficient of

thermal of material.

L at temperature TO

L L

L at temperature T1 > To

L-L

L at temperature < To

L

Expandon heating

Contracton cooling

TYPICAL CIVIL ENGINERING

STRUCTURES EXPERIENCING THERMAL

EXPANSION/CONTRACTION

TYPICAL CIVIL ENGINERING

STRUCTURES EXPERIENCING THERMAL

EXPANSION/CONTRACTION

APPROACH SLAB

BRIDGE

DECK

SMALL GAP

ALLOWING

EXPANSION

OF DECK

BRIDGE DECK SLAB APPROACH CONNECTION SYSTEM

EXPANSION JOINTS

STRAIN DUE TO

TEMPERATURE CHANGESDue to change in length by L, therefore the corresponding strain

= changes of length/original length

= L/L

L = original length x coefficient of thermal x temperature changes

= L x x ( T1 To)

EXAMPLE 1A steel rod, 20 mm diameter, and 1.5 m long, is constrained

between supports A and B. The material is stress-free at 270C.

Determine the stress in the material when the temperature

increases to 500C.

(Take E = 200 GPa. = 12 x 10-6/0C)

L = 1.5 m, = 12 x 10-6/0CTemperature changes = 50 27 = 230C

Changes in length, L = 1.5 x 12 x 10-6 x 23

= 0.000414 m = 0.414 mm

Strain in the bar, = L/L = 0.414/1500 = 0.000276

From Hookes Law, = E= 200 x 1o9 N/m2 x 0.000276

= 5.52 x 107 N/m2

PRINCIPLE OF

SUPERPOSITION

91

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 caused by each component

load acting separately on the member.

Resultant stress/displacement determined

algebraically by adding the contributions of

each component

4.3 PRINCIPLE OF SUPERPOSITIONPRINCIPLE OF

SUPERPOSITION

92

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?

Most loaded members will produce deformations so

small that change in position and direction of

loading will be insignificant and can be neglected

4.3 PRINCIPLE OF SUPERPOSITIONPRINCIPLE OF

SUPERPOSITION

STATICALLY

INDETERMINATE AXIALLY

LOADED MEMBER

For a bar fixed-supported at one end, equilibrium

equations is sufficient to find the reaction at the

support. Such a problem is statically determinate

If bar is fixed at both ends, then two unknown axial

reactions occur, and the bar is statically

indeterminate

+ F = 0;

FB + FA P = 0

STATICALLY INDETERMINATE AXIALLY LOADED MEMBER

To establish addition equation, consider geometry of

deformation. Such an equation is referred to as a

compatibility or kinematic condition

Since relative displacement of one end of bar to the

other end is equal to zero, since end supports fixed,

This equation can be expressed in terms of applied

loads using a load-displacement relationship, which depends on the material behavior

A/B = 0

STATICALLY INDETERMINATE AXIALLY LOADED MEMBER

Equilibrium

Draw a free-body diagram and write appropriate equations of

equilibrium for member using calculated result for redundant

force.

Solve the equations for other reactions

PROCEDURE FOR ANALYSIS

Compatibility

Choose one of the supports as redundant and write the equation of compatibility.

Known displacement at redundant support (usually zero), equated to displacement at support caused only by external loads acting

on the member plus the displacement at the support caused only

by the redundant reaction acting on the member.

Express external load and redundant displacements in terms of the loadings using load-displacement relationship

Use compatibility equation to solve for magnitude of redundant force

97

From free-body diagram, we can determine the

reaction at A

+=

PROCEDURE FOR ANALYSIS

EXAMPLE 1

A-36 steel rod shown has diameter of 5 mm. Its attached to fixed wall at A, and before it is loaded, theres a gap between wall at B and rod of 1 mm. Determine reactions at A and B.

SOLUTIONCompatibilityConsider support at B as redundant. Use principle of superposition,

0.001 m = P B Equation 1( + )

SOLUTIONCompatibility

Deflections P and B are determined from Eqn. 4-2

P =PLACAE = = 0.002037 m

B = FB LABAE

= = 0.3056(10-6)FB

Substituting into Equation 1, we get

0.001 m = 0.002037 m 0.3056(10-6)FBFB = 3.40(10

3) N = 3.40 kN

Equilibrium

From free-body diagram

FA + 20 kN 3.40 kN = 0FA = 16.6 kN

+ Fx = 0;

EXAMPLE 2

SOLUTION

EXAMPLE 3

SOLUTION

DEFORMATION OF A

STATICALLY

INDETERMINATE AXIALLY

LOADED MEMBER

COMPOSITE MATERIALS

INTRODUCTION Stresses, strains and deformations in a bar are easily

obtained by applying the equilibrium (static) conditions

alone called statically determinate problem.

However, certain problems cannot be solved by statics

alone where extra equations are required in order to be

solved. This is called statically indeterminate problem.

Composite bar subjected to an axial is an example of

statically indeterminate problem where the bar is made

of more than one material.

Consider the followings for solving:

1. Total force in the bar must equal to the applied load.

2. Axial deformation and strain in each material must

also be the same.

VISUALIZATION :

FORCE EQUILIBRIUMP

=

P1

+

P2

Force equilbrium , P = P1 + P2 (1)

= 1 A1 + 2 A2

A1

A2

VISUALIZATION :

COMPATIBILITY OF STRAIN P

=>

Shortening inside material 1 = shortening inside material 2

1 = 2

1/ E1 = 2/E2

Shortening at once due to

perfect bonding assumption

Concrete block

Steel

bar

cs

cs LL

VISUALIZATION :

COMPATIBILITY OF STRAIN

A weight of 300kN is supported by a short concrete column of

square shape with sides of 250mm length. The column is strengthen

by four steel bars with a total cross-sectional area of 50cm2

a) If the modulus of elasticity for steel is 15 times that of

concrete, find the stresses in the steel and the concrete

b) If the stress in the concrete must not exceed 4MN/m2,what

area of steel is required to allow column to support a load of 600kN?

EXAMPLE 1

250mm

250mm

SOLUTIONLc= Lsc = s = s = c

Es Ec

s = Es . c = 15c

Ec

Ac = (25 x 25) -50 =575cm2

Fc + Fs = 300kN

cAc + sAs = 300kN

c(0.0575) + s(0.005) =300kN

c= 2.27MN/m2

s=34MN/m2

Ac = (25 x 25) As

s = Es . c

= 15c

= 15(4)

=60MN/m2

cAc + sAs = 600 x 103N

c(0.0625-As) + sAs = 600 x 103N

As= 0.00625m2

EXAMPLE 2

SOLUTION

EXAMPLE 3

EXAMPLE 4

SOLUTION

TUTORIAL

TUTORIAL1. A steel bar of rectangular cross-section, 3cm by 2cm. Carries an axial

load of 30kN. Estimate the average tensile stress over a normal cross-

section of the bar.

2. A steel bolt, 2.5cm diameter, carries a tensile load of40kN. Estimate the

average tensile stress at the section a and at the screwed sectionb, where the diameter at the root of the thread is 2.10cm.

TUTORIAL1. A steel bar is 10mm diameter and 2m long. It is

stretched with a force of 20kN and extends by

0.2mm. Calculate the stress and strain of the steel

bar.

2. A rod is 0.5m long and 5mm diameter. It is stretched

0.006mm by a force of 3kN. Calculate the stress and

strain.

TUTORIAL

TUTORIALDetermine the stress at the base of a concrete cube of

1.2 m x 1.2 m x 1.2 m having a density of 2400 kg/m3. A

loading of 40 MN is imposed at the top centre of the

cube as shown.

40 MN

ground

TUTORIAL

THE END

ecs 238 - chapter 1 (stress and strain)

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