CES424 - Topic 6 (Deformation)

CES 424 - DETERMINATE STRUCTURES Topic 5 – DeformationSlide 1 of 95

(Pulau Pinang)

CES 424

DETERMINATE STRUCTURES

Topic 5

Deformation


(Pulau Pinang)

LEARNING OUTCOMES

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

1) Determine the elastic deflections of a structure by usinggeometrical method known as Moment-Area method.

2) Determine the elastic deflections of a structure by usingenergy method known as Virtual Work Method


(Pulau Pinang)

Deflection Diagrams & Elastic Curve

� Deflections of structures can come from loads, temperature,fabrication errors or settlement

� In designs, deflections must be limited in order to preventcracking of attached brittle materials

� A structure must not vibrate or deflect severely for the comfortof occupants

� Deflections at specified points must be determined if one is toanalyse statically indeterminate structures


(Pulau Pinang)


� In this topic, only linear elastic material response is considered

� This means a structure subjected to load will return to itsoriginal undeformed position after the load is removed

� It is useful to sketch the shape of the structure when it is loadedin order to visualise the computed results & to partially checkthe results


(Pulau Pinang)


This deflection diagram represent the elastic curve for the points atthe centroids of the cross-sectional areas along each of themembers.


(Pulau Pinang)


� If the elastic curve seems difficult toestablish, it is suggested that themoment diagram be drawn first andthen construct the curve

� Due to pin-and-roller support, thedisplacement at A & D must be zero

� Within the region of –ve moment, theelastic curve is concave downward

� Within the region of +ve moment, theelastic curve is concave upward

� There must be an inflection point wherethe curve changes from concave downto concave up


(Pulau Pinang)


� In Figure (a), the roller at A allowsfree rotation with no deflection whilethe fixed wall at B prevents bothrotation & deflection

� In Figure (b), no rotation or deflectionoccur at A & B

� In Figure (c), the couple moment willrotate end A, this will cause deflectionsat both ends of the beam since nodeflection is possible at B & C. Noticethat segment CD remains undeformedsince no internal load acts within


(Pulau Pinang)


� In Figure (d), the pin at B allowsrotation, so the slope of the deflectioncurve will suddenly change at this pointwhile the beam is constrained by itssupport

� In Figure (e), the compound beamdeflects as shown. The slope changesabruptly on each side of B

� In Figure (f), span BC will deflectconcave upwards due to load. Sincethe beam is continuous, the end spanswill deflect concave downwards


(Pulau Pinang)

Moment Area

Method


(Pulau Pinang)

Moment Area Theorem

� This method were initially developed by Otto Mohr and laterstated formally by Charles E. Greene in 1873.

� Concept: determine the slope of the elastic curve & deflectionthat due to bending.

� Advantage: deals with beams that subjected to series ofloadings or having different moments of inertia.

� Disadvantages: can only be used to determine the angles ordeviations between 2 tangents on the beam’s elastic curve.


(Pulau Pinang)

Moment Area Theorem

Theorem No. 1 : ∫=A

BB/A dx

EI

Mθ

� The change in slope between any twopoints on the elastic curve equals thearea of the M/EI diagram betweenthese two points.

� The notation θB/A is referred to as theangle of the tangent at B measuredwith respect to the tangent at A.

� The angle is measuredcounterclockwise from tangent A totangent B if the area of the M/EIdiagram is positive (clockwise if thearea of M/EI diagram is negative i.e.below the x axis)

� θB/A is measured in radians


(Pulau Pinang)

Moment Area Theorem

Theorem No. 2 :

The vertical deviation of the tangent at a point (A) on the elastic curve with respectto the tangent extended from another point (B) equals the “moment” of the areaunder the M/EI diagram between the two points (A and B). This moment iscomputed about point A (the point on the elastic curve), where the deviation is tobe determined.

A/BB/Att ≠

dxEI

Mxt

A

B

B/A ∫=


(Pulau Pinang)

Moment Area Theorem

Procedure of Analysis

Step 1: Draw M/EI diagram

� Determine the support reactions and draw the beam’s M/EI diagram

� If the beam is loaded with concentrated forces, the M/EI diagram will consist of aseries of straight line segments.

Step 2: Draw elastic curve diagram

� Draw an exaggerated view of the beam’s elastic curve. Recall that points of zeroslope occur at fixed supports and zero displacement occurs at all fixed, pin, androller supports.

Step 3: Apply Moment Area Theorems

� Apply Theorem 1 to determine the angle between two tangents, and Theorem 2to determine vertical deviations between these tangents.

� After applying either Theorem 1 or Theorem 2, the algebraic sign of the answercan be verified from the angle or deviation as indicated on the elastic curve.


(Pulau Pinang)

Example 1

Determine the slope at points B and C of the beam shown below.

Take E = 200GPa and I = 360(106)mm4. (Theorem 1).

AB

5 m

10 kN

C

5 m


(Pulau Pinang)

Example 1

Step 1.

Draw M/EI diagram. Draw normal BMD by assuming EI is a constant value.

Step 2.

Draw Elastic Curve diagram then draw a tangent line at point A, B and C.

ΣMX ↵+ = 0

MX = - 10 (X)

∴M0 = 0 kNm #

∴M5 = - 50 kNm #

∴M10 = - 100 kNm #

10 kN

MX

X


(Pulau Pinang)

Example 1

Step 3.

Apply Moment Area Theorem (Theorem No. 1)

θB/A can be determined by calculating the area from M/EI diagram between A and B

rad005210EI

kNm375

m5EI

kNm50

2

1m5

EI

kNm50

2

ABB

.

)()(/

−=−=

−

−=θ=θ

rad00694.0EI

kNm500

)m10(EI

kNm100

2

1

2

A/CC

−=−=

−=θ=θ

Note : The –ve sign indicates that the angle is measured clockwise from A

∫=A

BB/A dx

EI

Mθ


(Pulau Pinang)

Example 2

Determine the deflection at points B and C of the beam shown in the figure below. Values for the moment of inertia of each segment are indicated in the figure. Take E = 200 Gpa. (Theorem 2).

A B4 m

500 N.m

C

3 m

IAB = 8 x 106 mm4 IBC = 4 x 10

6 mm4


(Pulau Pinang)

Example 2

Step 1.


Take I = 4 x 106 mm4 then Member AB = 2I and member BC = I

Step 2.


500 N.m

2EI EI

X

M


(Pulau Pinang)

Example 2

Step 3.


t B/A can be determined by calculating the area from M/EI diagram between A and B

dxEI

Mxt

A

B

B/A ∫=

2.5mmEI

2000Nm∆

(2m) (4m)EI

250Nmt∆

BC

3

B

BC

B/AB

==

==

9.06mmEI

7250Nm ∆

(1.5m)(3m)EI

500Nm(5m)(4m)

EI

250Nmt∆

BC

3

C

BCBC

C/AC

==

+

==


(Pulau Pinang)

Example 3

Determine the slope at points C of the beam shown in the figure below. Take E = 200 GPa., I = 360 x 106 mm4

Note:When dealing with unsymmetrical loadings, you may need to combined both theorem 1 & 2.

A4 m

40 kN

C

2 m 2 mB


(Pulau Pinang)

Example 3

Step 1.


Step 2.


ΣMX ↵+ = 0

MX = 10 (X)

∴M0 = 0 kNm #

∴M2 = 20 kNm #

∴M6 = 60 kNm #

MX

X

10 kN

EI

M

X2m 2m4m

EI

20 EI

60

A BC

Since the deflection is actually very small

AB

AB

L

t /=φ

C/A

AB

B/A

C/AC θL

tθθ −=−φ=∴


(Pulau Pinang)

Example 3

Step 3.

Apply Moment Area Theorem (Theorem No. 1 & No. 2)

EI

20kN.m

EI

20kN.m(2m)

2

1 2

=

=C/Aθ

EI

800kN.m

EI

60kN.m(6m)

2

1(6m)

3

12m

EI

60kN.m(2m)

2

1(2m)

3

2

3

=

++

=B/A

t

∫=C

AC/A dx

EI

MθTheorem 1;

EI

M

X

2m 2m4m

EI

20 EI

60

A BC

12 3

C/A

AB

B/A

C θL

tθ −=Since

( ) EI

20kN.m

EI8m

800kN.m 23

−=∴Cθ

0.00111radEI

80kN.m 2

==∴ Cθ

Theorem 2; dxEI

Mxt

A

B

B/A ∫=


(Pulau Pinang)

Example 4

Determine the deflection at points C of the beam shown in the figure below. Take E = 200 GPa., I = 250 x 106 mm4

A8 m

6 kN/m

C8 m

B


(Pulau Pinang)

Example 4

Step 1.


Member AB From Left (0<x<8)

ΣMX ↵+ = 0

MX = -25 (X)

∴M0 = 0 kNm #

∴M8 = -192 kNm #

Member BC From Right (0<x<8)

ΣMX ↵+ = 0

MX = -6(X)(X/2)

MX = -3X2

∴M0 = 0 kNm #

∴M8 = -192 kNm #


(Pulau Pinang)

Example 4

Step 2.


B/AC/AC

C/AC

2tt∆

∆'t∆

−=

−=

=

8

t

16

∆' B/A


(Pulau Pinang)

Example 4

Step 3.


EI

11264kN.mt

EI

192kN.m(8m)

2

18m(8m)

3

1

EI

192kN.m(8m)

3

1(8m)

4

3t

3

C/A

C/A

−=

−

++

−

=

Theorem 2; dxEI

Mxt

A

C

C/A ∫= dxEI

Mxt

A

B

B/A ∫=

EI

2048kN.mt

EI

192kN.m(8m)

2

1(8m)

3

1t

3

B/A

B/A

−=

−

=

0.143mEI

7168-∆

EI

20482

EI

11264∆

C

C

−==

−−−=


(Pulau Pinang)

Review

Problems


(Pulau Pinang)

Problem 1

Calculate displacement and slope at point D of the beam shown in the Figure P1 using the Moment-Area Theorems. Take E = 200 GPa., I = 60 x 106 mm4 .

8 m5 m 7 m

20 kN

A B

C D

Figure P1


(Pulau Pinang)

Problem 2


Figure P2

3 m5 m 5 m

20 kN

A B

C D

7 m

E

20 kN


(Pulau Pinang)

Problem 3


Figure P3

8 m 5 m

20 kN.m

A B

C D

7 m


(Pulau Pinang)

Virtual Work

Method


(Pulau Pinang)

Virtual Work Method

Introduction to Principle of Virtual Work

� The principle of virtual work was developed by John Bernoulli in1717 and is sometimes referred to as the unit-load method. Itprovides a general means of obtaining the displacement and slopeat a specific point on a structure, be it a beam, frame, or truss.

� The concept is based on principle of conservation of energy.

� Consider the external work done by a unit virtual load applied to astructure in equilibrium that moves due to the deformationsassociated with a real-load system. By the principle ofconservation of energy, the external work is equal to the internalstrain energy done by the internal virtual forces under-going realdeformations.


(Pulau Pinang)

Virtual Work Method

Introduction to Principle of Virtual Work

� If we take a deformable structure of any shape or size & apply aseries of external loads P to it, it will cause internal loads U atpoints throughout the structure

� It is necessary that the external & internal loads be related by theequations of equilibrium

� As a consequence of these loadings, external displacement, ∆∆∆∆ willoccur at the P loads & internal displacement, δδδδ will occur at eachpoint of internal loads U

� In general, these displacement do not have to be elastic, & theymay not be related to the loads


(Pulau Pinang)

Virtual Work Method

Principle of Virtual Work

� In general, the principle states that:

Works of External Loads = Works of Internal Loads

∑P∆ = ∑Uδ

� Consider the structure (or body) to be of arbitraryshape as shown.

� Suppose it is necessary to determine thedisplacement ∆ of point A on the body caused bythe “real loads” P1, P2 and P3


(Pulau Pinang)

Virtual Work Method


� It is to be understood that these loads causeno movement of the supports

� They can strain the material beyond the elasticlimit

� Since no external load acts on the body at Aand in the direction of ∆, the displacement ∆,can be determined by first placing on the bodya “virtual” load such that this force P’ acts inthe same direction as ∆.


(Pulau Pinang)

Virtual Work Method


� We will choose P’ to have a unit magnitude,P’ =1

� Once the virtual loadings are applied, thenthe body is subjected to the real loads P1, P2and P3.

� Point A will be displaced an amount ∆causing the element to deform an amount dL

� As a result, the external virtual force P’ &internal load u “ride along” by ∆ and dL &therefore, perform external virtual work of 1.∆ on the body and internal virtual work ofu.dL on the element


(Pulau Pinang)

Virtual Work Method


In a similar manner, if the rotational displacement or slope of the tangent at a pointon a structure is to be determined.

Where

M’ = 1 = external virtual unit couple moment acting in the direction of θ.

uθ = internal virtual load acting on an element in the direction of dL.

θ = external rotational displacement or slope in radians caused by the real loads.

dL = internal deformation of the element caused by the real loads.

1 . θ = ∑uθ . dL

Virtual Loadings

Real Displacements


(Pulau Pinang)

Virtual Work Method

Applying Principle of Work & Energy (Bending)

Based on principle of conservation of energy Ue = Ui compatibly equation can bedeveloped.

therefore OR∫=∆L

0dx

EI

mM1. where∫ θ=θ

L

0dx

EI

Mm1.

Where

1 = external virtual unit load acting on the beam or frame in the stated direction of ∆.

m= internal virtual moment in the beam or frame, expressed as a function of x andcaused by the external virtual unit load.

∆ = external joint displacement of the point caused by the real loadsacting on thebeam or frame.

M = internal moment in the beam or frame , expressed as a function of x and causedby the real loads.

E = modulus of elasticity of a the material.

I = moment of inertia of cross-sectional area, computed about the neutral axis


(Pulau Pinang)

Virtual Work Method


Since Works for External Loads = Works for Internal Loads

Where

P’ = 1 = external virtual unit load acting in the direction of ∆.

u = internal virtual load acting on an element in the direction of dL.

∆ = external displacement caused by the real loads.

dL = internal deformation of the element caused by the real loads.

1 . ∆ = ∑u . dL

Virtual Loadings

Real Displacements


(Pulau Pinang)

Virtual Work Method

Applying Principle of Work & Energy (Axial Force)

Based on principle of conservation of energy Ue = Ui compatibly equation can bedeveloped.

AE

nNL1 ∑=∆.therefore where

Where

1 = external virtual unit load acting on the truss joint in the stated direction of ∆.

n = internal virtual normal force in a truss member caused by the external virtual unitload.

∆ = external joint displacement caused by the real loads on the truss.

N = internal normal force in a truss member caused by the real loads.

L = length of a member.

A = cross-sectional area of a member.

E = modulus of elasticity of a member.


(Pulau Pinang)

Virtual Work Method

Temperature Changes in Trusses

In some cases, truss members may change their length due to temperature. Thedisplacement of a selected truss joint may be written as;

1.∆ = ∑nα∆TL

Where

∆ = External joint displacement caused by temperature change

α = Coefficient of thermal expansion for member

∆T = Temperature changes in member


(Pulau Pinang)

Virtual Work Method

Fabrication Errors in Trusses

Errors in fabricating the lengths of the members of a truss may occur. Trussmembers may also be made slightly longer or shorter in order to give the truss acamber. The displacement of a truss joint from its expected position can be writtenas;

1.∆ = ∑n∆L

Where

∆ = External joint displacement caused by fabrications errors

∆T = Differences in length of member from its intended size as caused byfabrication error


(Pulau Pinang)

Virtual Work Method

Effect of Support Settlements

Settlement∆AE

nNL1.∆ +∑=

Note:+ve if the settlement direction is same with the virtual load direction

-ve if the settlement direction is opposite the virtual load direction.

Settlement

L

0

θ∆dx

EI

Mm1.θ += ∫

Axial Force

Bending

Settlement

L

0∆dx

EI

mM1.∆ += ∫


(Pulau Pinang)

Virtual Work

Method for Beam


(Pulau Pinang)

Example 1

Determine the displacement of point B of the steel beam as shown. Take E = 200 GPa, I = 500 x 106 mm4

A B

10 m

12 kN/m

Step 1.

Draw free body diagram for Real load and Virtual Load.

Real Load, M Vistual Load, m

12 kN/m

10 m

MA

VA

HA

1 kN

10 m

MA

VA

HA


(Pulau Pinang)

Example 1

Step 2.

Calculate Moment equation for the beam withreal load at all salient points

12 kN/m

X

MX

ΣMX ↵+ = 0

MX = -12 (X)(x/2)

∴MX = - 6X2#

1 kN

X

MX

ΣMX ↵+ = 0

MX = -1 (X)

∴MX = - X #

∫=∆L

0dx

EI

mM1.

Step 3:

Calculate Moment equation for the beam withvirtual load at all salient points

Step 4:

Calculate Displacement at point B using virtualwork equation

∫−−

=10

0

2

dxEI

)6xx)((1kN.∆

∫=10

0

3dx6xEI

11kN.∆

)150mm(0.15m∆

))(500x10(200x10

15x101kN.∆

B

66

3

B

↓==

=−

Limit Real Load (M) Virtual Load (m)

0<X<10 – 6X2 – X


(Pulau Pinang)

Example 2

Determine the displacement of point B of the steel beam if support A settle by 10mm. Take E = 200 GPa, I = 500 x 106 mm4

A B

10 m

12 kN/m

Step 1.



12 kN/m

10 m

MA

VA

HA

1 kN

10 m

MA

VA

HA


(Pulau Pinang)

Example 2

Step 2.

Calculate Moment equation for the beam withreal load at all salient points

12 kN/m

X

MX

ΣMX ↵+ = 0

MX = -12 (X)(x/2)

∴MX = - 6X2#

1 kN

X

MX

ΣMX ↵+ = 0

MX = -1 (X)

∴MX = - X #

Step 3:

Calculate Moment equation for the beam withvirtual load at all salient points

Step 4:

Calculate Displacement at point B if supportsettle by 10mm using virtual work equation

Settlement

L

0∆dx

EI

mM1.∆ += ∫

10mmdxEI

)6xx)((1kN.∆

10

0

2

+−−

= ∫

10mmdx6xEI

11kN.∆

10

0

3 += ∫

)160mm(0.01m0.15m∆

10mm))(500x10(200x10

15x101kN.∆

B

66

3

B

↓=+=

+=−


0<X<10 – 6X2 – X


(Pulau Pinang)

Example 3

Determine the slope, θ at point B of the steel beam as shown.

Take E = 200 GPa, I = 60 x 106 mm4

A B

5 m

3 kN

5 m

C

Step 1.



3 kN

10 m

MA

VA

HA

1 kN.m

5 m

MA

VA

HA

5 m


(Pulau Pinang)

Example 3

Step 2.

Calculate Moment equation for the beam with real load at all salient points

Step 3:

Calculate Moment equation for the beam with virtual load at all salient points

3 kN

X MX

ΣMX ↵+ = 0

MX = -3 (X)

∴MX = - 3X #

X

MX

ΣMX ↵+ = 0

MX = 0

∴MX = 0 #

0<x<10

0<x<5 5<x<10

ΣMX ↵+ = 0

MX = 1

∴MX = 1 #

1 kNm

X

MX5


(Pulau Pinang)

Example 3

Step 4.

Calculate Angular Displacement at point B using virtual work equation

∫=L

0

θ dxEI

Mm1.θ

dx EI

(1)(-3x)dx

EI

(0)(-3x)1kNm.θ

10

5

5

0 ∫∫ +=

∫=10

5dx3x -

EI

11kNm.θ

) ( rad 0.009375 rad 0.009375θ

))(60x10(200x10

112.5-1kNm.θ

B

66B

=−=

=−


0<X<5 – 3X 0

5<X<10 – 3X 1


(Pulau Pinang)

Example 4

Determine the vertical displacement of point C of the beam as shown. Take E = 200 GPa, I = 150 x 106 mm4

A4 m

20 kN

C

4 m

8 kN/m

B

Step 1.



8 kN/m

4 mVA

4 m

20 kN

VB

HB

4 mVA

4 m

1 kN

VB

HB


(Pulau Pinang)

Example 4

Step 2.


ΣMX ↵+ = 0

MX = 34 (X) – 8(X2/2)

∴MX = 34X – 4X2#

0<x<4

VA VB

HB

8 kN/m 20 kNΣFY↑+ = 0

VA + VB = 8(4) + 20

∴VA = 34 kN #

4<x<8

ΣMA ↵+ = 0

8VB = 8(4)(2) + 20(4)

∴VB = 18 kN #

ΣFX→+ = 0

- HB = 0

∴HB = 0 #

8 kN/m

X

MX34 kN

8 kN/m

X

MX34 kN

20 kNΣMX ↵+ = 0

MX = 34(X) – 8(4)(X – 2) – 20(X – 4)

∴MX = 144 – 18X #


(Pulau Pinang)

Example 4

Step 3.


ΣMX ↵+ = 0

MX = 0.5(X)

∴MX = 0.5X #

0<x<44<x<8

ΣMX ↵+ = 0

MX = 0.5(X) – 1(X – 4)

∴MX = 4 – 0.5X #

VA VB

HB

1 kN ΣFY↑+ = 0

VA + VB = 1

∴VA = 0.5 kN #

ΣMA ↵+ = 0

8VB = 1(4)

∴VB = 0.5 kN #

ΣFX→+ = 0

- HB = 0

∴HB = 0 #

X

MX0.5 kN MX

1 kNm

X

40.5 kN


(Pulau Pinang)

Example 4

Step 4.

Calculate Vertical Displacement at point C using virtual work equation

∫=L

0dx

EI

mM1.∆

∫∫ +−

=8

4

4

0

2

dxEI

18X)- 0.5x)(144-(4dx

EI

)4x(0.5x)(34X1kN.∆

)14.22mm(0.0142m∆

))(150x10(200x10

426.66671kN.∆

C

66C

↓==

=−

∫∫ +−=8

4

4

0

2 18X)dx- 0.5x)(144-(4EI

1dx )4x(0.5x)(34X

EI

11kN.∆


0<X<4 34X – 4X2 0.5X

4<X<8 144 – 18X 4 – 0.5X


(Pulau Pinang)

Example 5

Determine the slope at points C of the beam shown in the figure below. Take E = 200 GPa., I = 360 x 106 mm4

A4 m

40 kN

C

2 m B2 m

Step 1.



4 mVA 2 m

40 kN

VB

HA

2 m 4 mVA 2 m

1 kN.m

VB

HA

2 m


(Pulau Pinang)

Example 5

Step 2.


Step 3.


Step 4.

Calculate Angular Displacement at point B using virtual work equation

∫=L

0dx

EI

mM1.θ

∫∫∫+

++=8

6

6

2

2

0dx

EI

240)X/8)(-30X(1-dx

EI

X/8)(10X)(1-dx

EI

)(-X/8)(10X1kNm.θ

∫∫∫ +++=8

6

6

2

2

0240)dxX/8)(-30X(1-

EI

1xX/8)(10X)d(1-

EI

1)dx(-X/8)(10X

EI

11kNm.θ

) ( rad 0.001111θ

))(360x10(200x10

801kNm.θ

66

=

=−


(Pulau Pinang)

Example 6

Determine the slope at points B and C of the beam shown in the figure below. Take E = 200 GPa and I = 360 x 106 mm4

CB

5 m

30 kN

5 m

A

Step 1.



3 kN

10 m

MA

VA

HA

1 kN.m

5 m

MA

VA

HA

5 m

Vistual Load, m

1 kN.m

5 m

MA

VA

HA

5 m


(Pulau Pinang)

Example 6

Step 2.


Step 3.


Step 4.

Calculate Angular Displacement at point B and C using virtual work equation

∫=L

0dx

EI

mM1.θ


(Pulau Pinang)

Problem 1

Calculate displacement and slope at point D of the beam shown in the Figure P1 using the Virtual Work Method. Take E = 200 GPa., I = 60 x 106 mm4 .

8 m5 m 7 m

20 kN

A B

C D

Figure P1

Limit Real Load (M) Virtual Load (m∆∆∆∆) Virtual Load (mθθθθ)

0<X<5 15X 0.35X – 0.05X

5<X<13 – 5X + 100 0.35X – 0.05X

13<X<20 – 5X + 100 – 0.65X + 13 – 0.05X + 1


(Pulau Pinang)

Problem 2


Figure P2

3 m5 m 5 m

20 kN

A B

C D

7 m

E

20 kN


0<X<5 20X 0.6X – 0.05X

5<X<8 100 0.6X – 0.05X

8<X<15 100 – 0.4X + 8 – 0.05X + 1

15<X<20 – 20X + 400 – 0.4X + 8 – 0.05X + 1


(Pulau Pinang)

Problem 3


Figure P3

8 m 5 m

20 kN.m

A B

C D

7 m


0<X<8 – 10X2 + 128X 0.25X – 0.05X

8<X<15 – 32X + 640 0.25X – 0.05X

15<X<20 – 32X + 640 – 0.75X + 15 – 0.05X + 1


(Pulau Pinang)

Virtual Work

Method for Frame


(Pulau Pinang)

Example 7

Determine the horizontal displacement of point C of the beam as shown. Take E = 200 GPa, I = 150 x 106 mm4

C

B

5 m

15 kN

8 m

A

4 m

D

Step 1.



HA

15 kN

MA

VA

HA

1 kN

MA

VA


(Pulau Pinang)

Example 7

Step 2.

Calculate Moment equation for the frame with real load at all salient points

15 kN

X MX

ΣMX ↵+ = 0

MX = -15 (X)

∴MX = - 15X #

ΣMX ↵+ = 0

MX = - 15 (4)

∴MX = - 60 kNm #

0<x<4

0<x<5

0<x<8

ΣMX ↵+ = 0

MX = 15(X – 4 )

∴MX = 15X – 60#

15 kN

X

MX

4 m

15 kN

X

MX

4 m


(Pulau Pinang)

Example 7

Step 3.

Calculate Moment equation for the frame with virtual load at all salient points

ΣMX ↵+ = 0

MX = - 1 (X)

∴MX = - X #

0<x<5

0<x<8

ΣMX ↵+ = 0

MX = - 1(5)

∴MX = - 5 kNm#

X

1 kN

MX

4 m

X

MX

4 m1 kN

5 m


(Pulau Pinang)

Example 7

Step 4.

Calculate Horizontal Displacement at point C using virtual work equation

∫=L

0dx

EI

mM1.∆

∫∫∫ ++=8

0

5

0

4

0dx

EI

60)(-5)-(15Xdx

EI

(-60)(-X)dx

EI

(-15x)(0)1kN.∆

)25.0mm(0.025m∆

))(150x10(200x10

7501kN.∆

C

66C

←==

=−

++= ∫∫∫

8

0

5

0

4

060)(-5)dx-(15Xx(-60)(-X)dx(-15x)(0)d

EI

11kN.∆


0<X<4 – 15X 0

0<X<5 – 60 – X

0<X<8 15X – 60 – 5


(Pulau Pinang)

Example 8

Determine the vertical displacement at points C of the two-member frame shown in the figure below. Take E = 200 GPa., I = 160 x 106 mm4

CB

5 m

40 kN

2 m

A

3 m

20 kN/m

60 0


(Pulau Pinang)

Example 8

Step 1.



CB

5 m

40 kN

2 m

A

3 m

20 kN/m

60 0

VA

HA

VB

C

B

5 m

1 kN

2 m

A

3 m

60 0

VA

HA

VB


(Pulau Pinang)

Example 8

Step 2.


MX

x

103 MX

x

103

40 kN

Member BC (0<x<3)

ΣMX ↵ + = 0

MX - 103 (x)= 0

∴∴∴∴ MX = 103.0x #

ΣMX ↵ + = 0

MX + 40(x-3) - 103 (x)= 0

∴∴∴∴ MX = 63x + 120 #

Member BC (3<x<5)

173.21

63

MX

x ΣMX ↵ + = 0

- MX + 173.21 Sin 600(x) – 63 Sin 300(x) - 15 (x)(x/2)= 0

∴∴∴∴ MX = 118.50X - 7.5X2#

Member AB (0<x<10)


(Pulau Pinang)

Example 8

Step 3.


MX

x

0.5

Member BC (0<x<5)

ΣMX ↵ + = 0

MX - 0.5 (x)= 0

∴∴∴∴ MX = 0.5x #

0

0.5

MX

x

ΣMX ↵ + = 0

- MX + (0.5) X Cos 600(x)= 0

∴∴∴∴ MX = 0.25X #

Member AB (0<x<10)


(Pulau Pinang)

Example 8

Step 4.

Calculate Vertical Displacement at point C using virtual work equation

∫=L

0dx

EI

mM1.∆

∫∫∫+

++=5

3

3

0

10

0

2

dxEI

120)(0.5X)(63Xdx

EI

X)(0.5X)(103dx

EI

)7.5X-8.50X(0.25X)(111kN.∆

)22.375mm(0.22375m∆

))(160x10(200x10

71601kN.∆

C

66C

↓==

=−

∫∫∫ +++=5

3

3

0

10

0

2 120)dx(0.5X)(63XEI

1X)dx(0.5X)(103

EI

1)dx7.5X-8.50X(0.25X)(11

EI

11kN.∆


0<X<10 118.50X – 7.5X2 0.25X

0<X<3 103X 0.5X

3<X<5 63X + 120 0.5X


(Pulau Pinang)

Example 9

Determine the slope at points C of the two-member frame shown in the figure below. The support at A is fixed. Take E = 200 GPa., I = 235 x 106 mm4

CB

3.6 m

30 kN/m

60 0

A


(Pulau Pinang)

Example 9

Step 1.



CB

3.6 m

30 kN/m

60 0

HAMA

VA

CB

3.6 m

1 kN.m

60 0

HAMA

VA


(Pulau Pinang)

Example 9

Step 2.


Step 3.


Step 4.

Calculate Angular Displacement at point C using virtual work equation

∫=L

0dx

EI

mM1.θ

∫∫ +=3.6

0

23

0dx

EI

)(1)(-15Xdx

EI

356.4)(1)-(54X1kNm.θ

) ( rad 0.0225θ

))(235x10(200x10

1059.48-1kNm.θ

66

=

=−


0<X<3 54X – 356.4 1

0<X<3.6 – 15X2 1


(Pulau Pinang)

Review

Problems


(Pulau Pinang)

Problem 1

Calculate vertical displacement and slope at point D of the frame shown in the Figure P1 using the Virtual Work Method. Take E = 200 GPa., I = 60 x 106 mm4 .

Figure P1

A

D

B

4 m

6 m

C

15 kN/m

3 m

3 m

1 m

3 m


0<X<5 17.925X

5<X<10 –12.075X + 150

0<X<5 21X

5<X<8 –7.5X2 + 96X–187.5

0<X<6 –11.875X + 100.5


(Pulau Pinang)

Problem 2

Figure P2 shows a rigid-jointed frame that issubjected to wind loads. The wind load istransferred to the members at the girts andpurlins from the roof segments (AB) and simplysupported wall (BC). The frame is rollersupported at A and pinned at C. Using the Virtualwork method, determine horizontal displacementat B.

Take E = 200 GPa., I = 60 x 106 mm4 .

15 k

N/m

4 m

2 m

Figure P2

2 m30 0

4 m

A

B

C

Girts

Purlins


(Pulau Pinang)

Problem 3

Calculate vertical displacement, horizontal displacement and slope at point C of the frame shown in the Figure P3 using the Virtual Work Method.

Take E = 200 GPa., I = 60 x 106 mm4 .

Figure P3

40 kN

A

D

B

C

20 kN/m

3 m 3 m 6 m

8 m


(Pulau Pinang)

Virtual Work

Method for Truss


(Pulau Pinang)

Example 1

The cross-sectional area of each member of the truss is A = 400 mm2 and E = 200 GPa.

1. Determine the vertical displacement of joint C if a 8-kN force is applied to the truss at C.

2. If no loads act on the truss, what would be the vertical displacement of joint C if member AB were 5 mm too short?

8 kN


(Pulau Pinang)

Example 1

Step 1.

Calculate all member forces of the truss

8 kN

4 m 4 m

3 m

A B

C

VA VB

HA

ΣMA ↵+ = 0

8VB = 8 (3)

∴VB = 3 kN #

ΣFX→+ = 0

8 - HA= 0

∴∴∴∴ HA = 8 kN #

ΣFY↑+ = 0

VA + VB = 0

∴∴∴∴VA = - 3 kN #

A

-3 kN

8 kN

FAC

FAB

B

3 kN

FBC

FAB

ΣFY↑+ = 0

-3 + 0.6FAC = 0

∴∴∴∴FAC = 5 kN #

ΣFX→+ = 0

FAB + 0.8FAC – 8 = 0

∴∴∴∴ FAB = 4 kN #

ΣFY↑+ = 0

3 + 0.6FBC = 0

∴∴∴∴FBC = - 5 kN #


(Pulau Pinang)

Example 1

Step 2.

Apply 1 kN unit vertical load at C Calculate all member forces

1 kN

4 m 4 m

3 m

A B

C

VA VB

HA

ΣMA ↵+ = 0

8VB = 1 (4)

∴VB = 0.5 kN #

ΣFX→+ = 0

HA= 0

∴∴∴∴ HA = 0 kN #

ΣFY↑+ = 0

VA + VB = 1

∴∴∴∴VA = 0.5 kN #

A

0.5 kN

0 kN

FAC

FAB

B

0.5 kN

FBC

FAB

ΣFY↑+ = 0

0.5 + 0.6FAC = 0

∴∴∴∴FAC = -0.833 kN #

ΣFX→+ = 0

FAB + 0.8FAC = 0

∴∴∴∴ FAB = 0.667 kN #

ΣFY↑+ = 0

0.5 + 0.6FBC = 0

∴∴∴∴FBC = - 0.833 kN #


(Pulau Pinang)

Example 1

Step 3.

Apply Virtual Work Method in Tabulation formAE

nNL1 ∑=∆.

Member N (kN) n (kN) L (m) nNL(kN2.m)

AB 4 0.667 8 21.344

AC 5 -0.833 5 -20.825

BC -5 -0.833 5 20.825

Total = 21.344

)0.267mm(m2.668x10∆

)kN/m)(200x10m(400x10

.m21.344kN1kN.∆

4CV

2626

2

CV

↓==

=

−

−


(Pulau Pinang)

Example 1

Step 4.

If no loads act on the truss, what would be the vertical displacement of joint C if member AB were 5 mm too short?

Ln1 ∆∑=∆.

)3.335mm(m3.335x10∆

0.005m)(0.667kN)(1kN.∆

∆Ln1kN.∆

3VC

VC

ABABVC

↑=−=

−=

∑=

−

Step 5.

If applied loads act on the truss is considered, and member AB were 5 mm too short, then vertical displacement at C is,

)3.068mm(3.335mm0.267mm∆

LnAE

nNL1.∆

VC

VC

↑=−=

∆∑+∑=


(Pulau Pinang)

Example 2

A pin-jointed plane truss ABCDE, pinned supported at A and E as shown. The trussis subjected to a vertical concentrated load of 10 kN at B and 15 kN at C.

1. Used Method of Virtual work anddetermine the vertical deflection at C.Member EB has been fabricated 5mmtoo short. Take E = 200 GPa.

2. Remove the loads on the truss anddetermine the vertical displacement ofpoint B if members AB and BCexperienced a temperature increase of∆T = 1100C. Take E = 200GPa and α =1.8 x 10-6/ 0C

3. Remove the loads on the truss anddetermine the vertical displacement ofpoint B if member EB is fabricated 19mmtoo long.

15 kN

1.2 m

1.2 m

A

B C10 kN

DE

1.2 m

1200 mm2

1800 mm2 1800 mm2

1800 mm2


(Pulau Pinang)

Example 2

Step 1.

Used Method of Virtual work and determine the vertical deflection at C. Member EB has been fabricated5mm too short. Take E = 200 GPa.

Real Load FBD Virtual Load FBD

15 kN

1.2 m

1.2 m

A B C

10 kN

D

E

1.2 m

HA

VA

HE

VA

HA

1 kN

1.2 m

1.2 m

A B C

D

E

1.2 mVA

HE

VA


(Pulau Pinang)

Example 2

Step 2.

Tabulate the member forces (Real loads and Virtual Loads) in tabulation form.

Member L (m) E(kN/m2) A(m2) N (kN) n (kN) nNl/EA(kN.m)

)6.063mm(m6.063x10∆

)0(1.414x5x1kN.m1.007x101kN.∆

3-VC

33

VC

↑=−=

−= −−

LnAE

nNL1.∆VC ∆∑+∑=

AB 1.2 200X106 1.8X10-3 -40 -2 0.000267

BC 1.2 200X106 1.8X10-3 -15 -1 0.00005

BD 1.2 200X106 1.8X10-3 -15 -1 0.00005

BE 1.697 200X106 1.2X10-3 35.355 1.414 0.000353

CD 1.697 200X106 1.2X10-3 21.213 1.414 0.000212

DE 1.2 200X106 1.2X10-3 15 1 0.000075

Total = 0.001007


(Pulau Pinang)

Example 2

Step 3.

Remove the loads on the truss and determine the vertical displacement of point B if members AB and BCexperienced a temperature increase of ∆T = 1100C. Take E = 200GPa and α = 1.8 x 10-6/ 0C

Virtual Load FBD

HA

1 kN

1.2 m

1.2 m

A B C

D

E

1.2 mVA

HE

VATLn1 ∆α∑=∆.

BCBCABABBV TLnTLnkN ∆∑+∆∑=∆ αα.1

)0.2376mm(m2.376x10∆

0))(110)(1.21)(1.8x10(1kN.∆4

BV

6BV

↑=−=

+−=−

−


(Pulau Pinang)

Example 2

Step 4.

Remove the loads on the truss and determine the vertical displacement of point B if member EB isfabricated 19mm too long.

Virtual Load FBD

HA

1 kN

1.2 m

1.2 m

A B C

D

E

1.2 mVA

HE

VALn∆∑=∆.1

EBEB LnkN ∆∑=∆.1

)26.87mm(∆

19mm)(1.414kN)(1kN.∆

BV

BV

↓=

=


(Pulau Pinang)

Review

Problems


(Pulau Pinang)

Problem 1

A pin-jointed plane truss ABCDE, pinned supported at A and roller supported at E asshown. The truss is subjected to a vertical concentrated load of 40 kN andhorizontal concentrated load of 10 kN at C.

1. Used Method of Virtual work and determine

the vertical deflection at D when member AD

has been fabricated 5mm too short and

member BD experienced a temperature

increase of ∆T = 1100C. Take E = 200 GPa and

α = 1.8 x 10-6/ 0C

2. Remove the loads on the truss and determine

the horizontal displacement of point B if

members AB and BC experienced a

temperature increase of ∆T = 1100C. Take E =

200GPa and α = 1.8 x 10-6/ 0C

3. Remove the loads on the truss and determine

the horizontal displacement of point B if

member BD is fabricated 19mm too long.

40 kN

4 m4 m

A

B

C10 kN

D

E

3 m

1200 mm2 1200 mm21200 mm2

4 m

1

1

1200 mm2


(Pulau Pinang)

Problem 2

A pin-jointed plane truss ABCD, pinned at A and supported on rollers at D as shown.The truss is subjected to a uniformly distributed load of 5 kN/m acting verticallydownward on member BC and a horizontal concentrated load of 20 kN at B. AE isconstant for all members.

1. Determine the horizontal displacement ofthe truss at C using virtual work.

2. If in addition to the loads shown, memberBD is cooled 300C, re-calculate thehorizontal displacement at C. Given thethermal expansion coefficient, α = 1x10-

5/0C and the axial rigidity, AE = 12,000 kN.A

B

C

20 kN

D

3 m4 m

30 0

60 0


(Pulau Pinang)

Problem 3

A pin-jointed plane truss ABCDE, pinned at A and B as shown. The truss issubjected to 450 inclined loads at D and E with 10 kN and 20 kN respectively. GivenA = 1800 mm2 and E = 200 GPa.

1. Used Method of Virtual work and determine

the Determine the horizontal displacement of

the truss at E using virtual work.

2. Comment for the horizontal displacement at E

when the member CE is fabricated 10 mm too

long, while the external loads are still in place.

20 kN

1.8 m1.8 m

A B

C

10 kN

D

E

2.1 m

2.1 m

45 0

45 0


(Pulau Pinang)

Hibbeler R.C, (2012), “Structural Analysis, 8th Edition in S.I. Units”, Pearson,Singapore



UiTM Structural Division (2003), “Basic Structural Analysis”, Cerdik PublicationsSdn. Bhd.

UiTM Structural Division (2003), “Basic Structural Mechanics”, Cerdik PublicationsSdn. Bhd.

Mc Cormac N., (1999), “Structural Analysis”, 2nd Edition, John Wiley and Sons.

References

CES424 - Topic 6 (Deformation)

Documents

determinate structures

elastic deflections

theelastic curve

figure b

determinate structurestopic

pulau pinangces

figure d

figure f