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Chapter 8Chapters (8 & 9) Deflection• Chapter 8. Conjugate beam method

• Chapter 9. Virtual Work Method

Iqbal Marie

2018-2019

Hibbeler, R. C., Structural Analysis, 7th Edition,

8.1 Deflection of Beams and Elastic Curve

ASSUMPTIONLinear Elastic Material Response: ( A structure subjected to a load will return to its original un-deformed position after load is removed )

• Conjugate Beam• Virtual Work Method

METHODS to be considered in this course are:

Causes of deformation for structures are the internal forces:

• Axial forces for trusses

• Bending moment for beams

Deflections of structures can occur from:• loads,

• temperature,

• fabrication errors or

• settlement

In designs, deflections must be limited in order to prevent cracking of attached brittle materials such as concrete, glass and plaster and provide integrity and stability of roofs.

• A structure must not vibrate or deflect severely for the comfort of occupants.

• Deflections at specified points must be determined if one is to analyze statically indeterminate structures

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Deflected Shapes Depends on Support Conditions

Moment Diagrams are a Good Indication for the shape of the elastec cure ( deflection curve)

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Example 8.1 Draw the deflected shape ( elastic curve) of each of the beams and frames.

8.5 Conjugate Beam MethodDeveloped by H. Muller Breslau - 1865

Used to find slopes and deflection due to bending of beams . It is based on principles of Statics only

Internal Loadings Beam Theory

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A “fictitious” beam of the same length as the real beam loaded with the real beam’s M/EI diagram…

Real Beam = V Conjugate Beam

V Real Beam = M Conjugate Beam

Procedure for Analysis

Draw the conjugate beam of the real one with suitable supports as shown in the table

The conjugate beam is loaded with the M/EI diagram derived from the load w on the real beamFrom the above comparisons, we can state 2 theorems related to the conjugate beam

Theorem 1The slope at a point in the real beam is numerically equal to the shear at the corresponding point in the conjugate beam

Theorem 2The disp. of a point in the real beam is numerically equal to the moment at the corresponding point in the conjugate beam

When drawing the conjugate beam, it is important that the shear & moment developed at the supports of the conjugate beam account for the corresponding slope & displacement of the real beam at its supports

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Examples of Conjugate Beam Supports

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http://ocw.nthu.edu.tw/ocw/upload/8/258/Chapter_9-98.pdf

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Find the deflection at Point D

https://www.google.com/search?q=conjugate+beam+solve+example&tbm=isch&tbo=u&source=univ&sa=X&ved=2ahUKEwi_hebxu-_fAhXwhaYKHbGkDXEQsAR6BAgDEAE&biw=1600&bih=758#imgrc=oF_z46H3ZCCXaM:

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Chapter 8: Deflections

Determine the deflection of the steel beam at point C. The reactions have been computed. Take E = 200GPa, I = 60(106)mm4

C

+M/EI

18/EI

EIC =-162C = -162/(200E9 x 60E-6) = 0.0135 mm

MC + 45(9) -81(3)= 0 MC = -162kN.m

Prob. 8.29 use the conjugate beam method to determine the displacement at C and slope at C

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Chapter 9

Deflection

Using Energy Method To Calculate Slope and Deflection

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12

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2. Temperature

TLadL

LTna1

member L n T L nT L

Sum

If any members undergoes an increase in temperature, T will be positive, whereas a decrease in temperature results in a negative value

3. Fabrication Errors

udL1

VirtualLoads

Real Displ.

n

When a fabrication error increase in length of a member, L is positive, whereas a decrease in length is negative

member L n L n L

Sum

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4. Combined Effects ( external Load + temperature+ fabrication error)

LnLTnaAE

nNL1

Fabrication error

Ln

Example 9.12 The cross sectional area of each member of the truss show, is A = 400mm2

and E = 200GPa.Determine the vertical displacement of joint C

A virtual force of 1 kN is applied at C in the vertical direction

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1. Support reactions

2. Using the method of joints to determine the force in each member ( N), due to the applied loads

3. Apply the virtual load at the point of interest in the desired direction. Since the deflection at point G is required . Therefore, apply a unit load at point G. Then find reaction

35 k

25 k

Member

n(k) N(k) L(in) AE (in2-ksi)nNL/AE

(in-k)

AB -0.67 -33.33 48 58000 0.0184

BC -0.67 -33.33 48 58000 0.0184

CD -0.67 -46.66 48 58000 0.0257

DE -0.67 -46.66 48 58000 0.0257

AF 0.83 41.67 60 58000 0.0359

BF 0 -10 36 58000 0

CF -0.83 -25 60 58000 0.0216

FG 1.33 53.33 48 58000 0.0589

CG 1 0 36 58000 0

CH -0.83 -8.33 60 58000 0.0072

GH 1.33 53.33 48 58000 0.0589

DH 0 -30 36 58000 0

HE 0.83 58.33 60 58000 0.0503

Total 0.3209

Fill the results in table.

Determine the vertical deflection of point G

Determine the vertical deflection at joint C due to temperature drop of 8Cin members AB and BC and a temperature increase of 30C in members AF, FG, GH and EH ( = 1.2(10-5 )

N N

N

N

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Determine the vertical deflection at joint D if member CF is 15 mm too longer and member EF is 10 mm too short.

N N

N

L

dxxEI

xMxm0 )(

)()(1

1Slope at A

Deflection at A

9.5 Virtual Work for beams and Frames

Clockwise negative and counterclockwisepositive

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The integration to solve for the displacement or rotation can be carried out using either direct integration or by utilizing a visual integration method. With direct integration, the equations of M and m for each segment of the structure must be developed for use in the equation,

An alternative to this approach is to construct the moment diagrams by utilizing the following relationship

L

dxxEI

xMxm0 )(

)()(1

Where n is the number of segments in the M diagram. The segments are selected and numbered to simplify the integration of equation. A is the area of the moment diagram of each segment and h is the respective height of the m diagram at the centroid of each segment of the moment diagram, M. By using that the calculation of deflections and rotations becomes a simple matter of addition rather than integration. Or use (Mm) integration tables

IMPORTANT NOTES:

In performing the integration using visual integration, the following rules must be observed.

• Construct the moment diagram (M) due to the applied loads on the structure.• Divide the moment diagram, M, to segments that you can easily be able to calculate the

area and locate the center of each segment . Calculate the area and locate the center of each segment on the M-diagram. Project the location of the center of each area on the m-diagram.

• Draw the m-diagram due to a virtual load ( UNIT LOAD for displacement OR UNIT MOMENT for rotation). This load is applied at the point of interest and in the direction of which a displacement is to be calculated. Measure the height, hi, on the moment diagram of the virtual load.

• Both moment diagrams must be continuous over the length over which the integration being performed.

• If the moment diagram due the applied loads or the moment diagram due to the virtual load is not continuous, one MUST divide the integration into segments, each of which is continuous over the integration length.

• You can use tables for Mm integration tables

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Determine the vertical displacement at end C of the beam

1. calculate the support reactions

2. Moment diagram using superposition (M)

3. Apply the virtual load at the point of interest in the desired direction

4. calculate the support reactions with unit load applied

5. draw virtual moment diagram (m)

.

Since EI is constant throughout the structure, the total deflection at C = -1008 /EI.

the deflection is upward

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Determine the horizontal displacement at D of the frame M

Moment diagram due to 20 k load

Moment diagram due to 1 k/ft load

EI is constant throughout the structure, the total horizontal deflection at D equals -44296.875/EI

deflection is to the right

Moment diagram due to the unit load at D,

m

+

20

412

40

1

M

1.50.5

5

1/6*40*5*10 *123/ (29E3 * 53.8) = 0.369 inDeflection at C =

m

Example: 9.4 Determine the displacement at point B of a steel beamE = 200 GPa , I = 500(106) mm4

B = (600x10x10)/(4x200x106x500x10-6) =.15 m

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Example 9.6 Determine the slope at point B of the steel beam shown .

E = 200 GPa, I = 60(106)mm4.

B = (0.5x-1(-30-15)x5)/( 200(106 ) x60 x10-6) =0.00938 rad

Example: Determine the horizontal displacement at A

= .5(100x5x-5)/(200E6x200E6x10-12)= - 0.03125 m (- ) Means opposite to the assumed virtual load

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Example 9-8 Determine the horizontal displacement of point C on the frame. E = 200GPa, I= 235x106 mm4 for both members

112.5

112.5

60 k

N/m

180

C =[1/3(270x3x2.4) + 5/12(270x3x3)]/ (200x106 x 235x10-6 ) = 0.035 m