DCC20053

MECHANICS OF CIVIL

ENGINEERING

STRUCTURES

GROUP MEMBER’S NAME :

FATIN HUSNA BINTI MUSTAPHA (04DKA20F2027)

SITI NUR ATHIRAH BINTI MAHUSIN (04DKA20F2111)

LECTURER’S NAME :

PUAN NORHAIZAH BINTI AMBIAH

CONTENTS

❑ INTRODUCTION

❑ DEFINTION

❑ EXAMPLE QUESTION

❑ CONCLUSION

❑ REFERENCES

TOPIC 6

SLOPE AND DEFLECTION OF

BEAM DUE TO SYMMETRICAL

BENDING

INTRODUCTION

• The slope deflection method is a structural analysis method for

beams and frames introduced in 1915 by George A. Maney. The

slope deflection method was widely used for more than a decade

until the moment distribution method was developed.

• There are many methods to find out the slope and deflection at a

section in a loaded beam

• The amount of deflection depends on :-

• The magnitude of the load, and

• The stiffness of the member

MACAULAY METHOD DEFINTION MOMENT AREA METHOD

6.1 MACAULAY METHOD

❑ It is thus obvious, that after first integration the original, differential

equation, we get the value of slope at any point

❑ On further integrating, we get the value of deflection at any point

❑ The bending moment at a point:

M = EI d2y

dx2

❑ Integrating the above equation,

EI dy = ∫ M → slope equation dx

❑ And integrating the above equation once again,

EI.y = ∫∫ M → deflection equation

6.1.1 GUIDELINE TO USE MACAULAY METHOD

1. Find the reaction at support

2. Start x from the left to the right beam

3. Draw a cutting plane x-x across the beam before reach the end of beam at free body

diagram

4. Write the equation for the bending moment at x-x by considering the portion of the beam

from the origin to x-x.

5. Integrate once; Integrate second time.

6. Two boundary conditions:

i. For the simply supported beam

a) At the x-values of the two supports, deflection is zero, i.e. y = 0.

b) If the point (i.e the x-value) of maximum deflection is known, then at the

x-value of the point, the slope is zero, i.e dy/dx = 0.

iii. For the cantilever beams, at the x-value of the built-in end:

a) The deflection is zero, i.e. y = 0

b) The slope is zero, i.e. dy/dx = 0

7. Substituting the values of C1 and C2 in equations and we get the general equations.

6.1.2 MOMENT EQUATION AT SECTION – POINT LOAD

P M

aX

Mx = R[x] – P [x – a] 1

6.1.3 MOMENT EQUATION AT SECTION – MOMENT LOAD

M1 M2 Mx Mx = R[x] – M1 [x – a] 0 + M2 [x – b] 0

a

b

X

6.1.4 MOMENT EQUATION AT SECTION –UNIFORMLY DISTRIBUTED

LOAD

(i) W W M

M

(ii)

a X

R a

X

R

Mx = Rx – w [x – a]2 Mx = Rx – w [x]2 + w [x – a]2

2 22

(iii) W M

a b Mx = Rx – w [x – a]2 + w [x – b]2

R X 22

EXAMPLE QUESTION OF MACAULAY METHOD

A horizontal beam which is simply supported beam at its end, A and B, have a uniform cross section and is 8m

long. Concentrated load of 10 KN act 2m from A. Determine the slope and deflection of beam at point C.

10 KN

HA 2m 3m 3m B

VA

VA : MB = 0

A

(VA x 8) – (10 x 6) = 0

VA = 7.5 KN

10 KN

HA

VA 3m 3m X (

2m

Moment equation = Mx = 7.5(x) – 10(x – 1)

Macaulay equation = EI 2 = 7.5[x] – 10[x – 1]

2

= EI 2 = 7.5[x]1– 10[x – 1]

2 1

Slope equation : EI = 7.5[ ]2 - 10[ − 1]2 + 1

2 2

Deflection equation : EIY = 7.5[ ]3 - 10[ − 1]3 + 1 + 2

6 6

When x = 0 m and y = 0 , replace at deflection equation

EIY = 7.5[ ]3 - 10[ − 1]3 + 1 + 2

6 6

EI(0) = 7.5[0]3 - 10[0 − 1]3 + 1[0] + 2

6 6

2 = 0

When x = 8 m and y = 0 , replace at deflection equation (2)

EI(0) = 7.5[8]3 - 10[8 − 1]3 + 1[8] + 0

6 6

1 = 8.54

When 1 = -8.54 , replace at slope equation (1)

EIY = 7.5[ ]3 - 10[ − 1]3 - 8.54

2 2

When 1 = -8.54 and 2 = 0 , replace at deflection equation (2)

EIY = 7.5[ ]3 - 10[ − 1]3 + 12 ( 4)

6 6

When = 5 at point C , replace at slope equation (3)

EI = 7.5[ ]2 - 10[ − 1]2 - 8.54

2 2

EI = 7.5[5]2 - 10[5 − 1]2 - 8.54

2 2

= 5.21 radian

When = 5 at point C , replace at derflection equation (4)

EIY = 7.5[ ]3 - 10[ − 1]3 - 8.54 [x] (equation 4)

6 6

EIY = 7.5[5]3 - 10[5 − 1]3 - 8.54 [5]

66

Y = 6.88 m

6.2 MOMENT AREA METHOD

This is a method of determining the change in slope or the deflection between two points on a

beam. It is expressed as two theorems.

Theorem 1

If A and B are two points on a beam the change in angle (radians) between the tangent at A and the

tangent at B is equal to the area of the bending moment diagram between the points divided by the

relevant value of EI (the flexural rigidity constant).

Theorem 2

If A and B are two points on a beam the displacement of B relative to the tangent of the beam at A

is equal to the moment of the area of the bending moment diagram between A and B about the

ordinate through B divided by the relevant value of EI (the flexural rigidity constant).

Undeflected beam Tangent at B

y

L F

x A

B

A B VB YB

tBA θB

ΔθAB B

C

Elastic curve θA

Tangent at A

Figure 3.3 (a) : Slope and deflection for simply supported beam Figure 3.3(b): Slope and deflection for cantilever

beam

6.2.1 GUIDELINE TO USE MOMENT AREA METHOD

1. Find reaction at each support. Neglect reaction at right end for fixed

support.

2. Sketch possibility deflection curve. This curve must consistent with

condition at support eg. Deflection at support equal to zero.

3. Draw the tangent for deflection curve.

4. Draw bending moment for each load.

5. Apply Theorem 1 to determine the angle between any two tangents on

the elastic curve and Theorem 2 to determine the tangential deviation.

Area and Centroid Area, A Centroid, h

Shape bh M

b Fxb

h

WL2

b 2

h

b

h

Note:

M = Moment

F = Point Load

W = Uniformly Distributed Load

REFERENCES

✓https://youtu.be/YqxJlNZBILA

✓https://pkb.cidos.edu.my/

CONCLUSION

At the end of this title, we will be able to learn

macaulay and moment area method. In

addition, we will be able to solve the questions

of maucaulay and moment area method. In

addition, we can learn to work together

between our group members