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CHAPTER 7 DEFLECTIONS OF BEAMS

• Determine the deflection and slope at specific points on beams and shafts, using various analytical methods including:

o The integration method

o The use of discontinuity functions (McCaulay)

o The virtual unit-load method

OBJECTIVES

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• Deflection is a result from the load action to the beam (self weight, service load etc.)

• If the deflection value is too large, the beam will bend

and then fail. Therefore it is vital that deflection must be limited within the allowable values as stipulated in the Standards

• The theory and background of deflection comes from ‘curvature’

INTRODUCTION

• The deflection diagram of the longitudinal axis that passes through the centroid of each cross-sectional area of the beam is called curvature or elastic curve, which is characterized by the deflection and slope along the curve

CURVATURE

P

P

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• Moment-curvature relationship:

o Sign convention:

ELASTIC CURVATURE

From the figure , if DE = L ; AB = DE = L 𝐿′ = 𝐴′𝐵′ = 𝑅𝜃 − 𝑅 − 𝑦 𝜃 Displacement, 𝛿 = 𝐿 − 𝐿′ = 𝑅𝜃 − 𝑅 − 𝑦 𝜃 = y

From strain, 휀 =𝛿

𝐿=𝑦𝜃

𝑅𝜃=𝑦

𝑅

Therefore, curvature: 1

𝑅=𝑦

In elastic region; 𝐸 =𝜎

휀 =𝜎

𝐸

CURVATURE

y

R

L

C L’ A’

D

(R – y)

B’

E

A

B

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It is known that 𝜎 =𝑀𝑦

𝐼

Therefore : 1

𝑅=𝑦=

𝜎

𝐸𝑦=𝑀𝑦

𝐸𝐼𝑦

𝟏

𝑹=𝑴

𝑬𝑰

where: EI = Stiffness or Flexure Rigidity (The higher the EI value, the stiffer the material the smaller the curvature)

CURVATURE

• Deflection is influenced by I, E and L (and load)

• From 1

𝑅=𝑀

𝐸𝐼 …………. (1)

• Kinematic relationship between radius of curvature R and location

x: 1

𝑅=

𝑑2𝑦 𝑑𝑥2

1 +𝑑𝑦𝑑𝑥

2 3/2

• But, in the case of elastic curve, the slope (dy/dx) is too small ( 0)

and can be ignored. Then: 1

𝑅=𝑑2𝑦

𝑑𝑥2 …………. (2)

SLOPE & DEFLECTION BY DOUBLE INTEGRATION

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• Substitute Eq. (1) into Eq. (2): 𝑀

𝐸𝐼=𝑑2𝑦

𝑑𝑥2 or 𝐸𝐼

𝑑2𝑦

𝑑𝑥2= 𝑀

Elastic curve differential equation (Moment Equation)

• After integration:

𝐸𝐼𝑑𝑦

𝑑𝑥= 𝐸𝐼𝜃 = 𝑀𝑑𝑥 + 𝐶1𝑥

(Slope Equation)

• Double Integration will produce:

𝐸𝐼𝑦 = [𝑥𝑀𝑑𝑥]𝑑𝑥 + 𝐶1𝑥 + 𝐶2𝑥 (Deflection Equation)

where C1 and C2 are the constant to be determined from the boundary conditions

SLOPE & DEFLECTION BY DOUBLE INTEGRATION

• The integration constants can be determined by imposing the boundary conditions, or continuity condition at specific locations

• 3 beam cases are considered:

I. Simply Supported Beam

BOUNDARY CONDITIONS

L

x

y P

x = 0; y = 0 x = L; y = 0

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II. Overhanged Beam

III. Cantilever Beam

BOUNDARY CONDITIONS

L

x

y P

x = L; y = 0 x = 0; y = 0

L1

L

x

y

P

x = 0; y = 0

x = 0; = 0

• Deflection: + -

• Slope: + - • Bending moment:

SIGN CONVENTIONS

+ve -ve

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Determine the mid-span deflection of beam shown below. Given E = 20 kN/mm2 and I = 1600 x 106 mm4.

EXAMPLE 1 (Double Integration)

x

y

18 kN

6 m

2 m 4 m

A B C

Determine the Reaction Forces at A and C

Taking moment at C; MC = 0

VA (6) – 18(4) = 0

VA = 12 kN

Fy = 0

VA + VC = 0

VC = 6 kN

EXAMPLE 1: Solution

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Segment AB (0 x 2) Segment BC (2 x 6)

x

M A

12 B

18

EXAMPLE 1: Solution

x

M A

12

𝑀 = 12𝑥

𝐸𝐼𝑑2𝑦

𝑑𝑥2= 𝑀 = 12𝑥

𝐸𝐼𝑑𝑦

𝑑𝑥=12𝑥2

2+ 𝐶1

𝐸𝐼𝑦 =12𝑥3

6+ 𝐶1𝑥 + 𝐶2

𝐸𝐼𝑦𝐴𝐵 = 2𝑥3 + 𝐶1𝑥 + 𝐶2 ….(1)

𝑀 = 12𝑥 − 18 𝑥 − 2 = −6𝑥 + 36

𝐸𝐼𝑑2𝑦

𝑑𝑥2= 𝑀 = −6𝑥 + 36

𝐸𝐼𝑑𝑦

𝑑𝑥=−6𝑥2

2+ 36𝑥 + 𝐶3

𝐸𝐼𝑦 =−6𝑥3

6+36𝑥2

2+ 𝐶3𝑥 + 𝐶4

𝐸𝐼𝑦𝐵𝐶 = −𝑥3 + 18𝑥2 + 𝐶3𝑥 + 𝐶4 ….(2)

Boundary Conditions

When x = 0, y = 0 …. (A)

When x = 6, y = 0 …. (B)

Matching Conditions

At x = 2; 𝑑𝑦

𝑑𝑥 𝐴𝐵=

𝑑𝑦

𝑑𝑥 𝐵𝐶 and yAB = yBC

Substitute (A) into Eq. (1): EI(0) = 2(0) + C1(0) + C2 C2 = 0

Substitute (B) into Eq. (2): EI(0) = -(6)3 + 18(6)3 + 6C3 + C4

C4 = -432 – 6C3 ….(3)

EXAMPLE 1: Solution

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From the Matching Conditions:

12(2)2

2+ 𝐶1 =

−6(2)2

2+ 36(2) + 𝐶3

𝐶1 = 36 + 𝐶3 ….(4)

2(2)3+𝐶1 2 = − 2 3 + 18 2 2 + 𝐶3 2 + 𝐶4

2C1 = 48 + 2C3 + C4 ….(5)

EXAMPLE 1: Solution

Substitute Eq. (3) and Eq. (4) into Eq. (5):

2(36 + C3) = 48 + 2C3 + (– 432 – 6C3) C3 = -76

From Eq. (3): C4 = -432 – 6(-76) = 24

From Eq. (4): C1 = 36 – 76 = -40

Therefore:

𝑦 =1

𝐸𝐼2𝑥3 − 40𝑥 (0 x 2)

𝑦 =1

𝐸𝐼−𝑥3 + 18𝑥2 − 76𝑥 + 24 (2 x 6)

EXAMPLE 1: Solution

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To determine the deflection at mid-span, x = 3 m:

𝑦3𝑚 =1

𝐸𝐼−(3)3 + 18(3)2−76(3) + 24 =

−69

𝐸𝐼

=−69

20×106 1600×10−6 = 0.00215 m = 2.15 mm (ANS)

*Negative deflection value shows downward direction

Conclusion:

Each different load produces different section and 2 constant unknowns. Say if we have 4 sections (8 unknowns).Therefore, this method is not practical.

EXAMPLE 1: Solution

This is a simplified method based on the double integration concept. In this method, only ONE section will be considered which is at the last loading type.

MAC CAULAY METHOD

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• Mac Caulay functions

𝑥 − 𝑎 𝑛 = 0

(𝑥 − 𝑎)𝑛

• Integration of Mac Caulay functions:

𝑥 − 𝑎 𝑛 𝑑𝑥 =𝑥 − 𝑎 𝑛+1

𝑛 + 1+ 𝐶

USE OF CONTINUOUS FUNCTIONS

for x a

for x a

n a

The cantilever beam shown in the figure below is subjected to a vertical load P at its end. Determine the equation of the elastic curve. EI is constant.

EXAMPLE 2: Mac Caulay Method

x

y

A

P

B

yA

L x

A Elastic curve

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• From the free-body diagram, with M acting in the positive direction as shown in figure, we have M = -Px

• Integrating twice yields;

𝐸𝐼𝑑2𝑦

𝑑𝑥2= −𝑃𝑥 …. (1)

𝐸𝐼𝑑𝑦

𝑑𝑥= −

𝑃𝑥2

2+ 𝐶1 …. (2)

𝐸𝐼𝑦 = −𝑃𝑥3

6+ 𝐶1𝑥 + 𝐶2 …. (3)

EXAMPLE 2: Solution

x

P

V

M

• Using the boundary conditions dy/dx = 0 at x = L and y = 0 at x = L, Eq. (2) and Eq. (3) becomes;

0 = −𝑃𝐿2

2+ 𝐶1

0 = −𝑃𝐿3

6+ 𝐶1𝐿 + 𝐶2

Therefore, 𝐶1 =𝑃𝐿2

2 and 𝐶2 = −

𝑃𝐿3

3

• Substituting these results, we get;

𝜽 =𝑷

𝟐𝑬𝑰𝑳𝟐 − 𝒙𝟐 and 𝒚 =

𝑷

𝟔𝑬𝑰−𝒙𝟑 + 𝟑𝑳𝟐𝒙 − 𝟐𝑳𝟑 (ANS)

EXAMPLE 2: Solution

x

P

V

M

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Repeat Example 1 with using the Mac Caulay Method.

EXAMPLE 3

x

y

18 kN

6 m

2 m 4 m

A B C

EXAMPLE 3: Solution

Determine the Reaction Forces at A and C

Taking moment at C; MC = 0

VA (6) – 18(4) = 0

VA = 12 kN

Fy = 0

VA + VC = 0

VC = 6 kN

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Consider to make a section after last load, i.e. in region BC (section made from left to right). Therefore, the moment equation:

Mx = 12x – 18(x – 2)

EXAMPLE 3: Solution

x

M A

12

B

18

2 m

Moment: 𝑀 = 12𝑥 − 18 𝑥 − 2

𝐸𝐼𝑑2𝑦

𝑑𝑥2= 12𝑥 − 18 𝑥 − 2 …. (1)

Slope: 𝐸𝐼𝑑𝑦

𝑑𝑥=12𝑥2

2−18 𝑥−2 2

2+ 𝐶1

= 6𝑥2 − 9 𝑥 − 2 2 + 𝐶1 …. (2)

Deflection: 𝐸𝐼𝑦 =12𝑥3

6−18 𝑥−2 3

6+𝐶1 𝑥 + 𝐶2

= 2𝑥3 − 3 𝑥 − 2 3 ++𝐶1𝑥 + 𝐶2 …. (3)

EXAMPLE 3: Solution

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Boundary Conditions At x = 0, y = 0 and from Eq. (3): 𝐸𝐼(0) = 2 0 3 − 3 0 − 2 3 ++𝐶1(0) + 𝐶2 C2 = 0 At x = 6, y = 0 and from Eq. (3): 𝐸𝐼(6) = 2 6 3 − 3 6 − 2 3 ++𝐶1(6) + 0 C1 = 40

EXAMPLE 3: Solution

Therefore, Eq. (2) becomes:

𝐸𝐼𝑑𝑦

𝑑𝑥= 6𝑥2 − 9 𝑥 − 2 2 − 40 …. (4)

and Eq. (3) becomes: 𝐸𝐼𝑦 = 2𝑥3 − 3 𝑥 − 2 3 − 40𝑥 …. (5) At mid-span, x = 3 m: 𝐸𝐼𝑦3𝑚 = 2(3)

3 − 3 3 − 2 3 − 40 3 = 69

𝑦3𝑚 =−69

20×106 1600×10−6 = 0.00215 m = 2.15 m (ANS)

𝐸𝐼𝑑𝑦

𝑑𝑥 3𝑚= 6(3)2−9 3 − 2 2 − 40 = 5

𝑑𝑦

𝑑𝑥 3𝑚=

5

20×106 1600×10−6= 𝟏. 𝟓𝟔 × 𝟏𝟎−𝟒 rad (ANS)

EXAMPLE 3: Solution

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Determine the maximum deflection of the beam shown in the figure below. EI is constant.

EXAMPLE 4

10 m 20 m

8 kN

120 kNm

A

B yC

C

D

yD

The beam deflects as shown in the figure. The boundary conditions require zero displacement at A and B (yA = yB = 0).

EXAMPLE 4: Solution

• The moment equation section at x-x is: 𝑀 = −8𝑥 + 6 𝑥 − 10 = −8𝑥 + 6 𝑥 − 10 kNm

10 m 30 m

8 kN

120 kNm

A B

VB = 2 kN VA = 6 kN x

x

x

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Integrating twice yields:

𝐸𝐼𝑑2𝑦

𝑑𝑥2= −8 𝑥 1 + 6 𝑥 − 10 1 …. (1)

𝐸𝐼𝑑𝑦

𝑑𝑥= −4𝑥2 + 3 𝑥 − 10 2 + 𝐶1 …. (2)

𝐸𝐼𝑦 = −4

3𝑥3 + 𝑥 − 10 3 + 𝐶1𝑥 + 𝐶2 …. (3)

Boundary Conditions

y = 0 at x = 10 m and from Eq. (3): 0 = −1333 + 10 − 10 3 + 𝐶1 10 + 𝐶2

y = 0 at x = 30 m and from Eq. (3): 0 = −36000 + 30 − 10 3 + 𝐶1 30 + 𝐶2

C1 = 1333 and C2 = 12000

EXAMPLE 4: Solution

From Eq. (2):

𝐸𝐼𝑑𝑦

𝑑𝑥= −4𝑥2 + 3 𝑥 − 10 2 + 1333 …. (4)

From Eq. (3):

𝐸𝐼𝑦 = −4

3𝑥3 + 𝑥 − 10 3 + 1333𝑥 − 12000 …. (5)

To obtain the deflection at C, x = 0. Therefore, from Eq. (5):

𝒚𝑪 = −𝟏𝟐𝟎𝟎𝟎

𝑬𝑰 kNm3 (ANS)

*The negative sign indicates that deflection is downward

EXAMPLE 4: Solution

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To determine the length at point D, use Eq. (4) with x 10 and 𝑑𝑦

𝑑𝑥= 0

0 = −4𝑥𝐷

2 + 3 𝑥𝐷 − 102 + 1333

4𝑥𝐷2 + 60𝑥𝐷 − 1633 = 0

Solving for the positive root, xD = 20.3 m

EXAMPLE 4: Solution

Hence from Eq. (5):

𝐸𝐼𝑦𝐷 = −4

320.3 3 + 20.3 − 10 3 + 1333(20.3) − 12000

𝒚𝑫 =𝟓𝟎𝟎𝟔

𝑬𝑰 kNm3 (ANS)

*The positive sign indicates that deflection is upward

Comparing yD with yC, ymax = yC (ANS)

EXAMPLE 4: Solution

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Determine the slope and deflection at x = 3 m. Also, determine the location and magnitude of the maximum deflection. Given EI = 4000 kNm2.

EXAMPLE 5

2 m

10 kNm

A B

5 kN

2 kN/m

2 m 2 m 2 m 2 m

EXAMPLE 5: Solution Determine the reaction forces at A and B Taking moment at A, MA = 0 10 + (2)(2)(5) + (5)(8) VB(10) = 0 VB = 7 kN Fy = 0 VA – 2(2) – 5 + VB = 0 VA = 2 kN

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EXAMPLE 5: Solution Cut at section x-x as shown in the figure. Therefore, the moment equation is given as:

𝑀𝑥−𝑥 = 2 𝑥1 + 10 𝑥 − 2 0 −

2 𝑥 − 4 2

2+2 𝑥 − 6 2

2− 5 𝑥 − 8 1

10 kNm

A

5 kN

2 kN/m

2 m 2 m 2 m 2 m

2 kN

x

x

x

EXAMPLE 5: Solution

𝐸𝐼𝑑2𝑦

𝑑𝑥2= 𝑀 = 2 𝑥 1 + 10 𝑥 − 2 0 −

2 𝑥 − 4 2

2+2 𝑥 − 6 2

2− 5 𝑥 − 8 1

𝐸𝐼𝑑𝑦

𝑑𝑥= 𝜃 =

2 𝑥 2

2+10 𝑥 − 2 1

1−2 𝑥 − 4 3

6+2 𝑥 − 6 3

6−5 𝑥 − 8 2

2+ 𝐶1

𝐸𝐼𝑦 =2 𝑥 3

6+10 𝑥 − 2 2

2−2 𝑥 − 4 4

24+2 𝑥 − 6 4

24−5 𝑥 − 8 3

6+ 𝐶1𝑥 + 𝐶2

Boundary Conditions At x = 0 m, y = 0: C2 = 0 At x = 10 m, y = 0:

0 =2 10 3

6+10 8 2

2−2 6 4

24+2 4 4

24−5 2 3

6+ 10𝐶1

C1 = 56

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EXAMPLE 5: Solution

Slope Equation:

𝐸𝐼𝜃 = 𝑥 2 + 10 𝑥 − 2 1 −𝑥 − 4 3

3+𝑥 − 6 3

3−5 𝑥 − 8 2

2− 56

Deflection Equation:

𝐸𝐼𝑦 =𝑥 3

3+ 5 𝑥 − 2 2 −

𝑥 − 4 4

12+𝑥 − 6 4

12−5 𝑥 − 8 3

6− 56𝑥

Now, we can find the slope and deflection at x = 3 m. Given EI = 4000 kNm2

𝜃 =1

40003 2 + 10 3 − 2 1 − 56 = 0.00925 rad

𝑦 =1

4000=

3 3

3+ 5 3 − 2 2 − 56(3) = 0.00385 m

To determine the position and magnitude of the maximum

deflection, ymax when 𝑑𝑦

𝑑𝑥= 0:

Therefore, at xmax = 5.1 m; ymax = 48.3 mm (ANS)

x (m) 𝑬𝑰

𝒅𝒚

𝒅𝒙

5 -1.3

5.5 +8.13

5.2 +2.5

EXAMPLE 5: Solution

From interpolation; xmax = 5.1 m

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Find the deflection equation for the given beam. Then, determine the maximum deflection at mid-span along span AB. Given E = 200 kN/mm2 and I = 10 106 mm4.

EXAMPLE 6

4 m 2 m

A B

20 kN

16 kN/m

C

Determine the reactions forces at A and B Taking moment at B, MB = 0; 20(2) (16)(4)(2)+ VA(4) = 0 VA = 22 kN Fy = 0; VA + VB – 16(4) – 20 = 0 VB = 62 kN Cut at section x-x as shown in the figure.

EXAMPLE 6: Solution

x

22 62

16 kN/m

x

x

4 m

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EXAMPLE 6: Solution

Therefore, the moment equation is given as:

𝑀𝑥−𝑥 = 22 𝑥1 + 62 𝑥 − 4 1 − 16𝑥

𝑥

2+ 16 𝑥 − 4

𝑥 − 4

2

𝐸𝐼𝑑2𝑦

𝑑𝑥2= 𝑀 = 22 𝑥 1 + 62 𝑥 − 4 1 − 8 𝑥 2 + 8 𝑥 − 4 2

𝐸𝐼𝑑𝑦

𝑑𝑥= 𝐸𝐼𝜃 =

22 𝑥 2

2+62 𝑥 − 4 2

2−8 𝑥 3

3+8 𝑥 − 4 3

3+ 𝐶1

𝐸𝐼𝑦 =22 𝑥 3

6+62 𝑥 − 4 3

6−8 𝑥 4

12+8 𝑥 − 4 4

12+ 𝐶1𝑥 + 𝐶2

=11 𝑥 3

3+31 𝑥−4 3

3−2 𝑥 4

3+2 𝑥−4 4

3+ 𝐶1𝑥 + 𝐶2

Boundary Condition at A At x = 0 m, y = 0: C2 = 0 Boundary condition at B At x = 4 m, y = 0:

𝐸𝐼(0) =11 4 3

3+31 4 − 4 3

3−2 4 4

3+2 4 − 4 4

3+ 4𝐶1

C1 = 16 Therefore, the slope and deflection equations are:

𝐸𝐼𝑑𝑦

𝑑𝑥= 11 𝑥 2 + 31 𝑥 − 4 2 −

8 𝑥 3

3+8 𝑥 − 4 3

3− 16

𝐸𝐼𝑦 =11 𝑥 3

3+31 𝑥 − 4 3

3−2 𝑥 4

3+2 𝑥 − 4 4

3− 16𝑥

EXAMPLE 6: Solution

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To determine the maximum deflection and where it occurred, 𝑑𝑦

𝑑𝑥= 0. Therefore:

𝐸𝐼(0) = 11 𝑥 2 + 31 𝑥 − 4 2 −8 𝑥 3

3+8 𝑥 − 4 3

3− 16

Consider the maximum deflection occurs along span AB (0 x 4)

𝐸𝐼(0) = 11 𝑥 2 + 31 0 2 −8 𝑥 3

3+8 0 3

3− 16

0 = 11𝑥2 −8𝑥3

3− 16 or

8x3 – 33x2 + 48 = 0 By try and error: xmax = 1.52 m from A.

EXAMPLE 6: Solution

Therefore, the maximum deflection occurs when x = 1.52 m from A. To calculate the maximum deflection:

𝐸𝐼𝑦𝑚𝑎𝑥 =11 1.52 3

3+31 1.52 − 4 3

3−2 1.52 4

3+2 1.52 − 4 4

3− 16(1.52)

𝐸𝐼𝑦𝑚𝑎𝑥 = −15

𝑦𝑚𝑎𝑥 =−15

(200×106)(10×10−6) = 0.0075 m = 7.5 mm (downward)

EXAMPLE 6: Solution

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• Based on the properties of elastic curve and bending moment diagram.

• Suitable to use for determining deflection and slope at a particular point.

• Also suitable for beam with different cross-section.

• There are two important theorems used in this method.

MOMENT AREA METHOD

Theorem 1

• The angle between the tangents at any two points on the elastic

curve equals the area under the 𝑀

𝐸𝐼 diagram between these two

points.

EI𝑑2𝑦

𝑑𝑥2= 𝐸𝐼

𝑑

𝑑𝑥

𝑑𝑦

𝑑𝑥= 𝑀

𝑑𝜃 =𝑀

𝐸𝐼𝑑𝑥

• Since 𝜃 ≈𝑑𝑦

𝑑𝑥, so 𝑑𝜃 =

𝑀

𝐸𝐼𝑑𝑥

• Therefore, 𝜃𝐴𝐵 = 𝑀

𝐸𝐼𝑑𝑥

𝑥𝐵𝑥𝐴

=1

𝐸𝐼∙ 𝑀𝐴−𝐵 𝐴𝑟𝑒𝑎

MOMENT AREA METHOD

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MOMENT AREA METHOD Theorem 1 (Continued) • This equation forms the basis for the first moment-area

theorem:

𝜃𝐴𝐵 = 𝜃𝐵/𝐴 = 𝑀

𝐸𝐼𝑑𝑥

𝐵

𝐴

Theorem 2 • The vertical deviation of the tangent

at point A on the elastic curve with respect to the tangent extended from another point B equals the moment of the area under the 𝑴

𝑬𝑰diagram between these two

points (A and B). This moment is computed about point, A where the vertical deviation tA/B is to be determined.

MOMENT AREA METHOD

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Theorem 2 (Continued) • The vertical deviation of the tangent at A with respect to the

tangent at B is given as:

𝑡𝐴/𝐵 = 𝑥𝑀

𝐸𝐼𝑑𝑥

𝐵

𝐴

• Then:

𝑡𝐴/𝐵 = 𝑥 𝑀

𝐸𝐼𝑑𝑥

𝐵

𝐴

where x is the location of the centroid of the shaded area

𝑀

𝐸𝐼𝑑𝑥 between A and B.

MOMENT AREA METHOD

Theorem 2 (Continued) • Centroid and Area

𝐴𝑟𝑒𝑎, 𝐴 =𝑏ℎ

(𝑛 + 1)

𝐶𝑒𝑛𝑡𝑟𝑜𝑖𝑑, 𝑥 =𝑏

(𝑛 + 2)

*Notes: o Draw BMD separately for each load, with one reference point o Write the bending moment equation in f(x) terms o Upward force produces positive bending moment and vice versa

MOMENT AREA METHOD

x

b

h

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Procedures:

MOMENT AREA METHOD

The beam is subjected to the concentrated force shown in the figure. Determine the reactions at the supports. EI is constant.

EXAMPLE 7

L

P

L

A B

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• The free-body diagram is shown in Fig. (b).

EXAMPLE 7: Solution

L

P

L

A B

(b) VB

VA

HA

MA

EXAMPLE 7: Solution

• Using the method of superposition, the separate 𝑀

𝐸𝐼 diagrams for

the redundant reaction VB and the load P are shown in Fig. (c).

L A

(c)

+ 2L

𝑽𝑩𝑳

𝑬𝑰

B C

−𝑷𝑳

𝑬𝑰

−𝟐𝑷𝑳

𝑬𝑰

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30

• The elastic curve for the beam is shown in Fig. (d).

EXAMPLE 7: Solution

A

B

tB/A = 0

tan A

tan B (d)

• Applying Theorem 2, we have:

𝑡𝐵/𝐴 =2

3𝐿

1

2

𝑉𝐵𝐿

𝐸𝐼𝐿 +

𝐿

2

−𝑃𝐿

𝐸𝐼𝐿 +

2

3𝐿

1

2

−𝑃𝐿

𝐸𝐼𝐿 = 0

VB = 2.5P (ANS)

• Using this result, the reactions at A on the free-body diagram,

of Fig. (b), are:

+ Fx = 0; HA = 0 (ANS)

+ Fy = 0; VA + 2.5P – P = 0

VA = 1.5P (ANS)

+ MA = 0; MA + 2.5P(L) – P(2L) = 0

MA = 0.5PL (ANS)

EXAMPLE 7: Solution

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The beam shown in the figure is pin supported at A and roller at B. A point load of 60 kN is applied at 6 m from A. Determine:

i) Deflection at B

ii) Slope at B

iii) Maximum deflection

EXAMPLE 8

A

60 kN

3 m 3 m 3 m

B C

EXAMPLE 8: Solution

A

60 kN

3 m 3 m 3 m

B C

VA = 20 kN VB = 40 kN

B’ tC/A

tB/A

yB

The deflection and slope diagram of the beam. Reference point is taken from A.

From the figure, yB = BB’ – tB/A …. (1)

where 𝐵𝐵′

3=𝑡𝐶/𝐴

9 𝐵𝐵′ =

𝑡𝐶/𝐴

3 …. (2)

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The BMD for each loading i.e. VA = 20 kN and P = 60 kN

EXAMPLE 8: Solution

𝑡𝐵/𝐴 = +1

𝐸𝐼

1

2× 3 × 60

1

3× 3 = +

90

𝐸𝐼

𝑡𝐶/𝐴 = +1

𝐸𝐼

1

2× 9 × 180

1

3× 9 −

1

2× 3 × 180

1

3× 3 = +

2160

𝐸𝐼

A

180 kNm

+

180 kNm

3 m 6 m

B

3 m

60 kNm

C

EXAMPLE 8: Solution

𝐵𝐵′ =1

3

2160

𝐸𝐼=720

𝐸𝐼

From Eq. (1); 𝑦𝐵 =720

𝐸𝐼−90

𝐸𝐼=630

𝐸𝐼↓

Let say EI = 30,000 kNm2

𝒚𝑩 = 𝟐𝟏 𝒎𝒎 ↓ (ANS)

A = AB + B

B = A AB

𝜃𝐴 ≈ 𝑡𝑎𝑛𝜃𝐴 =𝑡𝐶/𝐴

9=2160

9𝐸𝐼

𝜃𝐴𝐵 = +1

𝐸𝐼

1

2× 3 × 60 = +

90

𝐸𝐼

𝜃𝐵 =2160

9𝐸𝐼−90

𝐸𝐼=150

𝐸𝐼

Let say EI = 30,000 kNm2

B = 0.005 rad (ANS) A

60 kN

B C

tC/A

B

AB A

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Say ymax between 0 x 6 as shown in the figure:

EXAMPLE 8: Solution

ymax = MM’ – tM/A

where 𝑀𝑀′

𝑥=𝑡𝐶/𝐴

9 𝑀𝑀′ =

𝑥

3𝑡𝐶/𝐴

𝑡𝑀/𝐴 = +1

𝐸𝐼

1

2∙ 𝑥 ∙ 20𝑥

1

3𝑥 =

10𝑥3

3𝐸𝐼

𝑦𝑚𝑎𝑥 =𝑥

9∙2160

𝐸𝐼−10𝑥3

3𝐸𝐼

= 1

𝐸𝐼240𝑥 −

10𝑥3

3 …. (3)

A

60 kN

M C

tC/A

M’

tM/A

ymax

x

EXAMPLE 8: Solution

Maximum deflection occurs when 𝑑𝑦

𝑑𝑥= 0

From Eq. (3): 𝑑𝑦𝑚𝑎𝑥

𝑑𝑥=

1

𝐸𝐼240𝑥 −

10𝑥3

3= 0

𝑥 = 24 = 4.9 𝑚 < 6 𝑚 OK as assumed

𝑦𝑚𝑎𝑥 =1

𝐸𝐼240(4.9) −

10(4.9)3

3=785

𝐸𝐼

Let say EI = 30,000 kNm2

𝒚𝒎𝒂𝒙 =𝟕𝟖𝟓

𝟑𝟎,𝟎𝟎𝟎= 𝟐𝟔 𝒎𝒎 ↓ (ANS)