Eb9bG Moment Area Example

Moment Area Examples

For the cantilever shown below calculate

vertical deflection at the tip of the cantilever.

E = 200 kN/mm2

I = 200 000 cm4

EI

8 m

EI

2 m

A B C

100 kN

Taking Moment of the area about C

DC = [(800x8/2) x 22/3] / EI = 70400/3EI

E =200 kN/mm2 = 200 x 106 kN/m2

I = 20000 cm4 = 200000x10-8 = 2000x10-6 m4

DC = [70400/(3x200x106x2000x10-6)] x 1000

DC = 58.66 mm

Solution

8 m, EIA BC

DC

2 m, EI100 kN

800

8/3 m 2 + 2x8/3 = 22/3 m

=+

10m, E, I

2

100 kN

1

D1

10m, E, I

2

V1

1

D1

10V1

20/3

500

5 +10/3 = 25/3

10m, E, I

2

100 kN

1 5m

D1 = [(500 x 5/2) x 25/3] / EI

D1 = 31250/3EI

D1 = [(10V1 x 10/2) x 20/3] / EI

D1 = 1000V1/3EI(Equate deflections)

1000 V1 = 31250

V1 = 31.25 kN V2 = 100 31.2 = 68.8 kN

M2 = 500 10V1 = 500 312.5 = 187.5 kN-m

MUnder the Load = 31.25 x 5 = 156.25 kN-m

Or 312.5/2 = 156.25 kN-m

187.5 kN-m

156.25 kN-m

Bending Moment Dia.

10V1=

312.5

20/3 500

5 +10/3 = 25/3

For the structure shown below:

Calculate support reactions

Sketch BMD showing values at critical locations

Calculate vertical deflection at C.

=

L, E, IA B

DB

qB

100 kN/m

VB

+

EI = 80000 kN-m2

VB

DB

There are 4 D.o.F and

only 3 Equations of

Equilibrium:

Remove reaction at B

to change the structure

to a determinate

structure

Apply VB such that the

net deflection at B is

Zero

8m, E, I B

100 kN/m C2m, E, I

VA

0

MA

A

Draw two tangents at A and B

DB = (1/3x3200x8) x 6/EI = 51200/EI


DB = (8VBx8/2) x 16/3EI = 512VB/3EI

8VB

16/3 m

VB

DB

L, E, IA

B

DB

qB

100 kN/m3200

+CG

2m 6m 2m

51200/EI = 512VB/3EI (No deflection at B)

VB = 300 kN

VA = 800 300 = 500 kN

MA = 3200 8 VB = 3200 2400 = 800 kN-m

Bending Moment Diagram

5 mB C

800

450

A3

20

0

+CG

2m 6m

8V

B=

24

00

16/3 m

Sagging Moment = 300*3/2 = 450kN-m

500

300

Shear Force Diagram

5m3m

Deflection calculations

DC1 = (1/3x3200x8) x 8/EI = 68267/EI

DC2 = (2400x8/2)x(16/3 + 2)/EI = 70400/EI

DC = DC1 DC2 = -2133/EI

DC =(-2133.33/80000) * 1000

DC = 26.7 mm Upward

32

00

+CG

2m 6m

8m

24

00

16/3 m

8m

2m

Alternative solution

Introduce a hinge at A to get red of a D.o.F

8m, E, IA B

100 kN/m C2m, E, I

VA

0

MA

=

8m, E, IA B

100 kN/m C2m, E, I

VA

0

MA

+

8m, E, IA BC2m, E, I

VA

0

MA

qA = 2/3 x 800 x 4 /EI =6400/3EI

MA = 3EI/8 qA = 3EI/8 x 6400/3EI = 800kN-m

VA = 400 + 800/8 = 500 kN

VB = 400 800/8 = 300 kN

8m, E, IA B

100 kN/m C2m, E, I

VA

0

MA

400

0

B

C

800

qA

100 kN/m

400

qA

MA/8

MA

A BC

MA/8

0MA

Now student try the following question

For the structure shown below:

Calculate support reactions

Sketch BMD showing values at critical

locations

Calculate vertical deflection at c

8m, E, I BC2m, E, I

VA

0

MA 100 kN

A

EI = 80000 kN-m2

4m

VB

DB

There are 4 D.o.F

Take reaction at B out

Apply VB such that the net

deflection at B is Zero

8m, E, I BC2m, E, I

VA

0

MA

A

100 kN

4m

400

A B

8/3 +4 = 20/3A B

DB

qB

100 kN4m


DB = (8VB*8/2) x 16/3EI = 512VB/3EI8V

B

16/3

8m

512VB/3EI = 16000/3EI VB = 31.25kN & VA= 68.75kN


DB = (400*4/2) x 20/3EI = 16000/3EI

4m

400

A B

8/3 +4 = 20/3

250

16/3

8m

31.25 *4 = 125 kN-m

Bending Moment Dia.

400-250 =150kNm68.7

5kN

31.2

5kN4m4m

Shear Force Dia.

Remove the support at 2V1

100 kN

5m, EI 5m, EI 5m, EI

1 32

V3V2

100 kN

5m, EI

100

100 kN

5m, EI 10m, EI1 32

100

100 kN

5m, EI

D2 D2


Moment of area at 3D2 =[(500x5/2) x (2/3)x5 + (500x5) x (5+2.5)]/EI

D2 = 12500/3EI + 18750/EI = 68750/3EI

500500

Now apply V2 back to push the beam to

its original position

V3 = V2/210m, EI 10m, EI1 32

V1 = V2/2 V2

D2 D2

(V2/2) x 10 = 5V2

D2 = (5V2x10/2) x 20/3EI =500V2/3EI

500V2/3EI = 68750/3EI V2 = 68750/500 = 137.5 kN

Moment of area at 3

V1 = 100 V2/2 = 31.25 kN

V3 = V1 = 31.25 kN

M2 = 5V2 500 = 5*137.5 -500 =187.5 kN-m

31.25x5

=156.25

187.5

156.25

Thinking outside the box

Students to work in groups of 2 or 3 and

discuss if they can use the conditions of

symmetry to simplify the structure to work

out the unknown parameters.

Hint: Draw deformed shape of the structure

& think about simplifying it.

V1

100 kN


1 32

V3V2

100 kN

5m, EI

Thinking outside the box

Due to symmetrical geometry and loading,

the structure can be simplified to:

At the central support the structure can not

move vertically or horizontally

Due to symmetry there is no rotation either.

This could be replaced to a fixed support.

10m, E, I10m, E, I

2

100 kN 100 kN

21 3

10m, E, I

2

100 kN

1

=

+10m, E, I

2

100 kN

1

D1

10m, E, I

2

V1

1

D1

10V1

20/3

500

5+10/3 = 25/3

From the first Structure:

D1 = [(500 x 5/2) x 25/3] / EI

D1 = 31250/3EI

From the Second Structure:

D1 = [(10V1 x 10/2) x 20/3] / EI

D1 = 1000V1/3EI

1000 V1 = 3125 (Equate deflections)

V1 = 31.25 kN

(which is the same as before)

Challenging Question

For the structure shown:

Calculate support reactions.

Sketch BMD and calculate values at critical location.

The structure is indeterminate as there are 4 variables and 3 equations of equilibrium.

120 kN


1 32

Solution

0

120 kN


1 32

V1 V3V2

=

+

120 kN2m, EI 4m, EI 6m, EI

1 32

100 20D2

D1 d2

V3 = V2/22m, EI 4m, EI 6m, EI1

32

V1 = V2/2 V2 = 2V1

D2D2

Draw three tangents at 1, 2 and 3.

D1 = [(200x2/2x4/3) + (200x10/2)(2+10/3)]/EI

D1 = [800/3+16000/3] / EI = 16800/3EI = 5600/EI

d2 + D2 = D1 / 2 = 2800/EI

d2= (120x6/2) x 6/3EI = 720/EI D2 = (2800-720)/EI

D2 = 2080/EI

120 kN

2m, EI 4m, EI 6m, EI1 32

100 20D2

D1 d2

200120

D12

D2 = (3V2x6/2) x 4/EI =36V3/EI 36V2/EI = 2080/EI V2 = 2080/36 = 57.78kN V1 = 100 57.78/2 = 71.11 kN V3 = 20 - 57.78/2 = -8.89 kN

V3 = V2/22m, EI 4m, EI 6m, EI1

32

V1 = V2/2 V2

D2D2

(V2/2) x 6 = 3V2

53.34

142.22

BMD

8.892m, EI 4m, EI 6m, EI

1 32

71.11 57.78

0120 kN


1 32

V1 V3V2

120 kN


1 32

V1 V2

D3

V3


1 32

V1 V2 D3

=

+

Alternative SolutionRemove V3

V32m, EI 4m, EI 6m, EI

1 32

V1D3A

V3

2m, EI 4m, EI

6m, EI1 32

V1

D3Bq2

M2 =6V3

V3


1 32

V3 2V3 D3

=

+

Draw two tangents at 1 and 2 take moment

of the area at 1

D3 =[(160x2/2) x (2x2/3) + (160x4/2) x (2+4/3)]/EI

D3 =(640/3 +320x10/3)/EI = 3840/3EI

120 kN


1 32

V1 V2

D3

6m, EI

1 32

80 40

2m 4m

160 kN-m BMD

D3

D3A =(6V3x6/2) x 4/EI = 72V3 / EI

M2 = (3EI/6) q2 = EI q2 /2

q2 = 2M2/EI = 2x(6V3)/EI = 12V3/EI

D3B =6x q2 = 6 x (12xV3/EI) = 72V3 / EI

D3 = D3A + D3B = 144V3 / EI

144V3 / EI = 3840/3EI V3 = 8.89 kN

6m, EI2 3

D3A

V3

BMD+

4m

6V3

6m, EI

1 32

V1

D3Bq2

M2 =6V36m, EI

V1 = 80 8.89 = 71.11 kN

V2 = 40 + 2x8.89 = 57.78 kN

53.34

142.22


71.11

48.89

8.89

Shear Force Diagram

Eb9bG Moment Area Example

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