chapter h4b

7/17/2019 Chapter h4b

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ANALYSIS OF DETERMINATE

STRUCTURES – FRAMECHAPTER 4b



OUTLINE

Shear force and bending moment diagram of frame, inclined frameand frame with hinge

Method of superposition



EXAMPLE 1

Draw shear force and bending moment diagram of the frame

L/2

A

C

B

L

P

L/2



Find the reactions (method similar to beam: equilibrium equations)

L/2

A

C

B

L

P

L/2Ay

Ax

Cy

Σ = 0; = 0

Σ = 0;

2 −

= 0

∴ = /2

Σ = 0; + − = 0

∴ = /2



Find internal reactions in each member. Ensure they are in equilibrium.

P

A

C

B

P/2

P/2

P/2

V=P/2

M=0

N=0

V=0

M=0

N=P/2

M=0

00

0



+P/2

- P/2

Draw SFD & BMD.

P

A

C

B

P/2

P/2

P/2

P/2

0

0

0

0

P/2

0

00

SFD



+PL/4

Draw SFD & BMD.

P

A

C

B

P/2

P/2

P/2

P/2

0

0

0

0

P/2

0

00

BMD



-P/2

Draw SFD & BMD.

P

A

C

B

P/2

P/2

P/2

P/2

0

0

0

0

P/2

0

00

AFD



EXAMPLE 2

Draw shear force and bending moment diagram of the frame

A

B

1.8 m

25 kN

C

1.5 m

4.5 m

15 kN



Find the support reactions

A

B

1.8 m

25 kN

C

1.5 m

4.5 m

15 kN

Cx=15 kN

Ay=30 kN

Cy=5 kN



Draw the FBD of each member & joint

A

B

1.8 m

25 kN

C

1.5 m

4.5 m

15 kN

15 kN

30 kN

5 kN

15

15

5

22.55

22.5

30

15

22.515

30



Draw the SFD & BMD

A

B

1.8 m

25 kN

C

1.5 m

4.5 m

15 kN

15 kN

30 kN

5 kN

15

15

5

22.5

5

22.5

30

15

22.515

30

SFD (kN)

-15

5



Draw the SFD & BMD

A

B

1.8 m

25 kN

C

1.5 m

4.5 m

15 kN

15 kN

30 kN

5 kN

15

15

5

22.5

5

22.5

30

15

22.515

30

BMD (kNm)

-22.5 -22.5



Usually these diagrams are drawn on the same graph. However, it isup to you!

SFD (kN)

-15

5

BMD (kNm)

-22.5

-22.5



EXAMPLE 3

Draw the SFD & BMD of the frame

30 kN/m

80 kN

4.5 m

6 m

9 m

A

B C

D



EXAMPLE 4

Draw the SFD & BMD of the frame

20 kNm

30 kN

1.5 m

3 mA

B C

D

2 m

12 kN/m

1 m

2 m

12 kN



2 m

A

CB

3 m

80 kN

2 m

40 kN/m

4 m

EXAMPLE 5

Draw the SFD & BMD of the inclined frame.



Draw FBD and find the support reactions.

2 m

A

CB

3 m

80 kN

2 m

40 kN/m

4 m

120 kN

82.5 kN

2.5 kN



Internal forces (shear and axial force) have to be along the inclinedface of the member.

To ease calculation, transform UDL & support reactions along theinclination.

Member A-B;

A

40 kN/m

4

120 kN

2.5 kN

V

M N

3

B

=

V

M N



Distributed load on inclined member:

=

40 kN/m

4

3

120 kN

4

3

4

3

120(4/5)

=96kN

120(3/5)

=72kN

==



Repeat for the support reactions at A. The final FBD as shown:

80 kN

97.5 kN

82.5 kN

70 kN

Next, calculate INTERNAL FORCES for EACH MEMBER. Finally, draw the SFD& BMD.

A

B C



Fine the internal forces for each member. Ensure the joint is in equilibrium.

80 kN

97.5 kN

82.5 kN

70 kN

A

BB C

B



Draw the SFD & BMD. 80 kN

82.5 kN

70 kN

2 kN

170 kNm

1.5 kN

2.5 kN

170 kNm

70

SFD (kN)

BMD (kNm)

170.1170

2

2.5

82.5

SFD (kN)

BMD (kNm)

170165



Usually the diagram of each member is combined for the frame.

SFD (kN)

BMD (kNm) 170.1 170

2.5

82.5

170165



3 m

3 m

Draw the SFD & BMD of the inclined frame.

A

C

B

3 kN/m

EXAMPLE 6

3 m



EXAMPLE 7

Draw the SFD & BMD of the hinged frame

15 kN/m

50 kN

1.5 m

4 m

A

B C

D

40 kN

2 m 1.5 m

6 m



EXAMPLE 8

Draw the SFD & BMD of the hinged frame

30 kNm

20 kN

4 m5 m

A

B C

D

2 m

5 kN/m1 m

2 m

E



METHOD OF SUPERPOSITION

On a linear elastic structure, the combined effect of several loadsacting simultaneously is equal to the algebraic sum of the effects ofeach load acting individually

Assumptions:

Small deformation

Linearly elastic material



EXAMPLE

Draw the SFD & BMD using method of superposition.

5 kN/m

20 kNm 40 kNm12 m



=

30 kN

5 kN/m

30 kN+

20 kNm

1.67 kN1.67

kN +

40 kNm

3.33

kN

3.33

kN

90

M

(kNm)

-20

-40

60.3

-20 -40

=

+

+

5 kN/m

20 kNm 40 kNm12 m



QUESTIONS?

chapter h4b

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