SA1 FRAME - Ch 3a - VTCtycnw01.vtc.edu.hk/CBE2027/SA1_FRAME.pdf · ... Take each member and joint as free-body, find the axial force, shear force and bending moment at the ends of

CBE2027 Structural Analysis I Chapter 3 – Analysis of Determinate Plane Frames

HD in Civil Engineering Page 3-1

PLANE FRAMES

A frame is a structure composed of straight-line members. The members

may be connected by rigid joints, pin-connected joints and semi-rigid joints.

If all the joints are pins (which transmit no bending moments), the frame is

commonly called a truss. Rigid joints are capable of transmitting both

forces and bending moments.

A rigid frame is one in which some or all of its joints are rigid. Rigid frames

are usually statically indeterminate. Our study will be confined to

determinate plane frames. In a plane frame, all the members and loading

must be in the same plane.

A frame is completely analyzed when its support reactions, and the

variations in axial forces, shear forces and bending moments along all its

members are found.

Rigid Frame

Frame with an

internal hinge

Truss (Pin-jointed Frame)



Rigid Joints

At rigid joints, the ends of connected members must not only move together

vertically and horizontally but must all rotate by the same amount. A rigid

joint preserves the angle between members connected to it.

In steel structures, rigid and pin-joints consist of welded connections and

simple bolted connections respectively. There is another type of connection

in between of these two extremes and it is called a semi-rigid connection

(the connection is neither rigid nor pinned, it is in between these two

extremes). In reinforced concrete structures, beams and columns are

usually formed together resulting in substantially rigid joints.

A

B C

D

L

H P.I.

P.I.

P.I.

B' C'



PROCEDURES FOR THE ANALYSIS OF DETERMINATE PLANE

FRAMES

1. Find the support reactions. (By using the equations of equilibrium

and the equations of condition if any).

2. Find the member end forces.

(a) Take each member and joint as free-body, find the axial force,

shear force and bending moment at the ends of the member

and joint.

(b) Each free-body diagram must show all external loads, support

reactions and possible internal forces acting on the member

and joint.

(c) A rigid joint can transmit two force components (V & N) and

a moment.

At a hinge, the internal moment is zero.

3. Plot the axial force, shear force and bending moment diagrams on an

outline of the frame.

(a) A common convention is to draw the bending moment

diagram on the tension side of the frame. (For reinforced

concrete frames, designers often draw the bending moment

diagram on the tension side. This allows the designer to tell at

a glance on which side of the frame steel reinforcement

should be placed.)

(b) The axial force diagram may be plotted on either side of the

member, with proper indication for tension and compression.

(Tension --- +ve, compression --- -ve)

(c) The shear force diagram may be plotted on either side of the

member but normally follows the convention used for the

bending moment diagram. (i.e. follows the convention of

plotting shear force and bending moment of beam). This can

be done by treating each individual member as a beam

element, plot the shear force diagram from the left end to the

right end of the member and draw the bending moments on

the tension side of the member.



Example 1

Determine the support reactions, and draw the axial force, shear force and

bending moment diagrams for the frame. Joint B is a rigid joint.

A

BC

4m

6m

10 kN

2 kN



Solution:

HA VA

MA

A

BC

4m

6m

10 kN

2 kN

∑X = 0, HA = 2 kN

∑Y = 0, VA = 10 kN

Take moment about A,

MA = 10*4 + 2*6 = 52 kNm



Free-body diagrams for the members and joint, and their member end

forces.

B

C

10 kN

B

B

A

10 kN

52 kNm

10

40

10

40

10

40

10

40

2 kN

22

2

2

2kN

Free-body Diagrams



Axial force, shear force and bending moment diagrams

A

B

C

2

-10

Axial Force (kN)

A

B

C2

10

Shear Force (kN)

A

B C

- 40

- 40

- 52

Bending Moment (kNm)

-

+

-10

2

10

2



Example 2


bending moment diagrams for the frame. Joint B is a rigid joint.

6m 3m

4m

12 kN

A B

C



Solution:

HA

VA

MA

6m 3m

4m

12 kN

A B

C

∑X = 0, HA = 0kN

∑Y = 0, VA = 12kN


MA = 12*9 = 108 kNm



Free-body diagrams for the members and joint, and their member end

forces.

12 kN

A

B

C

BB108 kNm

12 kN

12

36

36

12

12

3636

12

Free-body Diagrams



In order to draw the axial force and the shear force diagram of member BC,

we have to resolve the member end forces so that the member end forces are

perpendicular and along the member axis. This is because member BC is

an inclined member.

B

C

36

12

3

4 5φ

φ

φ

12∗4/5 = 9.612∗3/5 = 7.2

12

7.29.6




A B

C

9.6

9.6

Axial Force (kN)

A B

C

7.2

Shear Force (kN)

12 12

7.2

A B

C


-108

-36-36

0

+



Example 3


bending moment diagrams for the frame. Joints C and E are rigid joints.

A

B

CD

E

F

96 kN

48 kN

3m

5m

4.5

m3

m

3m



Solution:

HA V

A

VF

A

B

CD

E

F

96 kN

48 kN

3m

5m

4.5

m3

m

3m

∑X = 0, HA = 48 kN


48*4.5 + 96*3 = VF*6, ⇒ VF = 84 kN

∑Y = 0, VA + VF = 96 kN ⇒ VA = 12 kN



Free-body diagrams for the members and joints, and their member end

forces.

B

C

D

E

96 kN

48 kN

C12

216

12

216

E84

84

A

F

1248

84

12

216

12

216

84

84

Free-body Diagrams

C E




B

C

D

E

A

F

Axial Force (kN)

-12

-12

-84

-84

-

-

B

C

D E

A

F

Shear Force (kN)

48

12

-84

48

12

-84



B

C D E

A

F


216

252

216

216



Example 4


bending moment diagrams for the frame. Joints C and E are rigid joints.

A

B

CD

E

F

96 kN

48 kN

3m

5m

4.5

m3

m

3m



Solution:

A

B

CD

E

F

96 kN

48 kN

3m

5m

4.5

m3

m

3m

HA VA

MA

∑X = 0, HA = 48 kN

∑Y = 0, VA = 96 kN


48*4.5 + 96*3 = MA, ⇒ MA = 504 kNm




forces.

B

C

D

E

96 kN

48 kN

C96

288

96

288

E

A

F

96 48

96

288

96

288

Free-body Diagrams

504

C E




B

C

D

E

A

F

Axial Force (kN)

-

-96

-96

B

C D

E

A

F

Shear Force (kN)

48

96 96

48



B

C D

E

A

F

Bending Moment (kNm)-504

-288

-288

-288

0



Example 5


bending moment diagrams for the frame. Joints B and C are rigid joints.

20 kN

50 kN15 kN/m

A

B C

D

E

8m

6m

2m

2m



Solution:

HA V

A

VE

20 kN

50 kN15 kN/m

A

B C

D

E

8m

6m

2m

2m

∑X = 0, HA + 20 = 50, ⇒ HA = 30 kN


50*6 + 15*8*8/2 – 20*4 – VE*8 = 0

⇒ VE = 87.5 kN

∑Y = 0, VA + VE = 15*8 kN, ⇒ VA = 32.5 kN




forces.

50 kN15 kN/m

B C

20 kN

C

D

E

87.5

30A

B

32.5

B

32.5

180

32.5180

3020

87.5

40

87.540

2020

C

32.5

180

30

32.5

18020

87.5

40

20

40

20

87.5

Free-body Diagrams


B C

D

A

E

-32.5

-32.5

-20 -20-87.5

-87.5

-

-

-

Axial Force (kN)



B C

D

A

E

Shear Force (kN)

30

30

32.5

-87.5

20

20

2.17m

B C

D

A

E


-40 -40180

180

215.2

2.17m



Example 6


bending moment diagrams for the frame. Joints B and D are rigid joints.

30 kN

10 kN

5kN/m A

B

CD

E

F

3.5m 3.5m

3m

3m

8m



Solution:

HA V

A

VF

30 kN

10 kN

5kN/m A

B

CD

E

F

3.5m 3.5m3

m3

m

8m

∑X = 0, HA + 10 = 5*8/2, ⇒ HA = 10 kN


5*(8/2)*(8/3) + 30*3.5 - 10*(3+2) – VF*7 = 0

⇒ VF = 15.5 kN

∑Y = 0, VA + VF = 30 kN, ⇒ VA = 14.5 kN




forces.

30 kN

10 kN

5kN/m A

B

C

D

E

F

15.5kN

14.5kN10kN

DB

B

14.526.7

14.5

26.7

10 10

15.5

30

15.5

30

1010C

14.526.7

10

14.510

26.7

15.5

30

10

15.5

3010

Free-body Diagrams




Determination of the position of zero shear force and the maximum bending

moment of member AB.

B

14.5 26.7

10

x

5x/8 kN/m

14.5

M

V

Free-body Diagram for Part of Member AB

Consider the free-body diagram for part of member AB. The loading

intensity of the triangular load = (5x/8) kN/m. Set V = 0, and determine x.

To determine the position of zero shear force (i.e. V = 0 kN), take moment

about the “cut”,

∑Y = 0, (5x/8)*(x/2) – 10 = 0, ⇒ x = 5.66 m

To determine the value of maximum moment of member AB, consider

∑M = 0, M + 26.7 + (5*5.66/8)*(5.66/2)*(5.66/3) – 10*5.66 = 0

⇒ M = 11.0 kNm

To determine the position of zero moment of member AB, we have to use

“trial and error” method to solve a cubic equation. Consider the free-body

diagram for part of member AB. Set M = 0, to determine x

∑M = 0, 26.7 + (5*x/8)*(x/2)*(x/3) – 10*x = 0

⇒ 0.1042x3 – 10x + 26.7 = 0

Solving this equation by “trial and error”, we get x = 2.93m.



B

C

D

E

F

A

-

-

-

Axial Force (kN)-14.5

-14.5

-10 -10

-15.5

-15.5

B C D

E

F

A Shear Force (kN)

14.5 14.5

-15.5

10

10

10

2.3

4m

5.6

6m

-10



BC D

E

F

A Bending Moment (kNm)

2.3

4m

-30

-30-26.7

24.3

-26.7

2.9

3m

11.0



Example 7


bending moment diagrams for the frame. Joint B is a rigid joint and joint C

is an internal hinge of the frame.

A

B C

D

6m

4m

30 kN15 kN/m



Solution:

HA

VA

HD

VD

A

B C

D

6m

4m

30 kN15 kN/m


30*4 + 15*6*6/2 – VD*6 = 0

⇒ VD = 65 kN

∑Y = 0, VA + VD = 15*6 kN, ⇒ VA = 25 kN

Consider the free-body of CD,

65HD

65HC

C

D

Take moment about C,

HD*4 = 0, ⇒ HD = 0 kN

∑X = 0, HA + HD = 30, ⇒ HA = 30 kN




forces.

A

B C

D

30 kN15 kN/m

CB

B

25

120

25

120

30 65

65

C

2530 65

6525

120

30

6525

120

Free-body Diagram


A

B C

D

-25 -65

-65-25

Axial Force (kN)

- -



A

B C

D

Shear Force (kN)

30

30

25

-65

1.67m

A

B C

D


1.67m

120

120

140.9



Example 8


bending moment diagrams for the frame. Joints B and C are rigid joints,

and joint E is an internal hinge of the frame.

A

B C

D

EF

3m

4m

4m 4m

20 kN

3 k

N/m



Solution:

HAVA

HDV

D

A

B C

D

EF

3m

4m

4m 4m

20 kN

3 k

N/m


20*3 + 3*4*4/2 – VD*8 = 0

⇒ VD = 10.5 kN

∑Y = 0, VA + VD = 20 kN, ⇒ VA = 9.5 kN

Consider the free-body of ECD,

HD 10.5

C

D

EHE

VE

Take moment about E,

HD*4 = VD*4, ⇒ HD = 10.5 kN

Consider the whole frame,

∑X = 0, HA + HD = 3*4, ⇒ HA = 1.5 kN




forces.

A

B C

D

EF

20 kN

3 k

N/m

B

9.5

18

9.5

18

10.5 10.5

10.5

42

10.54210.5

10.5 C

9.51.510.5

10.5

9.5

18

10.5

10.5

42

10.5

10.54210.5

9.5

18 10.5

Free-body Diagrams

B C




C

D

EF

A

B

-10.5-10.5

-10.5

-10.5

-9.5

-9.5

-

--

Axial Force (kN)

C

D

E

F

A

B

Shear Force (kN)1.5

-10.5

9.5 9.5

-10.5 -10.5

10.5

10.5 0.5

m



C

D

E

F

A

B


0.5

m

-42

-42

-18

-18

10.5

0.375



Example 9


bending moment diagrams for the frame. Joints B and E are rigid joints, and

joint D is an internal hinge of the frame.

A

BC D

E

F

4m

6m

40 kN

5 k

N/m

1m 2m 3m



Solution:

HA V

A

HF V

FA

BC D

E

F

4m

6m

40 kN

5 k

N/m

1m 2m 3m

Take moment about F,

5*4*4/2 + 40*5 + HA*2 – VA*6 = 0

⇒ 3VA – HA - 120 = 0

⇒ 3VA = HA + 120 (1)



Consider the free-body diagram of ABCD,

HA VA

A

BC D

40 kN

H

V

D

D

Take moment about D,

40*2 + HA*6 – VA*3 = 0

⇒ 3VA – 6HA - 80 = 0 (2)

Sub. (1) into (2)

HA + 120 - 6HA – 80 = 0

⇒ HA = 8 kN (3)

Sub. (3) into (1),

⇒ 3VA = 8 + 120

⇒ VA = 42.7 kN


∑Y = 0, VA + VF = 40 kN, ⇒ VF = -2.7 kN

⇒VF = 2.7 kN (↓)


∑X = 0, HA + HF = 5*4, ⇒ HF = 12 kN




forces.

A

B

C D

E

F

40 kN5

kN

/m

E

B

42.78

2.712

B

42.7

48

42.7

48

8 8

2.7

8

2.7

8

88E

42.7

48

8

2.7

8

8

2.7

8

842.7

48

8

Free-body Diagrams


A

BC D

E

F

-8 -8-42.7

-42.7

2.7

2.7

Axial Force (kN)

-

+-



A

B C

D

E

F

Shear Force (kN)

-8

-8

42.7 42.7

2.7 2.78

-12

2.4

m

A

B C

D

E

F


2.4

m

-48

-48

-5.3

8

8

14.4



Tutorial 3 (Analysis of Frames)

Find the support reactions, and draw the axial force diagram, shear force

diagram and bending moment diagram for the frames shown below.

Q1.

6m

5m

2.5

m2

.5m

30 kN

6 kN/m

A

B

C D

E

Q2.

1.5m 1.5m2m

50 kN 40 kN

4m

6m

15

kN

/m

A

B C D E

F

E is an internal hinge



Q3.

3m 3m 3m

5m

60 kN 30 kN

6 k

N/m

A

B C D E

F

B & E are internal hinges

Q4.

3m 3m

4m

2m

20 kN24 kN

A

B CD

E

2 k

N/m



Q5.

2 k

N/m

20 kN

2m 1m 1m4

m

B

A

C D E

F

6 kNm 10 kNm

C is an internal hinge

Q6.

A

B

C D E F G

H

15 kN/m

65 kN

3m 5m 2m5m

3m

3m

E is an internal hinge



Q7.

3m 3m

3m

3m

A

B

C D E

F

G

30 kN10 kN/m

B, D & F are internal hinges

Q8.

3 kN/m

2 kN/m

3m 3m

4m

6m

A

B C D

E




Q9.

200 kN

100 kN

3m 4m

3m

3m

A

B

C

E

D


Q10.

4 kN/m

4m 4m

4m

6m

A

B C

D

Documents

SA1 FRAME - Ch 3a - VTCtycnw01.vtc.edu.hk/CBE2027/SA1_FRAME.pdf · ... Take each member and joint as free-body, find the axial force, shear force and bending moment at the ends of