CBE2027 Structural Analysis I
Chapter 3 – Analysis of Determinate Plane Frames
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.
Frame with an
internal hinge
Rigid Frame
Truss (Pin-jointed Frame)
HD in Civil Engineering
Page 3-1
CBE2027 Structural Analysis I
Chapter 3 – Analysis of Determinate Plane Frames
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.
B
B'
C
C'
P.I.
H
P.I.
P.I.
D
A
L
HD in Civil Engineering
Page 3-2
CBE2027 Structural Analysis I
Chapter 3 – Analysis of Determinate Plane Frames
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.
HD in Civil Engineering
Page 3-3
CBE2027 Structural Analysis I
Chapter 3 – Analysis of Determinate Plane Frames
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.
10 kN
2 kN
B
6m
C
A
4m
HD in Civil Engineering
Page 3-4
CBE2027 Structural Analysis I
Chapter 3 – Analysis of Determinate Plane Frames
Solution:
10 kN
2 kN
B
6m
C
A
HA
V
MA A
4m
∑X = 0, HA = 2 kN
∑Y = 0, VA = 10 kN
Take moment about A,
MA = 10*4 + 2*6 = 52 kNm
HD in Civil Engineering
Page 3-5
CBE2027 Structural Analysis I
Chapter 3 – Analysis of Determinate Plane Frames
Free-body diagrams for the members and joint, and their member end
forces.
10 kN
B
2
40
40
10
2
10
40
40
10
2 2
B
10
C
2kN
B
Free-body Diagrams
A
2 kN
10 kN
52 kNm
HD in Civil Engineering
Page 3-6
CBE2027 Structural Analysis I
Chapter 3 – Analysis of Determinate Plane Frames
Axial force, shear force and bending moment diagrams
10
10
2
2
B
-10
B
+
2
C
C
-
A
-10
2
Axial Force (kN)
A
Shear Force (kN)
- 40
- 40
B
C
Bending Moment (kNm)
A
- 52
HD in Civil Engineering
Page 3-7
CBE2027 Structural Analysis I
Chapter 3 – Analysis of Determinate Plane Frames
Example 2
Determine the support reactions, and draw the axial force, shear force and
bending moment diagrams for the frame. Joint B is a rigid joint.
B
4m
A
C
12 kN
6m
HD in Civil Engineering
3m
Page 3-8
CBE2027 Structural Analysis I
Chapter 3 – Analysis of Determinate Plane Frames
Solution:
MA
HA
A
B
4m
VA
C
12 kN
6m
3m
∑X = 0, HA = 0kN
∑Y = 0, VA = 12kN
Take moment about A,
MA = 12*9 = 108 kNm
HD in Civil Engineering
Page 3-9
CBE2027 Structural Analysis I
Chapter 3 – Analysis of Determinate Plane Frames
Free-body diagrams for the members and joint, and their member end
forces.
108 kNm
B
36
36
B
36
A
12 kN
12
12
12
36
B
12
Free-body Diagrams
C
12 kN
HD in Civil Engineering
Page 3-10
CBE2027 Structural Analysis I
Chapter 3 – Analysis of Determinate Plane Frames
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.
12
9.6
φ
7.2
B
36
4
φ
5
3
C
12∗3/5 = 7.2
φ 12∗4/5 = 9.6
12
HD in Civil Engineering
Page 3-11
CBE2027 Structural Analysis I
Chapter 3 – Analysis of Determinate Plane Frames
Axial force, shear force and bending moment diagrams
A
B
9.6
+
9.6
Axial Force (kN)
C
12
12
7.2
A
B
7.2
Shear Force (kN)
C
-108
-36
A
B
Bending Moment (kNm)
HD in Civil Engineering
-36
C
0
Page 3-12
CBE2027 Structural Analysis I
Chapter 3 – Analysis of Determinate Plane Frames
Example 3
Determine the support reactions, and draw the axial force, shear force and
bending moment diagrams for the frame. Joints C and E are rigid joints.
96 kN
C
E
B
F
4.5 m
48 kN
5m
3m
D
A
3m
HD in Civil Engineering
3m
Page 3-13
CBE2027 Structural Analysis I
Chapter 3 – Analysis of Determinate Plane Frames
Solution:
96 kN
C
E
48 kN
5m
3m
D
B
4.5 m
F
VF
A
HA
VA
3m
3m
∑X = 0, HA = 48 kN
Take moment about A,
48*4.5 + 96*3 = VF*6,
∑Y = 0, VA + VF = 96 kN
HD in Civil Engineering
⇒ VF = 84 kN
⇒ VA = 12 kN
Page 3-14
CBE2027 Structural Analysis I
Chapter 3 – Analysis of Determinate Plane Frames
Free-body diagrams for the members and joints, and their member end
forces.
216 216
C
12
12
216
216
12
C
12
96 kN
D
E
84
84
84
84
E
C
48 kN
E
B
F
Free-body Diagrams
A
84
48
12
HD in Civil Engineering
Page 3-15
CBE2027 Structural Analysis I
Chapter 3 – Analysis of Determinate Plane Frames
Axial force, shear force and bending moment diagrams
E
C
-12
-84
D
-
B
F
-12 A
-84
Axial Force (kN)
12
12 D
E
C
48
B
-84
A
48
HD in Civil Engineering
F
-84
Shear Force (kN)
Page 3-16
CBE2027 Structural Analysis I
Chapter 3 – Analysis of Determinate Plane Frames
C
216 D
216
B
E
252
216
F
A
Bending Moment (kNm)
HD in Civil Engineering
Page 3-17
CBE2027 Structural Analysis I
Chapter 3 – Analysis of Determinate Plane Frames
Example 4
Determine the support reactions, and draw the axial force, shear force and
bending moment diagrams for the frame. Joints C and E are rigid joints.
96 kN
C
E
B
F
4.5 m
48 kN
5m
3m
D
A
3m
HD in Civil Engineering
3m
Page 3-18
CBE2027 Structural Analysis I
Chapter 3 – Analysis of Determinate Plane Frames
Solution:
96 kN
C
E
48 kN
5m
3m
D
B
4.5 m
F
HA
A
V
MA A
3m
3m
∑X = 0, HA = 48 kN
∑Y = 0, VA = 96 kN
Take moment about A,
48*4.5 + 96*3 = MA, ⇒ MA = 504 kNm
HD in Civil Engineering
Page 3-19
CBE2027 Structural Analysis I
Chapter 3 – Analysis of Determinate Plane Frames
Free-body diagrams for the members and joints, and their member end
forces.
288 288
C
96
96
288
288
96
96 kN
C
96
E
D
C
E
E
48 kN
B
F
Free-body Diagrams
A
48
96
504
HD in Civil Engineering
Page 3-20
CBE2027 Structural Analysis I
Chapter 3 – Analysis of Determinate Plane Frames
Axial force, shear force and bending moment diagrams
-96
E
C
D
-
B
F
A
Axial Force (kN)
-96
96
96
E
C
48
D
B
F
48
HD in Civil Engineering
A Shear Force (kN)
Page 3-21
CBE2027 Structural Analysis I
Chapter 3 – Analysis of Determinate Plane Frames
-288
0
-288
C
-288
E
D
B
F
A
-504
HD in Civil Engineering
Bending Moment (kNm)
Page 3-22
CBE2027 Structural Analysis I
Chapter 3 – Analysis of Determinate Plane Frames
Example 5
Determine the support reactions, and draw the axial force, shear force and
bending moment diagrams for the frame. Joints B and C are rigid joints.
15 kN/m
B
C
20 kN
2m
50 kN
2m
6m
D
E
A
8m
HD in Civil Engineering
Page 3-23
CBE2027 Structural Analysis I
Chapter 3 – Analysis of Determinate Plane Frames
Solution:
15 kN/m
B
C
20 kN
2m
50 kN
2m
6m
D
E
VE
A
HA
VA
8m
∑X = 0, HA + 20 = 50,
⇒ HA = 30 kN
Take moment about A,
50*6 + 15*8*8/2 – 20*4 – VE*8 = 0
⇒ VE = 87.5 kN
∑Y = 0, VA + VE = 15*8 kN,
HD in Civil Engineering
⇒ VA = 32.5 kN
Page 3-24
CBE2027 Structural Analysis I
Chapter 3 – Analysis of Determinate Plane Frames
Free-body diagrams for the members and joints, and their member end
forces.
15 kN/m
180 180
50 kN B
20
32.5
20
32.5
180
180
30
30
32.5
B
40 87.5
40
C
20
87.5
20
C
20
87.5
40
40
32.5
87.5
20
B
C
20 kN
D
E
Free-body Diagrams
30
87.5
A
32.5
Axial force, shear force and bending moment diagrams
-20
-20
-87.5
-
-32.5
B
C
D
-
E
-32.5
HD in Civil Engineering
A
-87.5
Axial Force (kN)
Page 3-25
CBE2027 Structural Analysis I
Chapter 3 – Analysis of Determinate Plane Frames
2.17m
32.5
30
20
C
B
-87.5
D
20
E
30
Shear Force (kN)
A
2.17m
180
B
C
180
-40
-40
D
215.2
E
A Bending Moment (kNm)
HD in Civil Engineering
Page 3-26
CBE2027 Structural Analysis I
Chapter 3 – Analysis of Determinate Plane Frames
Example 6
Determine the support reactions, and draw the axial force, shear force and
bending moment diagrams for the frame. Joints B and D are rigid joints.
30 kN
B
D
3m
C
10 kN
3m
8m
E
F
5kN/m
A
3.5m
HD in Civil Engineering
3.5m
Page 3-27
CBE2027 Structural Analysis I
Chapter 3 – Analysis of Determinate Plane Frames
Solution:
30 kN
B
D
3m
C
3m
8m
E
10 kN
F
5kN/m
VF
A
HA V
A
3.5m
∑X = 0, HA + 10 = 5*8/2,
3.5m
⇒ HA = 10 kN
Take moment about A,
5*(8/2)*(8/3) + 30*3.5 - 10*(3+2) – VF*7 = 0
⇒ VF = 15.5 kN
∑Y = 0, VA + VF = 30 kN,
HD in Civil Engineering
⇒ VA = 14.5 kN
Page 3-28
CBE2027 Structural Analysis I
Chapter 3 – Analysis of Determinate Plane Frames
Free-body diagrams for the members and joints, and their member end
forces.
30 kN
B
10
26.7
14.5
14.5
10
26.7
10
D 10
B
14.5
C
15.5
30
30
C
10
15.5
26.7
26.7
14.5
10
15.5
10
D
B
E
Free-body Diagrams
5kN/m
10
15.5
30
30
10 kN
F
15.5kN
A
10kN
14.5kN
HD in Civil Engineering
Page 3-29
CBE2027 Structural Analysis I
Chapter 3 – Analysis of Determinate Plane Frames
Axial force, shear force and bending moment diagrams
Determination of the position of zero shear force and the maximum bending
moment of member AB.
5x/8 kN/m
14.5
M
14.5
B
V
26.7
10
x
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.
HD in Civil Engineering
Page 3-30
CBE2027 Structural Analysis I
Chapter 3 – Analysis of Determinate Plane Frames
-10
-10
-
B
-14.5
D
-15.5
C
E
-
F
-14.5
A
-15.5
Axial Force (kN)
14.5
14.5
B
D
C
-10
10
5.66m
-15.5
E
10
2.34m
F
10
HD in Civil Engineering
A
Shear Force (kN)
Page 3-31
CBE2027 Structural Analysis I
Chapter 3 – Analysis of Determinate Plane Frames
-30
D
-26.7
2.93m
-26.7
C
-30
B
24.3
E
2.34m
11.0
A
HD in Civil Engineering
F
Bending Moment (kNm)
Page 3-32
CBE2027 Structural Analysis I
Chapter 3 – Analysis of Determinate Plane Frames
Example 7
Determine the support reactions, and draw the axial force, shear force and
bending moment diagrams for the frame. Joint B is a rigid joint and joint C
is an internal hinge of the frame.
15 kN/m
30 kN
C
A
D
4m
B
6m
HD in Civil Engineering
Page 3-33
CBE2027 Structural Analysis I
Chapter 3 – Analysis of Determinate Plane Frames
Solution:
15 kN/m
30 kN
C
A
D
4m
B
VA
HA
VD
HD
6m
Take moment about A,
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,
HC
65
C
D
65
HD
Take moment about C,
HD*4 = 0, ⇒ HD = 0 kN
∑X = 0, HA + HD = 30,
HD in Civil Engineering
⇒ HA = 30 kN
Page 3-34
CBE2027 Structural Analysis I
Chapter 3 – Analysis of Determinate Plane Frames
Free-body diagrams for the members and joints, and their member end
forces.
30 kN B
15 kN/m
120
120
25
25 30
25
B
C
C
65
65
65
120
120
30
65
25
B
C
Free-body Diagram
D
A
30
25
65
Axial force, shear force and bending moment diagrams
-25
-65
B
C
-25
D
A
-65
Axial Force (kN)
HD in Civil Engineering
Page 3-35
CBE2027 Structural Analysis I
Chapter 3 – Analysis of Determinate Plane Frames
1.67m
25
30
30
B
C
A
D -65
Shear Force (kN)
1.67m
120
B
C
120
140.9
A
D
Bending Moment (kNm)
HD in Civil Engineering
Page 3-36
CBE2027 Structural Analysis I
Chapter 3 – Analysis of Determinate Plane Frames
Example 8
Determine the support reactions, and draw the axial force, shear force and
bending moment diagrams for the frame. Joints B and C are rigid joints,
and joint E is an internal hinge of the frame.
3m
F E
C
4m
3 kN/m
B
20 kN
D
A
4m
HD in Civil Engineering
4m
Page 3-37
CBE2027 Structural Analysis I
Chapter 3 – Analysis of Determinate Plane Frames
Solution:
3m
B
20 kN
C
4m
3 kN/m
F E
D
A
VA
HA
4m
4m
Take moment about A,
20*3 + 3*4*4/2 – VD*8 = 0
⇒ VD = 10.5 kN
∑Y = 0, VA + VD = 20 kN,
VD
HD
⇒ VA = 9.5 kN
Consider the free-body of ECD,
VE
HE E
C
D
10.5
HD
Take moment about E,
HD*4 = VD*4, ⇒ HD = 10.5 kN
Consider the whole frame,
∑X = 0, HA + HD = 3*4,
HD in Civil Engineering
⇒ HA = 1.5 kN
Page 3-38
CBE2027 Structural Analysis I
Chapter 3 – Analysis of Determinate Plane Frames
Free-body diagrams for the members and joints, and their member end
forces.
20 kN
B
10.5
9.5
9.5
18
18 10.5
B
10.5
9.5
C
F E
10.5
C10.5
10.5
10.5
42 42 10.5 10.5
42
18
42
18
9.5
10.5
3 kN/m
10.5
10.5
B
A
1.5 9.5
HD in Civil Engineering
C
Free-body Diagrams
D
10.5
Page 3-39
10.5
CBE2027 Structural Analysis I
Chapter 3 – Analysis of Determinate Plane Frames
Axial force, shear force and bending moment diagrams
-10.5
-10.5
-9.5
-10.5
B
F E
C
-
A
-9.5
9.5
D
Axial Force (kN)
9.5
-10.5 F
B
E
0.5m
-10.5
1.5
HD in Civil Engineering
A
-10.5
C
10.5
-10.5
Shear Force (kN)
D
10.5
Page 3-40
CBE2027 Structural Analysis I
Chapter 3 – Analysis of Determinate Plane Frames
-42
-18
F
-18
0.5m
B
10.5
0.375
A
HD in Civil Engineering
E
C
-42
D
Bending Moment (kNm)
Page 3-41
CBE2027 Structural Analysis I
Chapter 3 – Analysis of Determinate Plane Frames
Example 9
Determine the support reactions, and draw the axial force, shear force and
bending moment diagrams for the frame. Joints B and E are rigid joints, and
joint D is an internal hinge of the frame.
40 kN
5 kN/m
D
6m
C
4m
E
B
F
A
1m 2m
HD in Civil Engineering
3m
Page 3-42
CBE2027 Structural Analysis I
Chapter 3 – Analysis of Determinate Plane Frames
Solution:
40 kN
5 kN/m
D
6m
C
4m
E
B
F
HF
A
HA
VF
VA
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)
HD in Civil Engineering
Page 3-43
CBE2027 Structural Analysis I
Chapter 3 – Analysis of Determinate Plane Frames
Consider the free-body diagram of ABCD,
40 kN
VD
B
D
C
HD
A
HA
VA
Take moment about D,
40*2 + HA*6 – VA*3 = 0
⇒ 3VA – 6HA - 80 = 0
Sub. (1) into (2)
HA + 120 - 6HA – 80 = 0
⇒ HA = 8 kN
(2)
(3)
Sub. (3) into (1),
⇒ 3VA = 8 + 120
⇒ VA = 42.7 kN
Consider the whole frame,
∑Y = 0, VA + VF = 40 kN,
⇒VF = 2.7 kN (↓)
Consider the whole frame,
∑X = 0, HA + HF = 5*4,
HD in Civil Engineering
⇒ VF = -2.7 kN
⇒ HF = 12 kN
Page 3-44
CBE2027 Structural Analysis I
Chapter 3 – Analysis of Determinate Plane Frames
Free-body diagrams for the members and joints, and their member end
forces.
40 kN
B
8 42.7
2.7
48
48
8
B
8
E
42.7
42.7
C
8 8
8
8
2.7
D
48
48
42.7
8
E8
2.7
8
8
2.7
8
E
B
5 kN/m
F
Free-body Diagrams
12
2.7
A
8
42.7
Axial force, shear force and bending moment diagrams
-8
-42.7
-8
-
B
C
E
2.7
D
+
F
2.7
A Axial Force (kN)
-42.7
HD in Civil Engineering
Page 3-45
CBE2027 Structural Analysis I
42.7
Chapter 3 – Analysis of Determinate Plane Frames
42.7
2.7 E
2.7
B C
8
-8
2.4m
D
-12
A
F
-8
Shear Force (kN)
-48
-48
B C
-5.3
E
8
D
8
2.4m
14.4
F
A
Bending Moment (kNm)
HD in Civil Engineering
Page 3-46
CBE2027 Structural Analysis I
Chapter 3 – Analysis of Determinate Plane Frames
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.
6 kN/m
2.5m
C
D
30 kN
2.5m
5m
B
E
A
6m
Q2.
50 kN 40 kN
1.5m 2m 1.5m
D
E
15 kN/m
6m
C
4m
B
E is an internal hinge
A
F
HD in Civil Engineering
Page 3-47
CBE2027 Structural Analysis I
Chapter 3 – Analysis of Determinate Plane Frames
Q3.
60 kN
B
30 kN
D
6 kN/m
C
E
5m
B & E are internal hinges
F
A
3m
3m
3m
Q4.
E
20 kN
2m
24 kN
D
C
2 kN/m
4m
B
A
3m
HD in Civil Engineering
3m
Page 3-48
CBE2027 Structural Analysis I
Chapter 3 – Analysis of Determinate Plane Frames
Q5.
20 kN
10 kNm
B
C
6 kNm
D
E
4m
2 kN/m
C is an internal hinge
A
F
1m
2m
1m
Q6.
15 kN/m
D
F
E
3m
C
G
E is an internal hinge
65 kN
3m
B
H
A
3m
HD in Civil Engineering
5m
5m
2m
Page 3-49
CBE2027 Structural Analysis I
Chapter 3 – Analysis of Determinate Plane Frames
Q7.
10 kN/m
30 kN
C
E
3m
D
F
A
G
3m
B
3m
3m
B, D & F are internal hinges
Q8.
2 kN/m
B
D
C
6m
4m
C is an internal hinge
E
3 kN/m
A
3m
HD in Civil Engineering
3m
Page 3-50
CBE2027 Structural Analysis I
Chapter 3 – Analysis of Determinate Plane Frames
Q9.
200 kN
C
3m
B
100 kN
3m
D
C is an internal hinge
E
A
3m
4m
Q10.
4 kN/m
B
4m
C
A
D
4m
HD in Civil Engineering
6m
4m
Page 3-51