Chapter 4

Structure Analysis I
Chapter 4
١
Types of Structures & Loads
١Chapter

Chapter 4
Internal Loading
Developed in
Structural Members

Internal loading at a specified
Point
In General
• The loading for coplanar structure will
consist of a normal force N, shear force V,
and bending moment M.
• These loading actually represent the
resultants of the stress distribution acting over
the member’s cross-sectional sectional are

Sign Convention
+ve Sign

Procedure for analysis
• Support Reaction
• Free-Body Diagram
• Equation of Equilibrium

Example 1
Determine the internal shear and moment acting in the
cantilever beam shown in figure at sections passing through
h
points C & D

∑
M
M
kN
V
V
F
C
C
y
0
20
5(3)
5(2)
5(1)
0
15
0
5
5
5
0
=
−
−
−
−
⇒ −
=
=
=
−
−
−
⇒
=
∑
∑
kN m
M
M
M
c
c
C
.
50
0
20
5(3)
5(2)
5(1)
0
= −
=
⇒
=
∑

M
M
kN
V
V
F
C
D
y
0
20
5(3)
5(2)
5(1)
0
20
0
5
5
5
5
0
=
=
−
−
−
−
⇒
=
∑
∑
kN m
M
M
M
D
D
C
.
50
0
20
5(3)
5(2)
5(1)
0
= −
=
−
−
−
−
⇒ −
=
∑

Example 2
Determine the internal shear and moment acting in section 1 in
the beam as shown in figure
18kN
R
=
R
= 9
kN
A B
9
6kN
∑
V
∑
F y
= 0
=
3
kN
M D
= 12
kN.
m
⇒ −V
+ 9 − 6 = 0
M
at section
= 0 ⇒M
+ 6(1) − 9(2) =
0

Example 3
Determine the internal shear and moment acting in the
cantilever beam shown in figure at sections passing through
h
points C

∑
V
∑
M
F
y
= 6k
D
M
c
= 0
= 0
= 48k.
ft
⇒ −VC
+ 9 − 3 = 0
⇒M
+ 3(2) − 9(6)
c
=
0

Shear and Moment function
Procedure for Analysis:
1- Support reaction
2- Shear & Moment Function
• Specify separate coordinate x and associated origins,
extending into regions of the beam between concentrated forces
and/or couple moments or where there is a discontinuity of
distributed loading.
• Section the beam at x distance and from the free body
diagram determine V from , M at section x

Example 4
Determine the internal shear and moment Function

Example 5
Determine the internal shear and moment Function

15
1
30
2 =
=
x
w
w 2
x
30
2
2
1 0
15
30
0
x
V
F y =
−
+
⇒ −
=
∑
2
1
2
0
600
)
30(
0
0.033
30
x
x
x
M
M
x
V
=
+
⎥
⎤
⎢
⎡
+
−
⇒
=
−
=
∑
3
2
0.011
30
600
0
600
3
15
)
30(
0
x
x
M
x
M
M S
−
+
= −
=
+
⎥
⎦
⎢
⎣
+
⇒
=
∑

Example 6
Determine the internal shear and moment Function

0 < x
1
< 12
∑
V
∑
=
0
y
=F
108 − 4x
M
S
=
⇒ −V
1
+ 108 − 4x
(
x
0 1588 108 4 )
1
⇒M
+ − x + x
M = −
1588 +
108
x − 2
x
1
2
2
1
1
=
1
0
1
2
=
0

12 < x2
< 20
∑
F
y
V = 60
∑
= 0 ⇒ −V
+ 108 − 48 = 0
( x − 6
) 0
M
S
= 0 ⇒M
+ 1588 −108x
2
+ 48
2
=
M = 60x
2
−1300

Example 7
Determine the internal shear and moment Function

w 20
w
x
=
20
9
x
9
∑
V =
⎡ x ⎤
F y
= 0 ⇒ −V
+ 75 −10x
−
⎢
x =
9 ⎥ ⎣ ⎦
75 −10x
−1.11x
x
1
∑
M
S
=
0
⇒
M −
75
x −
10
x
( )
2
−
2
M =
75x
+ 5x
2
2
− 0.370x
3
1 0
2
(20)
x ⎡ x ⎤
x
⎢
(20)
x
3
=
9 ⎥
⎣ ⎦
0

Shear and Moment diagram for a
Beam

∑
ΔV
∑
ΔM
F y
= 0 ⇒ V + w(
x)
Δx
− ( V + ΔV
) = 0
= w(
x)
Δx
M
O
= 0 ⇒ −VΔx
− M
= VΔx
+ w(
x)
ε
( Δx) 2
− w(
x)
Δx
( εΔx
)
+ ( M
+ ΔM
)
=
0
for Δx
→ 0
dV
= w(
x)
⇒ ΔV
= ∫ w(
x)
dx
dx
dM
dx
= V ⇒ ΔM
= ∫V
( x)
dx

Example 1
Draw shear force
and Bending
moment Diagram
S.F.D
B.M.D

Example 2
Draw shear force
and Bending
moment Diagram
S.F.D
B.M.D

18 kN
Example 4
Draw shear force
and Bending moment
Diagram
Max. moment at x = L/2
then
wL
M =
2
M =
max
⎛ L
⎜ ⎝
2
wL
8
2
⎞
⎟ −
⎠
w ⎛
⎜ 2
⎝
L
2
⎞
⎟ ⎠
2

Example 3 Example 3
Draw shear force and Bending moment Diagram

S.F.D
B.M.D

Example 5
Draw shear force
and Bending
moment Diagram
2 x = 14
x = 7
∑
M
M
S
= 49
= −M
−14(3.5)
+ 14(7)

Example 6a
Draw shear force
and Bending
moment Diagram
S.F.D
B.M.D

Example 6b
Draw shear force
and Bending
moment Diagram
S.F.D
B.M.D

Example 6c
Draw shear force
and Bending
moment Diagram
S.F.D
B.M.D

Example 6d
Draw shear force
and Bending
moment Diagram

Group Work
Draw shear force and Bending moment Diagram

Example 1
Draw shear force and Bending moment Diagram

V(kN)

Example 2
Draw shear force and Bending moment Diagram

Example 2
Draw shear force
and Bending
moment Diagram

Example 3
Draw shear force
and Bending
moment Diagram

+
+
+
+

Example 4
Draw shear force
and Bending
moment Diagram

+
+
+

Problem 1
Draw shear force and Bending moment Diagram

30.5 23.5
+
-
-
+
+

Problem 2
Draw shear force and Bending moment Diagram

2
3
x
at →V
5 =
M
M
= 0
5 2
=
12
x ⇒ x
2
2
3
x( RA)
=
3
=
= 11.55
3.46m
(3.46)(5)

Example 1
Draw shear force and Bending moment Diagram
Hinge

∑
Reaction Calculation
( )
C
M
k
A
A
M
y
y
left
B
0
60
4(32)
20(27)
5(16)
18(6)
(12)
0
4
0
60
20(5)
10
0
=
=
−
+
⇒ −
=
∑
∑
E
F
k
C
C
M
y
y
E
0
0
45
0
60
4(32)
20(27)
5(16)
18(6)
(12)
0
=
⇒
=
=
=
−
−
+
+
+
⇒
=
∑
∑
k
E
E
E
F
y
y
x
x
6
0
45
4
20
5
18
0
F
0
0
y
=
=
−
−
+
+
+
⇒
=
=
⇒
=
∑
∑
y

Frames (Example 1)
Draw Bending moment Diagram

Support reaction & Free Body diagram

_
_
S.F.D
B.M.D

+ S.F.D
- - B.M.D

Frames (Example 2)
Draw shear force and Bending moment Diagram

+
NFD N.F.D
+
_
S.F.D
N.F.D
S.F.D
B.M.D
+
-
+
B.M.D
N.F.D
+ -

Frames (Example 3)
Draw shear force and Bending moment Diagram

N.F.D
S.F.D
B.M.D
-
-
-

_
NFD N.F.D
64
+
26
S.F.D
+
B.M.D
251.6

NFD N.F.D
S.F.D
BMD B.M.D
168

64
+
13.22
S.F.D
_
26
36
_
31.78
432
_
432 139.3
_
+
251.6
168+
B.M.D

Frames (Example 4)
Draw shear force and Bending moment Diagram

S.F.D
B.M.D
+
+

_
S.F.D
+
B.M.D

Frames (Example 5)
Draw shear force and Bending moment Diagram

Frames (Example 6)
Draw shear force and Bending moment Diagram

N.F.D S.F.D B.M.D
_
_
_

_
N.F.D
+
_
+ S.F.D
_
+
_
B.M.D

B.M.D
S.F.D
N.F.D
_
+
_

Frames (Example 7)
Draw Normal force, shear force and Bending moment
Diagram

10kN/m
60kN
20.8
47.7
110
53.7
26.56 o
43.2
26.8
10.5

N.F.D S.F.D B.M.D
S.F.D
B.M.D

N.F.D
S.F.D
B.M.D

B.M.D

Moment diagram constructed by the
Example 1
method of superposition

Example 2.a

Example 2.b

Problem 1
D N l f h f d B di t
Draw Normal force, shear force and Bending moment
Diagram

Problem 2
D N l f h f d B di t
Draw Normal force, shear force and Bending moment
Diagram