Structure Analysis
y I
Chapter 4
١
Types of Structures & Loads
١Chapter
Chapter 4
Internal Loading
D l
Developed
d in
i
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.
M
• These loading actually represent the
resultants of the stress distribution acting over
the member
member’ss cross-sectional
cross sectional are
Sign Convention
+ve Sign
Procedure for analysis
• Support Reaction
• Free
Free-Body
Body Diagram
• Equation of Equilibrium
Example 1
Determine the internal shear and moment acting in the
cantilever
il
b
beam shown
h
iin fi
figure at sections
i
passing
i through
h
h
points C & D
∑F
y
=0
⇒ VC − 5 − 5 − 5 = 0
VC = 15kN
∑M
C
= 0 ⇒ − M c − 5(1) − 5(2) − 5(3) − 20 = 0
M c = −50kN .m
∑F
y
=0
⇒ VD − 5 − 5 − 5 − 5 = 0
VC = 20kN
∑M
C
= 0 ⇒ − M D − 5(1) − 5(2) − 5(3) − 20 = 0
M D = −50kN .m
Example
p 2
Determine the internal shear and moment acting in section 1 in
the beam as shown in figure
18kN
R A = RB = 9kN
6kN
∑F
y
=0
⇒ −V + 9 − 6 = 0
V = 3kN
∑M
at section
= 0 ⇒M + 6(1) − 9(2) = 0
M D = 12kN .m
Example 3
Determine the internal shear and moment acting in the
cantilever
il
b
beam shown
h
iin fi
figure at sections
i
passing
i through
h
h
points C
∑F
=0
y
⇒ −VC + 9 − 3 = 0
V = 6k
∑M
c
= 0 ⇒M c + 3(2) − 9(6) = 0
M D = 48k . fft
Shear and Moment function
Procedure for Analysis:
1 Support
1S
reaction
i
2- Shear & Moment Function
ƒ Specify separate coordinate x and associated origins,
origins
extending into regions of the beam between concentrated forces
and/or couple
p moments or where there is a discontinuity
y of
distributed loading.
ƒ 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
w
w
x
=
2
30
= 151
2
x
30
2
x
∑ Fy = 0 ⇒ −V + 30 − 12 15 = 0
V = 30 − 0.033x 2
⎡ 1 x2 ⎤ x
∑ M S = 0 ⇒ M − 30( x) + ⎢ 2 15 ⎥ 3 + 600 = 0
⎣
⎦
M = −600 + 30 x − 0.011x 3
Example 6
Determine the internal shear and moment Function
0 < x1 < 12
∑F
y
=0
⇒ −V + 108 − 4 x1 = 0
V = 108 − 4 x1
∑ M S = 0 ⇒M + 1588 − 108x1 + 4 x1
M = −1588 + 108 x1 − 2 x1
2
( )= 0
x1
2
12 < x2 < 20
∑F
y
=0
⇒ −V + 108 − 48 = 0
V = 60
∑M
S
= 0 ⇒M + 1588 − 108 x2 + 48( x2 − 6) = 0
M = 60 x2 − 1300
Example 7
Determine the internal shear and moment Function
w
w
x
=
20
9
20
x
9
x ⎤
⎡
⇒ −V + 75 − 10 x − ⎢ 12 (20) x ⎥ = 0
9 ⎦
⎣
V = 75 − 10 x − 1.11x 2
∑ Fy = 0
x ⎤x
⎡1
x
(
)
M
=
0
⇒
M
−
75
x
−
10
x
−
(
20
)
x⎥ 3 = 0
∑ S
2
⎢2
9 ⎦
⎣
M = 75 x + 5 x 2 − 0.370 x 3
Shear and Moment diagram for a
Beam
∑F
y
=0
⇒ V + w( x)Δx − (V + ΔV ) = 0
ΔV = w( x)Δx
∑M
O
= 0 ⇒ − VΔx − M − w( x)Δx(εΔx ) + ( M + ΔM ) = 0
ΔM = VΔx + w( x)ε (Δx )
2
for Δx → 0
dV
= w( x)
dx
dM
=V
dx
⇒ ΔV = ∫ w( x) dx
⇒ 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
Example 4
Draw shear force
and Bending moment
Diagram
Max. moment at x
then
= L/2
M=
wL ⎛ L ⎞ w ⎛ L ⎞
⎜ ⎟− ⎜ ⎟
2 ⎝2⎠ 2⎝2⎠
M max
wL2
=
8
2
18 kN
Example 3
Draw shear force and Bending moment Diagram
S.F.D
B.M.D
Example
p 5
Draw shear force
and Bending
moment Diagram
Di
2 x = 14
x=7
∑M
S
= − M − 14(3.5) + 14(7)
M = 49
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
D
Draw
shear
h fforce and
dB
Bending
di momentt Di
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
D
Draw
shear
h fforce and
dB
Bending
di momentt Di
Diagram
30.5
23.5
+
-
+
+
-
Problem 2
D
Draw
shear
h fforce and
dB
Bending
di momentt Di
Diagram
2
3
x
at → V = 0
5 = 125 x 2 ⇒ x = 3.46m
M = 23 x( RA ) = 23 (3.46)(5)
M = 11.55
Example 1
D
Draw
shear
h fforce and
dB
Bending
di momentt Di
Diagram
Hinge
Reaction Calculation
∑M
B (left )
= 0 ⇒ −10 Ay + 20(5) − 60 = 0
Ay = 4k
∑M
E
= 0 ⇒ C y (12) + 18(6) + 5(16) + 20(27) − 4(32) − 60 = 0
C y = 45k
∑F
∑F
x
= 0 ⇒ Ex = 0
y
= 0 ⇒ E y + 18 + 5 + 20 − 4 − 45 = 0
E y = 6k
Frames (Example 1)
Draw Bending moment Diagram
S
Support
t reaction
ti & Free
F
Body
B d di
diagram
_
_
S.F.D
B.M.D
+
- -
S.F.D
B.M.D
Frames (Example 2)
D
Draw
shear
h fforce and
dB
Bending
di momentt Di
Diagram
+
+
N.F.D
S.F.D
B.M.D
NFD
N.F.D
_
S.F.D
B.M.D
+
N.F.D
+
+
-
Frames (Example 3)
D
Draw
shear
h fforce and
dB
Bending
di momentt Di
Diagram
N.F.D
S.F.D
B.M.D
-
-
_
64
NFD
N.F.D
+
26
+
251.6
S.F.D
B.M.D
NFD
N.F.D
S.F.D
BMD
B.M.D
168
64
S.F.D
13 22
13.22
+
26
_
_
36
31.78
432
432
_
_
+
251.6
139.3
168+
B.M.D
Frames (Example 4)
D
Draw
shear
h fforce and
dB
Bending
di moment Di
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)
D
Draw
shear
h fforce and
dB
Bending
di momentt Di
Diagram
S.F.D
B.M.D
_
_
_
N.F.D
N.F.D
_
+
+
_
_
_
+
S.F.D
B.M.D
S.F.D
N.F.D
_
+
_
B.M.D
Frames ((Example
p 7))
Draw Normal force, shear force and Bending moment
Diagram
g
10kN/m
60kN
53.7
110
47.7
26.56o
43.2
20.8
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
method of superposition
Example 1
Example 2.a
Example 2.b
Problem 1
Draw Normal
D
N
l force,
f
shear
h force
f
and
dB
Bending
di momentt
Diagram
Problem 2
Draw Normal
D
N
l force,
f
shear
h force
f
and
dB
Bending
di momentt
Diagram