Post on 11-Mar-2018
Structure Analysis IStructure Analysis IyChapter 4
yChapter 4
Chapter١ Types of Structures & Loads ١
Chapter 4Chapter 4
Internal Loading D l d i Developed in
Structural MembersStructural Members
Internal loading at a specified Point
In General• The loading for coplanar structure will The loading for coplanar structure will
consist of a normal force N, shear force V, and bending moment Mand bending moment M.
• These loading actually represent the resultants of the stress distribution acting over the member’s cross-sectional arethe member s cross sectional are
Sign Convention+ve Sign
Procedure for analysisProcedure for analysis
• Support Reaction• Free-Body DiagramFree Body Diagram• Equation of Equilibrium
Example 1Example 1Determine the internal shear and moment acting in the
il b h i fi i i h h cantilever beam shown in figure at sections passing through points C & D
∑
MMkNV
V F
C
Cy
020)3(5)2(5)1(5015
05550
=−−−−−⇒=
=
=−−−⇒=
∑
∑
mkNMMM
c
cC
.50020)3(5)2(5)1(5 0
−=
=⇒=∑
MMkNV
V F
C
Dy
020)3(5)2(5)1(5020
055550
=
=−−−−⇒=
∑
∑
mkNMMM
D
DC
.50020)3(5)2(5)1(5 0
−=
=−−−−−⇒=∑
Example 2pDetermine the internal shear and moment acting in section 1 in the beam as shown in figure
18kN
kNRR 9==
18kN
kNRR BA 9==
6kN
kNV
V Fy
3
0690
=
=−+−⇒=∑
mkNMMM
kNV
120)2(9)1(6 0
3
sectionat =−+⇒=
=
∑mkNM D .12=
Example 3Example 3Determine the internal shear and moment acting in the
il b h i fi i i h h cantilever beam shown in figure at sections passing through points C
kV
V F Cy
6
0390
=
=−+−⇒=∑
ftkMMM
D
c
.480)6(9)2(3 0c
=
=−+⇒=∑fD
Shear and Moment functionShear and Moment functionProcedure for Analysis:1 S i1- Support reaction2- Shear & Moment Function
Specify separate coordinate x and associated origins 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 p ydistributed loading.
Section the beam at x distance and from the free body
diagram determine V from , M at section x
Example 4Example 4Determine the internal shear and moment Function
Example 5Example 5Determine the internal shear and moment Function
151
302 ==x
ww 2
x30
2
21 0
15300 xV Fy =−+−⇒=∑
21
2
0600)(300
033.030
xxxMM
xV
=+⎥⎤
⎢⎡
+−⇒=
−=
∑3
2
011.030600
0600315
)(30 0
xxM
xMM S
−+−=
=+⎥⎦
⎢⎣
+⇒=∑
Example 6Example 6Determine the internal shear and moment Function
1 120 x <<
∑1
1
4108
041080
xV
xV Fy
−=
=−+−⇒=∑
( )2
211S
21081588
041081588 0 1
xxM
xxMM x
−+−=
=+−+⇒=∑11 21081588 xxM +=
04810802012 2
=−+−⇒=
<<
∑ V Fx
y
( ) 06481081588 060
22S =−+−+⇒=
=
∑ xxMMV
( )130060 2
22S
−=
∑xM
Example 7Example 7Determine the internal shear and moment Function
20w
w 20
x920=x
w9
x
21 0
9)20(10750 xxxV Fy =⎥⎦
⎤⎢⎣⎡−−+−⇒=∑
( ) 1
2
0)20(10750
11.11075
xxxxMM
xxV
xx =⎥⎤
⎢⎡⇒=
−−=
∑ ( )32
321
2
370.0575
09
)20(1075 0
xxxM
xxxMM xxS
−+=
=⎥⎦⎢⎣−−−⇒=∑
Shear and Moment diagram for a Beam
∑
( ))(
0)()(0
xxwV
VVxxwV Fy
Δ=Δ
=Δ+−Δ+⇒=
∑
∑
( )( )2
O
)(
0)()( 0
xxwxVM
MMxxxwMxVM
Δ+Δ=Δ
=Δ++ΔΔ−−Δ−⇒=∑ε
ε
→ΔdV
xfor 0
∫=Δ⇒= dxxwVxwdxdV )( )(
∫=Δ⇒= dxxVMVdx
dM )( ∫dx
Example 1Example 1Draw shear force and Bending and Bending moment Diagram
S.F.DS.F.D
B.M.D
Example 2Example 2Draw shear force and Bending and Bending moment Diagram
S.F.D
B.M.D
Example 418 kN
Example 4Draw shear force and Bending moment and Bending moment Diagram
Max. moment at x = L/2then
2222
2LwLwLM ⎟⎠⎞
⎜⎝⎛−⎟
⎠⎞
⎜⎝⎛=
8
22222
maxwLM =
⎠⎝⎠⎝
8
Example 3Example 3Draw shear force and Bending moment Diagram
S.F.D
B.M.D
Example 5Draw shear force and Bending
Di
p
moment Diagram
142 =x
)7(14)5.3(147
+−−=
=
∑ MMx
S
49=∑M
S
Example 6aExample 6aDraw shear force and Bending and Bending moment Diagram
S.F.D
B.M.D
Example 6bExample 6bDraw shear force and Bending and Bending moment Diagram
S.F.D
B.M.D
Example 6cExample 6cDraw shear force and Bending and Bending moment Diagram
S.F.D
B.M.D
Example 6dExample 6dDraw shear force and Bending and Bending moment Diagram
Group WorkGroup WorkD h f d B di t DiDraw shear force and Bending moment Diagram
Example 1Draw shear force and Bending moment Diagram
Example 1
V(kN)
Example 2Example 2Draw shear force and Bending moment Diagram
++
Example 2Example 2Draw shear force and Bending moment Diagram
Example 3Example 3Draw shear force and Bending moment Diagram
+++
+ +
Example 4Example 4Draw shear force and Bending moment Diagram
+++
+
Problem 1Problem 1D h f d B di t DiDraw shear force and Bending moment Diagram
30.5 23.5
+
-
+
-
+ +
Problem 2Problem 2D h f d B di t DiDraw shear force and Bending moment Diagram
x32
46.350
2125 =⇒=
=→
mxxVat
55.11)5)(46.3()( 3
232
=
==
MRxM A
Example 1Example 1D h f d B di t DiDraw shear force and Bending moment Diagram
HingeHinge
∑
Reaction Calculation
( )
CM
kA
AM
y
yleftB
060)32(4)27(20)16(5)6(18)12(0
4
060)5(20100
=
=−+−⇒=
∑
∑
EF
kC
CM
y
yE
00
45
060)32(4)27(20)16(5)6(18)12( 0
=⇒=
=
=−−+++⇒=
∑
∑
kE
E
EF
y
y
xx
6
045420518 0F
0 0
y
=
=−−+++⇒=
=⇒=
∑∑
y
Frames (Example 1)Frames (Example 1)Draw Bending moment Diagram
S t ti & F B d diSupport reaction & Free Body diagram
__ _
S.F.D B.M.DS.F.D B.M.D
++ S.F.D
- - B.M.D
Frames (Example 2)Frames (Example 2)D h f d B di t DiDraw shear force and Bending moment Diagram
N F D+ N.F.D+
S.F.D+
_
B.M.DN.F.D S.F.D B.M.DN.F.D S.F.DB.M.D
+
+-
N.F.D
+ -
Frames (Example 3)Frames (Example 3)D h f d B di t DiDraw shear force and Bending moment Diagram
B.M.DN.F.D S.F.D
-
--
N F D_
+
N.F.D
64 +S.F.D
26
+ B.M.D
251.6
N F D
B M D
N.F.D
S.F.D
B.M.D
168168
S.F.D6413 22
26
13.22+
36
_
_36
31.78
B.M.D432 139.3432
+
_
168251.6
+_
+
Frames (Example 4)Frames (Example 4)D h f d B di DiDraw shear force and Bending moment Diagram
S.F.DS.F.D
+
B.M.D
+
S.F.D_
B.M.D+
Frames (Example 5)Frames (Example 5)Draw shear force and Bending moment DiagramDraw shear force and Bending moment Diagram
Frames (Example 6)Frames (Example 6)D h f d B di t DiDraw shear force and Bending moment Diagram
N.F.D S.F.D B.M.D
_
_
_
_ N.F.D
+_ + S.F.D
__B.M.D
+
N.F.DS.F.DB.M.D
_
_
+
_
Frames (Example 7)( p )Draw Normal force, shear force and Bending moment Diagramg
10kN/m
60kN
26.8
20.8
53.726.8
10.5
11026.56o
47.743.2
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
ProblemProblem 11Problem Problem 11D N l f h f d B di t Draw Normal force, shear force and Bending moment Diagram
ProblemProblem 22Problem Problem 22D N l f h f d B di t Draw Normal force, shear force and Bending moment Diagram