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