structure analysis i - الصفحات الشخصية...

Structure Analysis IStructure Analysis IChapter 4Chapter 4

Chapter1 Types of Structures & Loads 1

Chapter 4Chapter 4

I t l L diInternal Loading Developed in Structural p

Members

Internal loading at a specified PointInternal loading at a specified Point

In GeneralIn General

• The loading for coplanar structure will consist of a normal force N shear force Vconsist of a normal force N, shear force V, and bending moment M.

Th l di ll h l• These loading actually represent the resultants of the stress distribution acting over the member’s

i lcross-sectional are

Sign Convention+ve Sign

Procedure for analysisProcedure for analysis

• Support ReactionSupport Reaction• Free-Body Diagram

E i f E ilib i• Equation of Equilibrium

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

kNV

V F

C

Cy

15

05550

=

=−−−⇒=∑

mkNMMM

c

cC

.50020)3(5)2(5)1(5 0

−=

=−−−−−⇒=∑

kNV

V F

C

Dy

20

055550

=

=−−−−⇒=∑

mkNMMM

kNV

D

DC

C

.50020)3(5)2(5)1(5 0

20

−=

=−−−−−⇒=∑

Example 2D i h i l h d i i i 1 i hDetermine the internal shear and moment acting in section 1 in the beam as shown in figure

18kN

kNRR BA 9==

6kN

V Fy 0690 =−+−⇒=∑

MMkNV

y

0)2(9)1(6 03

sectionat =−+⇒=

=

∑

∑

mkNM D .12=

Example 3Example 3Determine the internal shear and moment acting in the cantilever beam shown in figure at sections passing through g p g gpoints C

V F Cy 0390 =−+−⇒=∑

ftkMMM

kV

c

480)6(9)2(3 0

6

c

=

=−+⇒=

=

∑ftkM D .48=

Shear and Moment functionShear and Moment functionProcedure for Analysis:1- Support reactionpp2- Shear & Moment Function

Specify separate coordinate x and associated origins, extending into regions of the beam between concentrated forces and/orinto 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 4Example 4Determine the internal shear and moment Function

w 2

151

302 ==x

ww

30x

21 0300 xVF =−+−⇒=∑

21

2

2

)(

033.030

015

300

xx

xV

V Fy

⎤⎡

−=

+⇒

∑

∑

3

21

011.030600

0600315

)(30 0

xxM

xxxMM S

−+−=

=+⎥⎦

⎤⎢⎣

⎡+−⇒=∑

120 x <<

1

1

4108

041080120

xV

xV Fx

y

=

=−+−⇒=

<<

∑

( )2

211S

1

21081588

041081588 0

41081

M

xxMM

xVx =+−+⇒=

−=

∑2

11 21081588 xxM −+−=

2012 2 <<

∑x

( )60

0481080

=

=−+−⇒=

∑

∑V

V Fy

( )130060

06481081588 0

2

22S

−=

=−+−+⇒=∑xM

xxMM

w 20

920=x

w

w

9x

21 0)20(10750 xxxV Fy =⎥⎦

⎤⎢⎣⎡−−+−⇒=∑

( )

2

2

11.110759

)(

xxxV

y

⎤⎡

−−=

⎥⎦⎢⎣

∑

∑

( )32

321

2

370.0575

09

)20(1075 0

xxxM

xxxxMM xxS

−+=

=⎥⎦⎤

⎢⎣⎡−−−⇒=∑

Shear and Moment diagram for a BeamShear and Moment diagram for a Beam

0)()(0 VVxxwVF =∆+∆+⇒=∑

( )O 0)()( 0)(

0)()(0

MMxxxwMxVMxxwV

VVxxwV Fy

=∆++∆∆−−∆−⇒=

∆=∆

=∆+−∆+⇒=

∑

∑

ε( )( )2

O

)(

)()(

xxwxVM ∆+∆=∆

∑ε

∫=∆⇒=

→∆

dxxwVxwdxdV

xfor

)( )(

0

∫

∫

∆⇒ dVMVdM

dx

)(∫=∆⇒= dxxVMVdx

)(

Example 1Example 1Draw shear force and Bending moment Diagram

S.F.D

B.M.D


S.F.D

B.M.D

Example 418 kN

Example 4Draw shear force and Bending moment Diagram

Max. moment at x = L/2/then

2LwLwLM ⎞⎜⎛⎞

⎜⎛

22222

maxwLM

M

=

⎠⎜⎝

−⎠

⎜⎝

=

8max

S.F.D

B.M.D

D h fExample 5

Draw shear force and Bending moment Diagramg

)7(14)53(147142

+−−=

==

∑ MMx

x

49

)7(14)5.3(14

=

+∑M

MMS

Example 6aExample 6aDraw shear force and Bending moment Diagram

S.F.D

B.M.D

Example 6bExample 6bDraw shear force and Bending moment Diagram

S.F.D

B.M.D

Example 6cExample 6cDraw shear force and Bending moment Diagram

S.F.DS.F.D

B M DB.M.D

Example 6dExample 6dDraw shear force and Bending moment Diagram

Group WorkGroup Work

Draw shear force and Bending moment Diagram

Example 1Draw shear force and Bending moment Diagram

Example 1

Example 2Example 2Draw shear force and Bendingand Bending moment Diagram

++

+ ++

Problem 1Problem 1


30.5 23.5

+

--

+ ++

Problem 2Problem 2


x32

46350

25

=→Vat

55.11)5)(46.3()(

46.35

32

32

2125

=

==

=⇒=

MRxM

mxx

A

Example 1Example 1


Hinge

Reaction Calculation

( )

kA

AM yleftB

4

060)5(20100 =−+−⇒=∑

Reaction Calculation

kC

CM

kA

y

yE

y

45

060)32(4)27(20)16(5)6(18)12( 0

4

=

=−−+++⇒=

=

∑

kE

E

EF

y

xx

6

045420518 0F

0 0

y =−−+++⇒=

=⇒=

∑∑

kEy 6=

Frames (Example 1)Frames (Example 1)Draw Bending moment Diagram

Support reaction & Free Body diagramSupport reaction & Free Body diagram

_ _

S.F.D B.M.D

++ S.F.D

- B M D- B.M.D

Frames (Example 2)Frames (Example 2)


N.F.D+

S.F.D+

B M DN FD S FD B M DN FD S FD

S.F.D

B.M.D

_

B.M.DN.F.D S.F.D B.M.DN.F.D S.F.D

+

+N.F.D

+

-

-

B.M.DN.F.D S.F.D

-

--

_N.F.D

+S.F.D

64

26

+ B.M.D+

251.6

N.F.D

B.M.D

S.F.D

168

S.F.D6413.22+

26

+

_

36_

432

31.78

B.M.D

168

432 139.3

251.6+

_

_

+


Draw shear force and Bending moment Diagramg g

S.F.D

B.M.D

+

+

S.F.D_

B.M.D+

Frames (Example 5)a es ( a p e 5)


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)Draw Normal force, shear force and Bending moment Diagram

10kN/m

60kN 20.8

53.726.8

10.5

26.56o

47.743.2

110

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

S.F.D

B M DB.M.D

N.F.D

S.F.D

B.M.D

B M DB.M.D

Moment diagram constructed by the h d f i imethod of superposition

Example 1Example 1

Example 2.a

Example 2.b

ProblemProblem 11Problem Problem 11Draw Normal force, shear force and Bending moment Diagram

ProblemProblem 22Problem Problem 22Draw Normal force, shear force and Bending moment Diagram

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