Chapter 4Chapter 4

Internal Loading Developed in Structural Members

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 are

Sign Convention+ve Sign

Procedure for analysis

• Support Reaction

• Free-Body Diagram

• Equation of Equilibrium

Example 1Determine the internal shear and moment acting in the

cantilever beam shown in figure at sections passing through

points C & D

mkNM

MM

kNV

V F

c

cC

C

Cy

.50

020)3(5)2(5)1(5 0

15

05550

mkNM

MM

kNV

V F

D

DC

C

Dy

.50

020)3(5)2(5)1(5 0

20

055550

Example 2

kNRR BA 9

mkNM

MM

kNV

V F

D

y

.12

0)2(9)1(6 0

3

0690

sectionat

6kN

Determine the internal shear and moment acting in section 1 in the

beam as shown in figure18kN

Example 3Determine the internal shear and moment acting in the

cantilever beam shown in figure at sections passing through

points C

ftkM

MM

kV

V F

D

c

Cy

.48

0)6(9)2(3 0

6

0390

c

Shear and Moment functionProcedure 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 4Determine the internal shear and moment Function

Example 5Determine the internal shear and moment Function

151

302

xw

2

12

2

2

12

3

0 30 015

30 0.033

0 30( ) 600 015 3

600 30 0.011

y

S

xF V

V x

x xM M x

M x x

w 2

30x

Example 6Determine the internal shear and moment Function

2

11

211S

1

1

1

21081588

041081588 0

4108

041080

120

1

xxM

xxMM

xV

xV F

x

x

y

130060

06481081588 0

60

0481080

2012

2

22S

2

xM

xxMM

V

V F

x

y

Example 7Determine the internal shear and moment Function

920

xw

12

2

12 2

2 3

0 75 10 (20) 09

75 10 1.11

0 75 10 (20) 09 3

75 5 0.370

y

xS

xF V x x

V x x

x xM M x x x

M x x x

w 20

9x

Shear and Moment diagram for a Beam

2O

)(

0)()( 0

)(

0)()(0

xxwxVM

MMxxxwMxVM

xxwV

VVxxwV Fy

dxxVMVdx

dM

dxxwVxwdx

dV

xfor

)(

)( )(

0

When F acts downward on the beam, ∆𝑉 is negative so that the

shear diagram shows a “jump” downward.

Likewise, if F acts upward, the jump is upward.

Internal Shear due to concentrated Load

If an external couple moment M’ is applied clockwise, ∆𝑀 is positive, so

that the moment diagram jumps downward,

and

when M’ acts counterclockwise, the jump must be upward.

Internal Moment due to concentrated moment

Example 8Draw shear force

and Bending

moment Diagram

B.M.D

S.F.D

Example 9Draw shear force

and Bending

moment Diagram

S.F.D

B.M.D

Example 10Draw shear force

and Bending moment

Diagram

18 kN

Max. moment at x = L/2

then

8

2222

2

max

2

wLM

LwLwLM

Example 11Draw shear force and Bending moment Diagram

S.F.D

B.M.D

Draw shear force

and Bending

moment Diagram

Example 12

49

)7(14)5.3(14

7

142

M

MM

x

x

S

Example 13aDraw shear force

and Bending

moment Diagram

S.F.D

B.M.D

Example 13bDraw shear force

and Bending

moment Diagram

S.F.D

B.M.D

Example 13cDraw shear force

and Bending

moment Diagram

S.F.D

B.M.D

Example 13dDraw shear force

and Bending

moment Diagram

Draw shear force and Bending moment Diagram

Example 14

32

Draw shear force and Bending moment Diagram

Example 15

V(kN)

Example 16Draw shear force and Bending moment Diagram

+

Example 17Draw shear force

and Bending

moment Diagram

99

Example 18Draw shear force and Bending moment Diagram

Sketch BMD

Sketch BMD

Sketch BMD

Example 19Draw shear force and Bending moment Diagram

++

+

+

Problem 1

Draw shear force and Bending moment Diagram

Example 20

Draw shear force and Bending moment Diagram

486

30.5 23.5

+

+

-

-

Example 21

Draw shear force and Bending moment Diagram

kE

E

EF

kC

CM

kA

AM

y

y

xx

y

yE

y

yleftB

6

045420518 0F

0 0

45

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

4

060)5(20100

y

Reaction Calculation

Frame Structures (Example 1)

Draw Bending moment Diagram

Support reaction & Free Body diagram

_ _

S.F.D B.M.D

+

-

S.F.D

B.M.D

-

B.M.D

-

15k.ft

15+

-

S.F.D

3

1

-

N.F.D

3-

Frames (Example 2)

Draw shear force and Bending moment Diagram

N.F.D

S.F.D

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

N.F.D

+

+

+

_

+

+

-

-

Frames (Example 3)

Draw shear force and Bending moment Diagram

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

-

--

_

+

+

N.F.D

S.F.D

B.M.D

251.6

64

26

B.M.D

N.F.D

S.F.D

168

S.F.D

B.M.D

168

432 139.3

251.6

432

36

64

26

13.22

31.78

+

_

_

+

_

_

+

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

_

+

_

N.F.DS.F.DB.M.D

Problem 1

Draw Normal force, shear force and Bending moment Diagram

Problem 2

Draw Normal force, shear force and Bending moment Diagram

Problem 3

Draw Normal force, shear force and Bending moment Diagram

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

10kN/m

60kN

26.56o

53.726.8

47.7

43.2

10.5

20.8

110

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 4

Draw Normal force, shear force and Bending moment Diagram

Problem 5

Draw Normal force, shear force and Bending moment Diagram

Problem 6

Draw Normal force, shear force and Bending moment Diagram