CH 4 (Structural Analysis)-V-M-N Diagrams

1

CHAPTER 4 Internal Loads developed in

Structural Members

0xF

0yF

0M

1

•Statics (CH 5-6-7)

•To get support reactions &

Internal forces

•Strength of Materials (CH 5)

drawing V, N, M diagrams

•Structural Analysis (CH 2)

To get support reactions2

STUDY:

2

3

Sign conventions for internal loadings

N: (+) in tension [or (-) in compression]

V: (+) rotating the segment in clockwise

M: (+) bending the segment concave upward

(or creating compression on the top face) +

+ +

+

+

FRAMES : Sign convention &

reference linesBased on sing convection, identify (+) & (-) moments shown in the figure below??

(+or -)

4

Moment diagram for a frame

Rigid connection of

a steel frame

We need to identify which areas of

the moment distributions will be

(+) or (-) so as to make them

consistent and universal. That is

why we consider reference lines.

Reference

lines

Reference lines

the (-) V,M,N must be

plotted on the face of the

member where the

reference line is located.

3

5

Reference line

(placed on the bottom face)

++

+

+

Reference line

(placed on the

right face)

Reference lines for V, M & N

the (negative) V,M,N diagrams must be plotted on the face of the

member where the reference lines are located.

45 k

2k/ft20 k

10 k5 k 10 k

Beam

Frame

Reference lines

Reference lines for V & M

6

the (negative) V,M,N diagrams must be plotted on the face of the


4

7

A

Reference lines

Reference lines for (negative) V, M & N

the (-) V,M,N diagrams must be plotted on the face of the


Reference

lines

+

++

+

+

- -

(M)

8

Ay

Dy

Dx

Reference lines

Reference lines for V, M & N

the (negative) V,M,N diagrams must be plotted on the face

of the member where the reference lines are located.

5

9

Shear & moment at a point• Shear force and bending couple at a point

are determined by passing a section

through the beam and applying an

equilibrium analysis on the beam portions

on either side of the section.

0yF 0M0xF

0yF 0M0xF

FBD-left

FBD-right

(+) positive

directions

(+) positive

Directions

Equations of equilibrium (E-of-E)

9

Example 4.1 (V & M at a point)

20 kN.m

20 kN.m

20 kN.m

20 kN.m

FBD-|CB|

Determine the internal forces (V and M)

at points C & D?

Note:

Point D just to the left of 5kN

Point C just to the right of 5kN

E-of-E for Segment |CB|

E-of-E for Segment |DB| FBD-|DB|

10

Study this !

6

Example 4.3 (V & M at a point)

Determine the V & M at point C?

Equations of equilibrium (E-of-E)

(+)

(+)

FBD-entire

FBD-|AC|

11

CA

CA B

Study this !

Relations between Distributed Load (w) & Shear

Force (V)

DERIVATIVE of V (-) x distributed

load (slope)

INTEGRATION of load V

[ (-) x area under distributed loading ]

xwV

xwVVV

Fy

0

0

wdx

dV

'

'

C

C

CC dxwVV

12

or

Load acting vertically

7

Relations between Shear (V) & Bending (M)

2

21

02

0

xwxVM

xxwxVMMM

MC

Vdx

dMDERIVATIVE (slope) of M V

INTEGRATION of V [area

under shear] M

'

'

C

C

CC dxVMM

13

or

Load acting vertically

- -

Page 151 (correction !!!!!) for the loads acting vertically

Summary

Relations between Load&Shear & Shear&Bending

14

8

L

dxwV0

Relationship between loading, shear & moment

wdx

dVV

dx

dM

derivative derivative

Integration

under w (distributed load)Integration

under V diagram

linear 1st

Load=0

Shear=0 Linear

15

Load=0

1st

linear

Shear=0

L

dxVM0

Distributed load

Relationship between loading, shear & moment

D

C

x

xCD dxwVV

wdx

dV

16

+1st -2nd 3rd

-1st

+2nd

3rd

-1st

-2nd= constant

Cubic equation

Cubic equation parabola

Distributed load wdx

dVV

dx

dM

parabola

parabola

L

dxwV0

Integration

under w (distributed load)x(-1)=ΔVIntegration

under V diagram=ΔM

L

dxVM0

9

Load condition: LOAD SHEAR MOMENT

No distributed

load between

|BC|

Linear function.

(y=a)

1st degree eq.

(y=ax+b)

[linear (increasing)

equation]

Uniformly

distributed

(y= w)

-1st degree eq.

(y= -wx+b)

[linear (degreasing)

equation]

2nd degree eq.

(y= -wx2+bx)

[- Parabola, or

quadratic equation]

+1st degree

(y=ax+b)

[linear equation]

+2nd degree eq.

(y=ax2+bx+c)

[Parabola, or quadratic

equation]

3rd degree eq.

(y=ax3+bx2+cx)

[cubic equation]

(integration)(-1)x(integration)

Relations between Load&Shear & Shear&Bending

17

SUMMARY: General shapes for M curves

based on V curves

(M)

(V)

18

++

- -

+

decreasing

(-) slope

increasing

(+) slope constant

(0) slope (+) parabola (-) parabola (-) parabola (+) parabola

(+) area

(0) slope

(0) area

(0)slope

(-) area

(0) slope (+) area

(+) slope

(+) area

(-) slope

(-) area

(-) slope

(-) area

(+) slope

+2nd

-2nd

+2nd+

+1st

-1st

+1st

+1st: linear increasing equation

-1st: linear decreasing equation

+2nd: positive parabola

-2nd: negative parabola

V=0

Mpeak

-2nd

10

SUMMARY: General shapes for M curves

based on V curves

(M)

(V)

++

- -

--

increasing

(+) slope

decreasing

(-) slope constant

(0) slope

decreasing

(-) parabola

(+) parabola (+) parabola

(-) parabola

(-) area

(0) slope

(0) area

(0)slope

(+) area

(0) slope (-) area

(-) slope

(-) area

(+) slope

(+) area

(+) slope

(+) area

(-) slope

-2nd

+2nd+2nd

-2nd

-1st +1st

V=0

Mpeak

+1st-1st

+1st: linear increasing equation

-1st: linear decreasing equation

+2nd: positive parabola

-2nd: negative parabola 19

20

(-)

(+)

(+)

Shear-force and bending-

moment diagrams for a simple

beam with several concentrated

loads

20

IMPORTANT NOTE: When you plot the

diagrams, you MUST indicate the

variations along the curves.

11

21

Mmax

Vmax

(-)

(-)

Shear-force and bending-

moment diagrams for a

cantilever beam with several

concentrated loads

IMPORTANT NOTE: When you plot the

diagrams, you MUST indicate the

variations along the curves.

21

4kN.m

A CB

1m 1m

4kN.m

A CB

1m 1m

2kN2kN 2kN 2kN

-

+

Jump in moment in (+) direction Jump in moment in (-) direction

22

+ +

--

2kN

2kN

2kN.m

2kN.m

2kN.m

2kN.m

Clockwise Counterclockwise

Concentrated moment at a point & M-diagram

12

23

Draw the shear and bending moment diagrams

for the beam and loading shown.

x

y

-y

x

y = ax2+bx+c

y= - ax2+bx+c (+)

(-)

(+)

Exercise # 3

23

x

y

-y

x

y = ax2+bx+c

y= - ax2+bx+c

Example 4.7 (V & M diagrams)

Draw the V & M diagrams.

309

10 0 2xVFy

xxMM 3027

10 0 3

9m

Using E-of-Es:

24

13

wdx

dV

Vdx

dM

(+)

(+)

(-)

L

dxVM0

L

dxwV0

309

10 2xV

xxM 3027

10 3

xw9

20

11.55

25

(+)

(-)

309

10 2xV

How to find the location (point of) of V=0

Equation for V:

when V=0 find xx

5.20m~ x

0309

10 2xV

V=0

26

14

wdx

dV

(+)

(-)

w=11.55

309

10 2xV

Equation for V:

How to find the magnitude of w at a point

xw )9

10(2

5.20m xfor

mkNw /55.11

x=5.20m

27

wdx

dV

(+)

(-)

309

10 2xV

Equation for V:

How to find the magnitude of w at a point

xw )9

10(2

m9 xfor

mkNw /20

x=9m

28

15

How to find the Mpeak when V=0

29

(+)


30

Draw the shear and bending moment diagrams

for the beams and loadings shown.

(-) (+)

(+)

(-) (-)

(+)

16

31

Draw the shear and bending moment

diagrams for the beam and loading

shown.

(-)

(+)


31

(+)

(-)

(+)

(-)

(+)

(-)

(-)

(-)

(+)

Draw the V & M diagrams for the beam and

loading shown (assume supports A & C are

rollers, and E is a pin, B is a pin joint).


Draw the FBDs of each member, and then find the reaction and joint forces

32

E: pin support

B: pin joint

17

33

2k/ft

10ft 6ft 4ft 6ft 6ft

60k.ft

EC

A B

Reactions :

Ax=0k

Ay=4k

Cy=45k

Ey=-6k

Joint forces:

Bx=0k

By=16kN

34

(-)

(+)

(+)

(-)

3k/ft2k/ft 5k

10ft 6ft 4ft 6ft 6ft

60k.ft

E

C

A B

Reactions :

Ax=0k

Ay=4k

Cy=45k

Ey=-6k

Joint forces:

Bx=0k

By=16kN

18

Draw the V & M diagrams for the beam.Example 4.10 (V & M diagrams)

FBD of |ABC|

35

Idealized beam

(+)

(-)

(-)

(+)

Exercise # 5 Draw the V and M diagrams for the beam.

36

Areas of shear diagram(+700)

(-200)

(-500)

700

500

(V)

(M)

19

Draw the V and M diagrams for the beam.

37

(V)

(M)

(+428.6

)(+107.2)

21.28

(-425.6)

Areas of shear diagram

428.6535.8

425.6

365.6

(-170.2)

Exercise # 6

38

Exercise # 7

(V)

(M)


38

(+1702) (+800)

(-2266)

1702

(-234)

1468

43.3

73.3

153.3

40Areas of shear diagram

20

2m 3m 4m 3m 2m

6kN 5kN 5kN 4kN2kN/m

A B

39

5.8

6

10

0.8

4.2

4 4+ +

16

1.4

4.6

4.6

(V)

(M)

Exercise # 8Draw the V and M diagrams for the beam.

+0

0

Support Reactions:

Ay=15.8 kN

By=8.2 kN

Not: Arrows shown are

the real directions of the

forces and moments.

100k

3k/ft

AB C

D

200ft

60k 50k 90k 80k

100ft 120ft 60ft 75ft 75ft 50ft

60ft 30ft30ft

Draw the V and M diagrams for the beam.Exercise # 9

40ft

Support reactions:

Ay=255k

By=435k

Cy=241.2k

Dy=48.8k


the real directions of the

forces and moments.

21

100k

3k/ft

AB C

D

200ft

60k 50k 90k 90k

100ft 120ft40ft 75ft 75ft 50ft

60ft 30ft30ft

345

90

48.8

(10,840)

(9,090)

(-2,440)

10,840

9000

2700 2100

4200

9000

902430

0 00 0

(-19,840)

(-600)

(4,800)

12070

10

(2,340)

Exercise#9 (cont.)

255 435 48.48241.2

0

85tf 115tf

(V)

(M)

121.231.2

255

90 70

+

+++

+

-

-

- -

(-6,300)

(11,700)

41

-

60ft

Support reactions:

Ay=255k

By=435k

Cy=241.2k

Dy=48.8k

Exercise# 10

42

10kN

8kN6kN 3kN/m

2kN/m

5m 3m 2m 5m 5m 2m 4m 4m

Work on this

problem !


A

B C

D

E F

Support

reactions:

Ay=3.38kN

By=33.944kN

Cy=22.676kN

Dy=4.0kN

Not: Arrows shown are the

real directions of the forces

and moments.

Joint forces:

Ey=8.62kN

Fy=4.0kN

22

Exercise #1110 kN2 kN 2kN/m

1.5kN/m

A B C DG1 G2

8m8m 8m2m 2m 2m

44

Draw the V & M diagrams for the gerber beam.

Work on this

problem !

Support reactions:

Ay=2.5 kN

By=21.25 kN

Cy=16.56 kN

Dy=6.69 kN



and moments.

46

3m

3m

3m

3m

3m

3m

3kN

3kN

3kN

3kN

3kN

3kN

50kN

130kN

A

Ay=180kN

Ax=0kN

M=0kN.m

M [kN.m]V [kN] N [kN]

50

180

(-)

180

3

6

9

9

6

3130kN.m

81

72

54

27

9

(+)

(+)

Exercise # 12Draw the V, M &N diagrams for the column.

23

Internal force diagrams for sloping members

47

They can be more challenging

48

V, M, N of sloping (inclined) members

Example 4.14

24

Exercise # 13 Draw the V, M & N diagrams for the sloping

beam.

Ay

Ax

4m

3m

SOLUTION:

• Determine the support reaction forces:

A

B

By

48kips48kips

FDB :1.5m

kN

x

kN

yy

kN

y

A

BA

B

0)0F(or 0 M

24A 0480F

24 0)(1.548)(3B0 M

xB

kN

yy

mkNm

yA

A

B

49

1.5m

)53(24

)54(24

)53(48)54(48

4

3

5

54sin

53cos

24

24

48kN

4m

3m

28.8

38.4

14.419.2

14.4kN19.2

24

24

Exercise #13 (cont.)

50

Adjust the support reactions and

external loads based on the axes shown

(i.e; y axis perpendicular to the beam)

25

Exercise # 14Draw the V, M & N diagrams for the sloping

beam.

Ay

Ax

3m

4m

SOLUTION:


A

B

Bx48kN

48kN

FDB:2m 2m

kN

x

m

x

kN

y

kN

x

AA

A

B

32 0)(248)3()(4A0 M

48A 0480F

32 0)(248)(3B0 M

mkNm

yB

kN

yy

mkNm

xA

51

kN6.25)(cos32

48

32

32

48

32

32

kN2.19)(sin32

kN4.38)(cos48kN8.28)(sin48

kN6.25)(cos32

kN2.19)(sin32

Exercise 14 (cont.) Adjusting the support reactions based on the

new x-y axes along the beam :

3

4

5

53sin

54cos

52

26

48

32

32

Exercise 14 (cont.)

25.6kN

19.2kN

48kN48kN

kN4.38)(cos48

kN8.28)(sin48

19.2kN54.4kN

=

Adjusted FBD (based on the new axis):

3

4

5

53sin

54cos

53

Exercise 14 (cont.)

25.6kN

19.2kN

38.4kN 28.8kN

19.2kN

54.4kN

54

27

55

Exercise #15Compare the drawings from Exercise #14 &

Exercise# 14?

A

B48kN

# 14

# 13

# 14

4m

3m

A

B48kips

1.5m 1.5m

# 13

CONCLUSION: Types

of support significantly

influence the V, M, N

diagrams

Exercise # 16The roof beam of a building is subjected to a

snow loading as shown. Draw the V, M & N

diagrams for the beam.

Ay

Ax

12kN/m

3m

4m

SOLUTION:


12kN/m

A

B

Bx

48kN

Snow load:

kN

x

m

x

kN

y

kN

x

AA

A

B

32 0)(248)3()(4A0 M

48A 0480F

32 0)(248)(3B0 M

mkNm

yB

kN

yy

mkNm

xA

NOTE: this problem is very similar to Example 13, except there is a distributed load, which has

the same magnitude, acting vertically on the beam.

57

FBD:

2m 2m

28

kN6.25)(cos32

48

32

32

48

32

32

kN2.19)(sin32

kN4.38)(cos48kN8.28)(sin48

kN6.25)(cos32

kN2.19)(sin32

Exercise #16 (cont.)Adjust the support reactions based on the new x-y

axes that is selected based on beam’s slope

3

4

5

53sin

54cos

58

19.2

54.4

12kN/m

48kN

19.2

48kN

kN4.38)54(48

kN8.28)53(48

25.6

3

4

5

53sin

54cos


59

29

12kN/m

48kN

48kN

38.4kN

28.8kN

38.4kN / 5m = 7.68kN/m

28.8kN / 5m = 5.76kN/m

Equivalent distributed load due to the component of the load

acting perpendicular to the beam’s centroidal axis:

Equivalent distributed load due to the component of the load

acting along the beam’s centroidal axis:

3

4

5

53sin

54cos

38.4kN

28.8kN


60

19.254.4

19.2

25.6

48

32

12kN/m

32

=


61

30

48

32

48kN / 5m = 9.6kN/m

48kN

32

62

25.6kN

19.2kN

19.2kN54.4kN

=

SUMMARY: Representing the loads &

reactions in various forms of axes:


Equivalent distributed loads

perpendicular & along the

centroidal axis of the beam

Equivalent distributed

load in vertical direction:

48

32

12kN/m48kN

32

4m

=3m

)(cos6.9 /mkN

)(sin6.9 /mkN


25.6kN

19.2kN

19.2kN

54.4kN

(V)

(M)

(N)

63

31


64

25.6kN

19.2kN19.2kN

54.4kN

(V)

(M)

(N)

NOTE: you can also look at the same beam & diagrams as rotated

below, which is more familiar to you in a way you are used to see..

NOTE: when

V=0 M=MMAX

A

B48kN

Exercise # 17 Compare the drawings from Exercise #14 &

Exercise# 16?

# 14

65

# 16

# 14

CONCLUSION: Types

of loading significantly

influence the V, M, N

diagrams

B12kN/m

48kN

# 16

32

66

Draw the V&M diagrams for the beam.

Assume support A (rolling) & B (pinned).Example 4.13 (V & M diagrams)

SOLUTION: The frame is statically determinate Find the support

reactions Bx, By & Ay.

0xF 0yF 0BM

0y , orxF 0AM 0BM

FBD of entire frame

Note:

(E-of-E: Equations of Equilibrium)

Use one of these combinations of

E-of-E to find the support reactions:

66

Example 4.13 (cont.) (ii) Draw the FBDs of members |AC| & |CB|

and calculate the internal forces.

FBD of Joint C

FBD of |CB|

FBD of |AC|

CHECK POINT: You may check joint C to verify if

those calculated internal forces from |AC| & |CB|

are correct. Use the E-of-E to check if joint C is in

equilibrium:

0xF 0yF 0CM

V, M, N are the same

V, M, N are the same

67

C

C

33

68

Example 4.13 (cont.) (iii) Draw the corresponding V & M diagrams based on

the internal forces shown on the FBDs of |AC|& CB|. FBD of Joint C

FBD of |CB|

FBD of |AC|

(-)

(-)

(M)

(M)

68

y

xy

x

(-)

(-)

(-)

(-)

NOYES

Example 4.13 (cont.) (iv) show the V & M diagrams on the entire

frame separately as shown.

NOTE: only moment diagram is shown in this example. Think

about how to draw the V and N diagrams for the given FBD’s.

(M) [-k.ft]

A

B

A

B

NOTE: make sure the (-) values of V , M & N are

plotted on the side where the doted lines are shown.

Summary of the M diagram:

69

34

Exercise # 18Draw the V, M, N diagrams for the frame.

70

NOTE: The support reactions are shown

with their real directions on the frame.

A

B

Ay=30

Ax=60

By=45

C

Support

reactions:

Ax=30kips

Ay=60kips

By=45kips

Exercise # 18 (cont.)

(V) (M) 71

A

B

Ay=30

Ax=60

By=45

C

A

CB

(M)

(V)

C

Member |CB|

Member |AC|

35


Reactions:

Ax=2 kN

Ay=6 kN

MA=8 kN.m

72

2kN

2kN

1kN/m 2m

2m 2m

A B

CD

Work on this problem !

FBDs of each member

Exercise # 20

FBD of the entire frame

Draw the V, M&N diagrams for the frame.

74

36

Exercise # 21


(V)

(M)

Draw the V, M &N diagrams for the frame.

76

100

(V)

(M)

(-)

(-) (-)

(-)

(-)

(+)

(+) (+)

(-)

Exercise # 22

FBDs of the members


Draw the V and M diagrams for the frame.

77

(+)

37

(1)

(2)(3)

(5)

(4)

78


79


38

80


+

81


39

82


83

Draw the shear and bending moment diagrams for

the frame shown. Assume supports A (pinned) & C

(rolling). Joint B is fixed.


FBD- entire framePrototype=one-to-one

SOLUTION: The frame is statically determinate (there are three unknowns). (i) using

the E-of-Es for the FBD of the entire framing system, first, find the support reactions

Ax, Ay & Cy .

0xF 0yF 0AM

0y , orxF 0AM 0CM

Note:


Use one of these combinations of E-of-E

to find the support reactions:

B C

83

40

Example 4.14 (cont.)

FBD of entire system

170 kN.mFBD of Joint B

FBD of |BC|

FBD of |AB|

V, M, N are the same here

CHECK POINT: You may check joint C to

verify if those calculated internal forces are

correct. Use the E-of-E to check if joint C is in

equilibrium:

0xF 0yF 0BM

(ii) Draw the FBDs of members |AB| & |CB| and

calculate the internal forces.

84

85


FBD of |AB|

(M) [kN.m]

(V) [kN]

Member |AB|

85

41


(-)

(+)

FBD of |BC|:

(M) [kN.m]

(V) [kN]

86

Member |BC|

(-)

(-)

(+)

(V)(M)

(+)


Summary of the V & M diagrams:

NOTE: make sure the (-) values of V & M are located

on the side where the doted lines are shown.

NOTE: only the V & M are drawn here. Think about how to

draw the N diagram for the given frame.87

(+)

42

Draw the V and M diagrams for the frame shown.

Assume supports A (pinned) & C (rolling). Joint B

(rigid).


88

ftk /142.0

Correct this in

your textbook

89

SOLUTION: The frame is statically determinate (there are three unknowns).

(i) Using the FBD of the entire framing system and E-of-Es, first, find the

support reactionsAx, Ay & Cx.

0yF 0AM

0y , orxF 0AM 0CM

Note:


Use one of these combinations of E-of-E

to find the support reactions: 89


ftk /142.0

B

C

A

kftftk 0.2)14.14(142.0 /


43

90

FBD of entire frame

Example 4.15 (cont.) FBD of |BC|

FBD of |BC|

(FBD of joint B)

B

V, M, N’s the same

CHECK POINT:

(ii) Draw the FBDs of

members |AB| & |BC|, and

calculate the internal forces.

90

Summary: the V , M & N diagrams:

NOTE: make sure the (-) values of V and M are

located on the side where the dashed lines are shown.

NOTE: only the V & M are drawn here. Draw the N diagram

for the given frame.

5

0.6250.354

1.06

0.5

(-)

(+)

(+)

(-)


91

(V) (M) (N)

44

Exercise # 23Draw the V, M &N diagrams for the frame.

Reactions:

Ax=36kips

Ay=64.04kips

By=52.96kips

Ay=64.04

Ax=36

By=52.96

92

Ay=64.04

Ax=36

By=52.96

Exercise # 23 (cont.) FBDs of each member:

45

FBDs of each member

94

(-)

(-)

(-)

(+)

(+)

(+)

(-)

(+)

(-)


(V)

(M)

NOTE: Draw the N diagram by yourself

Exercise #24Draw the V, M & N diagrams for the frame

shown (support A is fixed).

4m

2m

3m

3kN

2kN

4m

3kN

3kN

3kN

1kN

4kN.m

A

95

NOTE: the solution to this problem is very detailed in a way that the

internal forces and moments are determined for each section cuts shown.

The section cuts are determined based on the variations in the geometry

and the presence of loadings.

(1)

4m

2m

3m

3kN

2kN 3kN

3kN

3kN

1kN

4kN.m

A

(2)

(5) (4)

(3)

(6)

(8)

(7)

4m

46

Exercise# 24 (cont.)

kN

xxx AAF 303330

0)9(3)4(3)5(3)4(14)3(30 . mmmmmkNm

AA MM

kN

yyy AAF 001320

4m

2m

3m

3kN

2kN 3kN

3kN

3kN

1kN4kN.m

AAx

Ay

MA (1)

4m

2m

3m

3kN

2kN

4m

3kN

3kN

3kN

1kN

4kN.m

A

(2)

(5) (4)

(3)

(6)

(8) (7)

3kN

0kNMA=9kN.m

real

directions

(reversed)

mkN

AM .9

96

initially predicted

reactions

Determine the support reaction forces:

a b

NOT TO SCALE

3(-)

3

1(+)

3

3

3

(+)

(+)

(+)

(+)

(-)

N [kN]M [kN.m]V [kN]

1

3

33

3

1

(+)

(+)

(+)

(-)9

4

12

12

97


47


99

kN

kN

mkN

NN

VV

MM

303

1 01

404

:5Section

55

55

.

55

kN

kN

mkNm

NN

VV

MM

1 01

1 01

00)4(14

:6Section

66

66

.

66

kN

kN

mkN

NN

VV

M

101

3 03

0

:7Section

77

77

.

7

2kN

8kN5kN

3m

4m

3m 3m 3m 2m

A

B

G

Exercise # 25Draw the V, M & N diagrams for the frame

shown. Supports: A (fixed) & D (sliding).

104

NOTE: the solution to this problem is very detailed in a way that the

internal forces and moments are determined for each section cuts shown.

The section cuts are determined based on the variations in the geometry

and the presence of loadings.

48

2kN

8kN5kN

3m

4m

3m 3m 3m 2m

(1) (2)

(3)

(5)

(6) (7) (8)(9)

(11)

(12)

(4)

(10)

A G

B


105

Mark the critical locations (sections) where the internal

forces may be varying. We will determine the V, M & N

values for each of these sections:

2kN

8kN5kN

3m

4m

3m 3m 3m 2m

(1) (2)

(3)

(5)

(6) (7) (8)(9)

(11)

(12)

(4)

(10)Gx

BGy

Gx

MA

Ax

Ay

By

A

G


106

Find support reactions:

49

2kN

8kN5kN

3m

4m

3m 3m 2m

(3)

(5)

(6) (7) (8)(9)

(11)

(12)

(4)

(10)Gx

B

Gy

By

G

kN

xx GF 20

kN

yG BM 875.60

kN

yy GF 125.60

FIND SUPPORT REACTIONS:

at G &B


107

3m

(1) (2)

Gy

Gx

MA

Ax

Ay

A

kN

xxx GAF 20

mkNmkN

yAA mGMM .375.18)3(125.6)3(0

kN

yyy GAF 125.60

FIND SUPPORT REACTIONS at A:


108

50

109

2kN

8kN5kN

3m

4m

3m 3m 3m 2m

(1) (2)

(3)

(5)

(6) (7) (8)(9)

(11)

(12)

(4)

(10)

G

B

18.375kN.m

2

6.125kN

6.875 kN


Show the FBD of the entire system with known values

of the support reactions (make sure that your support

reactions show real directions):

kN

kN

kN

N

V

M

2

125.6

375.18

:1Section

1

1

1

kN

kN

mkN

N

V

M

2

125.6

0

:2Section

2

2

.

2

kN

kN

mkN

N

V

M

125.6

2

0

:3Section

3

3

.

3

kN

kN

mkNm

N

V

M

125.6

2

8)4(2

:4Section

4

4

.

4

kN

kN

mkNm

N

V

M

i

2

125.6

8)4(2

)(:5Section

5

5

.

5

Real

directions

Real

directions(+) sing

directions


4m

110

Calculate internal forces for each section using E-of-E:

51

kN

kN

mkN

NN

VV

MM

2 02

125.6 0125.6

375.100)4(2)3(125.6

:6Section

66

66

.

66

kN

kN

mkNmm

NN

VV

MM

2 02

125.1 05125.6

375.100)4(2)3(125.6

:7Section

27

77

.

77

kN

kN

mkN

NN

VV

MM

i

2 02

125.1 05125.6

75.130)3(5)4(2)6(125.6

)(:8Section

88

88

.

88

kN

kN

mkNm

NN

VV

MM

ii

2 02

125.1 08875.6

75.130)2(875.6

)(:8Section

88

88

.

88


111

kN

kN

mkN

N

V

M

146.6)(sin875.6

073.3)(cos875.6

75.13)2(875.6

:9Section

9

9

.

9

kN

kN

mkN

N

V

M

146.6)(sin875.6

073.3)(cos875.6

0)0(875.6

:10Section

10

10

.

10

kN

kN

mkN

N

V

M

875.6

0

0)0(875.6

:11Section

11

11

.

11

kN

kN

mkN

N

V

M

875.6

0

0)0(875.6

:12Section

12

12

.

12

Exercise# 25(cont.)

112

52

kN

kN

mkNm

NN

VV

MM

ii

2 02

125.6 0125.6

375.100)3(125.68

)(:6Section

66

66

.

66

2kN

8kN5kN

(3)

(5)

(6) (7) (8)(9)

(11)

(12)

(4)

(10)

B

Gy

By

G

kN

kN

mkNm

NN

VV

MM

ii

125.6 0125.6

2 02

80)4(2

)(:4Section

44

44

.

44

kN

kN

mkN

NN

VV

MM

ii

2 02

125.60125.6

8 08

)(:5Section

55

55

.

55

113

Exercise# 25(cont.)

2kN

8kN5kN

(3)

(5)

(6) (7) (8)(9)

(11)

(12)

(4)

(10)

B

Gy

By

G

kN

kN

mkN

NN

VV

MM

ii

2 02

125.1 05125.6

375.10 0375.10

)(:7Section

27

77

.

77

kN

kN

mkNm

NN

VV

MM

iii

2 02

125.1 0125.1

75.130)3(125.1375.10

)(:8Section

88

88

.

88

114

Exercise# 25(cont.)

NOTE: You may use different FBDs for the same

section cut to determine V, M,N at that section. For

instance, for Section 8, THREE different FBDs are

used. All are presented in this example as Section

8(i), (ii) & (iii)

53

V [kN]2

(-)

(+)

(-)

3.073

6.1251.125

6.125

(+)

18.375

8

810.375 13.75

(-)

(+)

(+)

(-)

M [kN.m]

NOT TO SCALE

Exercise# 25(cont.)

2

2

6.125

6.1252

6.146

6.875

N [kN](-)

(-)

(-)

(-)

(-)

6.146

2

115


6kN 1 kN/m

4kN

2m

8m2m

2m

A

E

B

DF

116

Reactions:

Ax=4 kN

Ay=10.5 kN

Ey=3.5 kN




and moments.

54


118

1m 2m 2m 1m 1m

5kN 3kN 1kN

4kN

2m

2m

A

E

D G C

H

F

B

2 kN/m

Reactions:

Ax=4 kN

Ay=3 kN

By=6 kN

MA=1kN.m




and moments.


Reactions:

Ay=16 kN

By=7.67 kN

Cx=3 kN

Cy=1.33 kN.m

120

3m6m10m2m

4m

2kN/m5kN

3kN

A B

C




and moments.

55

1kN/m

4m4m

4mI

4I 4I

I

A B

C DG

Exercise # 29Draw the V, M &N diagrams for the three-

hinge frame.

122

Reactions:

Ax=2 kN

Ay=4 kN

Bx=2 kN

By=4 kN

NOTE: I is the moment of inertia. Moment of inertias of members are used to

calculate the deflections (displacements) to be covered in the future chapters




and moments.


124

2m

4m

8m2m8m

6kN4kN1kN/m

I I 4I I

4I 2I

2I

AG D

EF

B C

5kN

9kN

4kN

H

Reactions:

Ay=4 kN

Bx=0 kN

By=9 kN

Cy=5 kN

NOTE: I is the moment of inertia. Moment of inertias

of members are used to calculate the deflections

(displacements) to be covered in the future chapters




and moments.

56

2m

4m

8m2m8m

6kN4kN1kN/m

AG

DE

F

B C

5kN

9kN

4kN

H

(1)

(2)

(3)

(4) (5)

(6)

(7)

(8)

Hint for moments:

10kN.m

18kN.m

8kN.m

D

(1)

(2) (3)8kN.m

10kN.m

2kN.m

E(4) (5)

(6) H

10kN.m

10kN.m(7)

(8)


125


Reactions:

Ax=3 kN

Ay=10 kN

Ey=2 kN

127

4m

A

BC

D

6m 3m 3m

2kN/m

3kN

4I

2I 2I

IE

NOTE: I is the moment of inertia. Moment of inertias of members are used to

calculate the deflections (displacements) to be covered in the future chapters



the real directions of

the forces and

moments.

57

128

4m

A

BC

D

6m 3m 3m

2kN/m

3kN

E

Hint for moments:

(2) (3) (4)(5)

(6)

(7)

(1)(8)

12kN.m

B

12kN.m

(3)

(2)

12kN.m

12kN.m

C 3kN

(4)

(5)

6kN.m

6kN.m

D

(7)

(6)

Exercise #31(cont.)

3

10 2


130

Reactions:

Ax=60 kN

Ay=4.7 kN

Dy=74.7 kN


the real directions of

the forces and

moments.

58

131

59

133

134

60

135

61

137

138

62

139

14

0

Exercise #32Draw the M diagrams resulted from

each of the unit forces & moments

(X1=…=X7=1) shown, individually.

X1

X2

X3

X4

X5

X6

X7

L1 L2 L3h

63

+

+

L1 L2 L3

h

1

1/L1 1/L1

-

1

hh

1

X1=1

X2=1

X2

X1

X3

X4

X5

X6

X7

- -

++

(L1+L2)/L1

1

-

hh

1 1

X4=1

X3=1

---

L1 L2 L3

h

X2

X1

X3

X4

X5

X6

X7

(L1+L2)/L1-1

64

- -

++

1

hh

L3

X6=1

X5=1

---

L1 L2 L3

h

X2

X1

X3

X4

X5

X6

X7

(L1+L2+L3)/L1(L1+L2+L3)/L1-1

L2 +L3

hh

+

+

++

X7=1

L1 L2 L3

h

X2

X1

X3

X4

X5

X6

X7

(L1+L2+L3)/L1

1/L1

11

1/L1

1

65

145


145

30k

30k

60k

20ft

20ft 10ft

10ft

10ft

10ft

Hint:

B C

D

A

M=-300k-ft

V= 10k

N=-10k


Homework - CH 4*

• 4.49

• 4.54

• 4.58

• 4.59

• 4.62

• 4.64

• 4.65• + Study all of the problems in the lecture slides

and example problems from your test book

*: suggested147

CH 4 (Structural Analysis)-V-M-N Diagrams

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Transcript of CH 4 (Structural Analysis)-V-M-N Diagrams