Analysis of Beams and Frames Theory of Structure - I

Analysis of Beams and Frames

Theory of Structure - I

Department of Civil EngineeringUniversity of Engineering and Technology, Taxila, Pakistan

2

Lecture Outlines

Shear and Moment Diagrams for Beams Shear and Moment Diagrams for a Frames Moment Diagrams Constructed by the

Method of Superposition Deflected Curves


3

. .B C

A D

F1 F3F2

w = w(x)

M1

.M2

.

w

x x

w(x)x

x

x

w(x)

O . M + M

V + V

Fy = 0:+

0)()( VVxxwV

+ MO= 0:

0)()()( MMxxxwMxV

2)()( xxwxVM

xxwV )(

M

V

Shear and Moment Diagrams for a Beam


4

----------(4-2)

----------(4-1)

2)()( xxwxVM

xxwV )(

Dividing by x and taking the limit as x 0, these equation become

Vdx

dM

Slope of Moment Diagram = Shear

)(xwdx

dV

Slope of Shear Diagram = -Intensity of Distributed Load


5

----------(4-4)

----------(4-3)

Equations (4-1) and (4-2) can be “integrated” from one point to another between concentrated forces or couples, in which case

dxxwV )(

Change in Shear = -Area under Distributed Loading Diagram

and

dxxVM )(

Change in Moment = Area under Shear Diagram


6

x

M

V

M + M

V + V

.M´

O

F

x

M

V

M + M

V + V

+ MO= 0: 'MM

Fy = 0:+ FV

Thus, 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 (V) is upward.

In this case, if an external couple moment M´ is applied clockwise, M is positive, so that themoment diagram jumps upward, and when M acts counterclockwise, the jump (M) must bedownward.


7

ML MRM´0

P

VLVR

ML MR

VLVR

ML MR

w0

VL VR

Slope = VL

Slope = VR0

0

ML

MR

ML MR

0

0

VL

VR

MR-wo Slope = VL

Slope = VR

ML


8

VLVR

ML MR

w1w2

VLVR

ML MR

w1 w2

VR

VL

ML MR

Slope = -w1

Slope = -w2

Slope = VR

Slope = VL

ML

VL

VR

Slope = w1

Slope = -w2 MR

Slope = VR

Slope = VL


9

Example 4-7

Draw the shear and moment diagrams for the beam shown in the figure.

9 m

20 kN/m


10

9 m

20 kN/m

+

SOLUTION

(2/3)9 = 6 m(1/2)(9)(20) = 90 kN

30 kN 60 kN

x

V (kN)

30

+

60

-

x

M (kN•m)

V = 0

M

)9

20)()(2

1(

xx

3

x

Fy = 0:+

x = 5.20 m

0)9

20)()(2

1(30

xxx

+ Mx = 0:

0)2.5(30)3

2.5)](

9

2.520)(2.5)(

2

1[( M

M = 104 kN•m

104

V = 0

= 5.20 m

x

)9

20(x

30 kN


11

Example 4-8

Draw the shear and moment diagrams for each of the beam shown in the figure.

P

L

L

MO

L

wo

L

wo


12

+

L

Mo

SOLUTION

P

LPPL

0

0

Mo

0

x

V P

+

x

M

-PL

-

x

V

x

M Mo


13

L

wo

wo L

0

wo L2

2L

wo

(wo L)/2

0

wo L2

6

x

V wo L

+

xM

-wo L2

2

-

x

V (wo L)/2

+

xM

-wo L2

6

-


14

Example 4-9

Draw the shear and moment diagrams for the beam shown in the figure.

3 kN

5 kN•m

A B

C D

3 m 1.5 m 1.5 m


15

3 kN

5 kN•m

A B

C D

3 m 1.5 m 1.5 m

SOLUTION

0.67 kN 2.33 kN

V (N)x (m)

0.67+

-2.33

-

M (kN•m)x (m)

2.01

+

-1.49

3.52

-+


16

Example 4-10

Draw the shear and moment diagrams for the compound beam shown in the figure. Assume the supports at A is fix C is roller and B is pin connections.

12 m12 m 15 m

8 kN 30 kN•m

A B C

hinge


17

Ay

Ax

MA

SOLUTION

Cy

By

Bx

Bx

By

0 =

= 2 kN

= 2 kN

= 2 kN

= 0= 0

= 6 kN

8 kN

30 kN•m

= 48 kN•m

12 m12 m 15 m

8 kN 30 kN•m

A B C

hinge


18

V (kN)x (m)

6 6

-2 -2

x (m)M (kN•m)

-48

24

-30

8 m

-

+

-

12 m12 m 15 m

8 kN

A B C

30 kN•m30 kN•m

6 kN

48 kN•m

2 kN


19

Example 4-11

Draw the shear and moment diagrams for the compound beam shown in the figure. Assume the supports at A and C are rollers and B and D are pin connections.

5 kN 3 kN/m2 kN/m60 kN • m

Hinge

10 m 6 m 4 m 6 m 6 m

AB C D


20

By

Bx

Bx

By

Ay

Cy Dy

Dx

= 0 kN

= 16 kN

= 4 kN

= 16 kN

0 == 0 kN

= 45 kN = -6 kN

SOLUTION

5 kN 9 kN 9 kN

4 m 4 m

60 kN • m

20 kN

5 kN 3 kN/m2 kN/m60 kN • m

Hinge

10 m 6 m 4 m 6 m 6 m

AB C D


21

V (kN) x (m)

-16-21 -21

24

6

M (kN • m)

x (m)

2 m

60

64

-96

-180

+

-

4

5 kN2 kN/m60 kN • m

Hinge

10 m 6 m 4 m 6 m 6 mB C D

3 kN/m

A

4 kN

45 kN

6 kN


22

P1

B

P2

Ay

Cx

CyBx

By

MB

Bx

By

MBBx

By

MB

Bx

By

MB

P1

P2

Ay

Cx

Cy

P1

P2

A

BC

Shear and Moment Diagrams for a Frame


23

Cx

Cy

Bx

By

MB

MB

P2

P2 = Bx

P2

Ay

Bx

By

MB

By By = Cy

MBMB

MB

A

B C


24

Example 4-12

Draw the shear and moment diagrams for the frame shown . Assume A, C and D are pinned and B is a fixed joint.

3 kN/m

15 kN4 m 4 m

12 m

60 kN•mA

B C

D


25

Find the Reaction

15 kN

4 m 4 m

B

36 kN

60 kN•m

Ax

Ay

Dx

Dy

= 5 kN

= 42 kN

= 41 kN

= -27 kN

Cx

Cy

Cx

Cy

= 5 kN

= 5 kN

= 42 kN

= 42 kN

3 kN/m

15 kN4 m 4 m

12 m

60 kN•mA

B C

D


26

V (kN)

x (m)Bx

By

MB

Member AB

= 276 kN•m

12 m

3 kN

/m

A

B

Ax=41 kN

Ay= 27 kN

= 5 kN

= 27 kN

41

5

M (kN•m)

x (m)

276


27

--

V (kN)

x (m)

15 kN

4 m 4 m

B

5 kN

42 kN

C

Member BC

41 kN

27 kN

Bx

By

MB

12 m

3 kN

/m

A

B= 5 kN

= 27 kN

= 276 kN•m

5 kN

27 kN

276 kN•m

-27-42 -42

M (kN•m)

x (m)

276168


28

-

V (kN)

x (m)

60 kN•m

5 kN

5 kN

42 kN

D

C

42 kN

12 m

Member CD

-5

-5

M (kN•m)

x (m)

60

+


29

Bending moment diagram of frame

276

A

B

+

276168

C

+

60 D

+

3 kN/m

15 kN4 m 4 m

12 m

60 kN•mA

B C

D


30

Example 4-13

Draw the moment diagram for the frame shown . Assume A is pin, C is a roller, and B is a fixed joint.

40 kN/m

80 kN

4 m 4 m

2 m3 m

A

B C


31

120 kN

1.5 m36.87o

80 kN

4 m 4 m

2 m3 m

B

Ay

Ax

Cy

+ MA = 0;

+ Fx = 0;

+Fy = 0;

82.5 kN =

- (120)(1.5) - (80)(6) + 8Cy = 0 Cy = 82.5 kN, ญ

120 kN =

-Ax + 120 = 0; Ax = 120 kN , ญ

= 2.5 kN

- Ay - 80 + 82.5 = 0; Ay = 2.5 kN , ญ


32

Bx

By

MB

Bx

By

MBB

80 kN

82.5 kN

B

C2 m 2 m

2.5 kN

120 kN

120 kN

A

B

36.87o 1.5 m

1.5 m

By´ Bx´

MB´

By´ Bx´

MB´

+ MB = 0:

Member BC

+ Fy = 0:

2.5 kN =

-By - 80 + 82.5 = 0, By = 2.5 kN , ญ

36.87o

By´ cos 36.87By´ sin 36.87

-Bxćos 36.87 + By´sin 36.87 + 0 = 0 -----(1)

Joint B

+ Fx = 0;

36.87o

Bxćos 36.87

Bx´sin 36.87

-Bx´sin 36.87 - Byćos 36.87 + 2.5 = 0 -----(2)

+ Fy = 0;

=2.5 kN

= 0 kN

=170 kN•m

170 kN•m=

From eq. (1) and (2): Bx´ = 1.5 kN By´ = 2 kN

1.5 kN = = 2 kN

=1.5 kN

=170 kN•m2 kN=

0 kN =

170 kN•m =

-MB -80(2) + 82.5(4) = 0, MB = 170 kN•m


33

1.5 kN

170 kN•m2 kNB

120 kN36.87o

2.5 kN

120 kN

36.87o

53.13

o5 m

120s

in36

.87o

1.5 kN

170 kN•m2 kN

70 kN97.5 kN

120 s

in 36

.87/5

=14

.4 kN

/m

70

-2

x (m)

M (kN•m)

4.86m

+

-

170.1

170

x (m)

V (kN)


34

40 kN/m

80 kN

A

B C

-

-

80 kN

82.5 kN

BC170 kN•m

2.5 kN

V (kN) x (m)-2.5

-82.5

M (kN•m) x (m)

170 165

170

+

170 165

+

A

CB

M (kN•m)


35

L

P

Mx

-PL

L

wo

Mx

2

2Lwo Parabolic curve

Most loading on beams in structural analysis will be a combination of the loadings shown in the figure below:

Moment Diagrams Constructed by the Method of Superposition


36

M

x

Mo

L

Mo

L

wo

Mx

6

2Lwo Cubic curve


37

M (kN•m)

x (m)

-20

70

=

M (kN•m)

x (m)

90

M (kN•m)

x (m)

-20

+

M (kN•m)

x (m)

-20

+

12 m

20 kN•m20 kN•m 5 kN/m

12 m

5 kN/m=

12 m

20 kN•m +

12 m

20 kN•m+


38

Example 4-14

Draw the moment diagrams for the beam shown at the top of the figure below using the method of superposition. Consider the beam to be cantilevered from the support at B.

6 m

20 kN•m5 kN/m

2 m


39

6 m

20 kN•m5 kN/m

2 m

SOLUTION

8.33 kN 6.67 kN

6 m

8.33 kN

+

6 m

5 kN/m+

20 kN•m=

M (kN•m)

x (m)

49.98+

M (kN•m)

x (m)

-20 -20

-20

M (kN•m)

x (m) 4.84

=

M (kN•m)

x (m)

-30

+


40

+ M + M

positive moment,concave upward

- M - M

negative moment,concave downward

P1

P2M

x

inflection point

M

x

inflection point

P1

P2

Deflected Curve


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Documents

Analysis of Beams and Frames Theory of Structure - I