Analysis of Beams and Frames

Theory of Structure - I

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Lecture Outlines

Internal Loadings at a Specified Point Shear and Moment Functions

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

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N N

V

V

M M

N

M

V

V

N

M

Positive Sign Convention

Internal Loadings at a Specified Point

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

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Example 4-1

Determine the internal shear and moment acting in the cantilever beam shown in the figure at sections passing through points C and D.

5 kN 5 kN 5 kN 5 kN 5 kN

1 m 1 m 1 m 1 m 1 m20 kN•m

A BCD

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SOLUTION

5 kN 5 kN 5 kN

1 m 1 m 1 m20 kN•m

BCVC

MC

NC0 =

Fy = 0:+ VC -5 - 5 - 5 = 0, VC = 15 kN

+ MC = 0: -MC -5(1) - 5(2) - 5(3) -20 = 0, MC = -50 kN

VD

MD

ND0 =

5 kN 5 kN 5 kN

1 m 1 m 1 m20 kN•m

BD

5 kN

Fy = 0:+ VD -5 - 5 - 5 -5 = 0, VD = 20 kN

+ MD = 0: -MD -5(1) - 5(2) - 5(3) -20 = 0, MD = -50 kN

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

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Example 4-2

Determine the internal shear and moment acting at a section passing through point C in the beam shown in the figure.

9 m

20 kN/m

3 mC

A B

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9 m

20 kN/m

C

AB

VC

MC

NC

3 m

)209

3(

30 kNC

1 m

10 kN

= 0

SOLUTION

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

ByAy

Ax

+ MA = 0: By(9) - 90(6), By = 60 kN

= 60 kN

Fy = 0:+ Ay - 90 + 60 = 0 Ay = 30 kN

30 kN =

0 =

Fy = 0:+-VC - 10 + 30 = 0, VC = 20 kN

= 20 kN

+ MC = 0:

MC + 10(1) - 30(3) = 0, MC = 80 kN•m

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

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Example 4-3

The 40 kN force in the figure is supported by the floor panel DE, which in turn is simply supported at its ends by floor beams. These beams transmit their loads to the sumply supported girder AB. Determine the internal shear and moment acting at point C in the girder.

B

40 kN

2 m 2 m 2 m0.6 m 1.4 m

A

C2.5 m

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C

SOLUTION

40 kN

0.6 m 1.4 m

12 kN28 kN

12 kN28 kN4 m

6 m

8 m 23 kN17 kN

B

40 kN

2 m 2 m 2 m0.6 m 1.4 m

A

C2.5 m

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23 kNC

28 kN 12 kN2.5 m

17 kN

C

2.5 m

17 kN VC

MC

NC = 0

Fy = 0:+

-VC + 17 = 0, VC = 17 kN

+ MC = 0:

MC - 17(2.5) = 0, MC = 42.5 kN•m

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

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x1

x2

x3

x1 x2 x3

w

P

AB C

D

w

P

AB C

D

OR

Shear and Moment Functions

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

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Example 4-4

Determine the shear and moment in the beam shown in the figure as a function of x.

9 m

10 kN/m

x

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x

109

x

45 kN

270 kN•m

SOLUTION (1/2)(9)(10) = 45 kN

(2/3)(9) = 6 m45 kN

270 kN•m

95)10

9)()(

2

1(

2xxx

3

xV

M

N = 0

Fy = 0:+

0459

5 2

x

V

459

5 2

x

V

+ Mx= 0:

027045)3

(9

5 2

xxx

M

27045)3

(9

5 2

xxx

M

9 m

10 kN/m

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

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Example 4-5

Determine the shear and moment in the beam shown in the figure as a function of x1, x3, x3, and x4.

x2

x3x1

x4

60 kN/m265 kN

135 kN•m4 m

6 m

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SOLUTION60 kN/m

265 kN

1350 kN•m4 m

6 m

Fy = 0:+ 505 - 60x1 - V = 0V = 505 - 60x1

Fy = 0:+ 505 -240 - V = 0V = 265 kN

+ Mx2= 0: 03420505)2(240 22 xxM

2940265 2 xM

505 kN

3420 kN•m

MV505 kN

3420 kN•m

x1

60x1

+ Mx1= 0: 03420505)2

(60 11

1 xx

xM

342050530 12

1 xxM

2 m x2 -2

240

x2

505 kN

3420 kN•m

MV

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

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60 kN/m265 kN

1350 kN•m4 m

6 m505 kN

3420 kN•m

Fy = 0:+ 505 - 240 - V = 0

V = 265 kN

+ Mx3= 0: 03420)4(505)2(240 33 xxM

1880265 3 xM

M

V

240 kN/m

2 m505 kN

3420 kN•m2 m x3

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

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60 kN/m265 kN

1350 kN•m4 m

6 m505 kN

3420 kN•m

265 kN

1350 kN•mx4

MV

Fy = 0:+ V - 265 = 0

V = 265 kN

+ Mx3= 0: 01350265 4 xM

1350265 4 xM

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

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Example 4-6

Determine the shear and moment in the beam shown in the figure as a function of x.

30 kN/m

10 kN/m

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

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9 m

4.5 m

Fy = 0:+

0])9

)(20(2

1[1075 Vx

xx

+ Mx= 0:

20 kN/m

90 kN

6 m

90 kN

10 kN/m

2

x

2

x3

x

10x

10 kN/m

xx

)9

)(20(2

1

mkNx

/9

20

M

V

105 kN75 kN

211.11075 xxV

0)3

]()9

)(20(2

1[)

2)(10(75 M

xx

xxxx

32 37.0575 xxxM 75 kN

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

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