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Engineering Mechanics (Statics) 7-1 Analysis of Beams & Compound Beams

Ex. 7.6 : A weight W rests on beam AB of negligible weight.

The string connecting the load is passing over a

smooth pulley. Show that reaction at A is

W (l a)

(l + a). (Refer Fig. Ex. 7.6).

Fig. Ex. 7.6

Soln. : Draw F.B.D of block and beam. (Refer Fig. Ex. 7.6(a)).

For block : Fy = 0 T W + RC= 0R

C = (W T) (1)

For beam : Fy = 0 T RC+ Ay = 0Substituting value of RC

We get, Ay = RC T = (W T) T

Ay = W 2T (2)MA = 0 RCa T l = 0RCa = T l

(W T) a = T l Wa = T (a + l)

T =Wa

a + l(3)

Substituting this value in Equation (2)

Ay = W 2[ ]Wa

a + l

=W(a + l) 2 Wa

a + l=

Wa + Wl

a + l

Ay =W (l a )

l + a

Reaction at A RA =W (l a )

l + aproved. Fig. Ex. 7.6(a)

Ex.7.8 : A beam 10 m long is loaded as shown in Fig. Ex.7.8. Find reactions at supports.

Fig. Ex.7.8

Soln. : Here given loading is a varying load so to find out total load and its point of application (centroid) using

integration method.

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Engineering Mechanics (Statics) 7-2 Analysis of Beams & Compound Beams

Fig. Ex.7.8(a) Fig. Ex.7.8(b)

Select an elementary strip of thickness dx as shown in Fig. Ex.7.8(a).

The area of strip dA = y dx

Total area bounded = 0

10

ydx =

0

10

(100 + 0.3 x

3

) dx = [ ]100 x + 0.3x4

(4)

10

0 = 100 (10) +

0.3 (10)4

4 = 1750 N

The distance of centroid is given by,

Ax = x

e1dA =

0

10x y dx =

0

10(100 x + 0.3 x4) dx

= [ ]100 x2

2+ 0.3

x5

5

10

0= 100

(10)2

2+ 0.3

(10)5

5

Ax = 11000 m3

x =

11000

1750= 6.285 m from y-axis

Conditions of equilibrium, [Refer Fig. Ex.7.8(b)].MA = 0 RB 10 1750 6.285 = 0 RB = 1100 N Fy = 0 RA + RB = 1750 RA = 650 N Ans.

Ex.7.13 : Determine the reaction components at A, B, D, E and F. Also find value of load W if reaction at support

G is found to be 25 kN. Refer Fig. Ex.7.13.

Fig. Ex.7.13

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Engineering Mechanics (Statics) 7-3 Analysis of Beams & Compound Beams

Soln. : F.B.D. of beams

Fig. Ex.7.13(a)

Since beam does not carry any horizontal force, all horizontal components are equal to zero.

Ax = Dx = 0Consider beam GF

Use, MF = 0 (RE 3) + (6 4) (25 5) = 0 RE = 33.67 kN

Fy= 0 25 + R

F 6 R

E= 0

RF = 6 + 33.67 25 RF = 14.67 kN Consider beam ED,

Use, MD = 0 RE 7 + (7.5 5) + RB 1 = 0 ( 33.67 7) + (7.5 5) + RB = 0

RB = 198.19 kNFy= 0 RE + Dy 7.5 RB = 033.67 + Dy 7.5 198.19 = 0

Dy = 172.02 kN Consider beam CA,

Use,

MA = 0 (36

2) 12 (RB

4) + W (6) = 0 72 12 198.19 4 + 6W = 0

W = 122.13 kN Fy= 0 W 36 + RB + Ay = 0

122.13 36 + 198.19 + Ay = 0

Ay = 40.06 kN

Ay = 40.06 kN

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Engineering Mechanics (Statics) 7-4 Analysis of Beams & Compound Beams

Ex.7.17 : When closed vertical sluice gate is supported alongthe lines AA and BB normal to the plane of

Fig. Ex.7.17. It is subjected to water pressure

from sides. If w = 1000 kg/m3 for water find

reaction at A and B.

Soln. : Water pressure

P1 =1

2wh

2

1=

1

2 1000 9.81 42

= 78480 N = 78.48 kN

acting at 4/3 m from B

P2 =1

2wh

2

2

=1

2 1000 9.81 (6)2

= 176580 N = 176.58 kN Fig. Ex.7.17

acting at 6/3 = 2 m from B

Apply conditions of equilibrium

MB = 0 P1 (4/3) + P2 (2) RA (6) = 0 78.48 ( )

4

3+ 176.58 (2) = 6 RA

RA = 41.46 kN Ans.Fx = 0 RA + P1 + RB P2 = 0

RB = P2 P1 RA

= 176.58 78.48 41.46

RB = 56.64 kN Ans. Fig. Ex.7.17(a)

Analysis of Beams & Compound Beams Ends.