Analysis Of Beams And Compound Beams 1s6e1q

Engineering Mechanics (Statics) Ex. 7.6 :

7-1

Analysis of Beams & Compound Beams

A weight W rests on beam AB of negligible weight. The string connecting the load is ing 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 = 0

RC = (W – T) For beam :

 Fy = 0

…(1)

T – RC + Ay = 0

Substituting value of RC We get,

Ay = RC – T = (W – T) – T Ay = W – 2T

 MA = 0

…(2)

RC  a – T  l = 0 RC  a = T  l (W – T) a = T  l  Wa = T (a + l) Wa  T = a+l

…(3)

Substituting this value in Equation (2) Wa Ay = W – 2 a + l

[ ]

W(a + l) – 2 Wa – Wa + Wl = a+l a+l W (l – a ) = l+a W (l – a ) = proved. l+a

= Ay  Reaction at A

RA

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 s.

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.

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 10 10 x4  Total area bounded = ⌠ y  dx = ⌠ (100 + 0.3 x3 ) dx = 100 x + 0.3 (4)

⌡

[

⌡

10

]

0

= 100 (10) +

0.3 (10)4 = 1750 N 4

0 0 The distance of centroid is given by, – Ax

10 10 = ∫ –x e1 dA = ⌠ x  y  dx = ⌠ (100 x + 0.3 x4) dx

⌡

⌡

0

[100 x2 + 0.3 x5 ] 2

=

5

10 0

0

= 100

(10)2 (10)5 2 + 0.3 5

– Ax = 11000 m3 11000 –  x = 1750 = 6.285 m from y-axis Conditions of equilibrium, [Refer Fig. Ex.7.8(b)].

 MA = 0

RB  10 – 1750  6.285 = 0 

 Fy = 0

RB = 1100 N 

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

Fig. Ex.7.13

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 = 0 Consider beam GF Use,

MF = 0

(RE  3) + (6  4) – (25  5) = 0  RE = 33.67 kN

Fy = 0

25 + RF – 6 – RE = 0 RF = 6 + 33.67 – 25

RF

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 kN

Fy = 0

RE + Dy – 7.5 – RB = 0 33.67 + 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 

Fy = 0

W = 122.13 kN 

– W – 36 + RB + Ay = 0 – 122.13 – 36 + 198.19 + Ay = 0 

Ay = – 40.06 kN Ay = 40.06 kN 

= 14.67 kN 

Engineering Mechanics (Statics)

7-4

Analysis of Beams & Compound Beams

Ex.7.17 : When closed vertical sluice gate is ed along the 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 1 2 1 P1 = 2 wh1 = 2  1000  9.81  42 = 78480 N = 78.48 kN  acting at 4/3 m from B 1 2 1 P2 = 2 wh2 = 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

(43) + 176.58 (2) = 6 R

A

 RA = 41.46 kN 

Fx = 0

…Ans.

– 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….