Three spheres A, B and C having their diameters
500 mm, 500 mm and 800 mm, respectively, are placed
in a trench with smooth side walls and floor as shown in
Fig. 3.54. The centre-to-centre distance of spheres A and
B is 600 mm. The weights of the cylinders A, B and C
are 4 kN, 8 kN and 4 kN respectively. Determine the
reactions at P, Q, R and S.
[Ans. Rp = 2.1545 kN; R = 7.4424 kN; Rp = 7.0296 kN;
Rs = 2.2962 kN)
T
TTTT
A
75 (
65
Timur
600mm
Fig. 3.54
Answers
Answer:
From triangle ABC in fig
Cosα = AD/AC = 300/(250 + 400)
Cosα = 62.51°
Consider FREEE BODY DIAGRAMof sphere C
Consider equilibrium of block C
∑H = R1Cosα – R2 Cosα = 0
R1 = R2
∑H = R1 sinα – R2 sin α – 8 = 0
putting R1 = R2 ∑V = R1 sinα – R1 sinα = 8 ⇒ 2R1 = 8/sina = R1 = 8/2 sinα =4.509
R1 = R2 = 4.509 KN
Consider equilibrium of block
A ∑H = Rp sin 75° – R1 cosα = 0
RP = R1 cosα/sin 75° = 4.5 cos 62.51°/sin 75° Rp
= 2.15KN .
∑V = Rp cos75° – R1 sin 62.50 + RQ – WA = 0 RQ = 7.44KN
AFTER THIS THE PROBLEM GET SIMPLER
Consider equilibrium of block B
∑H = RS sin 65° – R2 cosα = 0 =
RS = R2 cosα/sin 65° = 4.5 cos 62.51°/sin 65°
RS = 2.29KN ∑V = RS cos 65° – R2 sinα + RR – WB = 0
∑V = 2.29 cos 65° – 4.509 sin 62.5° + RR – 4 = 0
RR =7.02KN
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