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. The centre-to-centre distance of spheres A and B is 600 mm. The cylinders A, B and C weigh 4 kN, 4 kN and 8 KN respectively. Determine the reactions developed at contact points P, Q, R and S.
Answers
Answer:
Explanation:
From triangle ABC in fig(I)
Cosα = AD/AC = 300/(250 + 400)
Cosα = 62.51°
Consider FBD of sphere C(Fig (ii)(a))
Consider equilibrium of block C ∑H = R1Cosα – R2 Cosα = 0
i.e., R1 = R2 …..(I)
∑H = R1 sinα – R2 sin α – 8 = 0
⇒ putting R1 = R2
∑V = R1 sinα – R1 sinα = 8 ⇒ 2R1 = 8/sin a
= R1 = 8/2 sinα =4.509 i.e.,
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
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.