Solved problems on mechanics and engineering

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Que-1
Solution
Considering FBD AT point C
∑Fx = 0
-1500 COS 90° + PCB COS 180 -POC COS 225 = 0
PCB + 1
√ 2 POC = 0
POC = √ 2PDB
Again
∑FY = 0
-1500 SIN90° + PCB COS 180 -POC COS 225 = 0
-1500 + 1
√ 2 PDC = 0
45°
45°
45°
45°
1500N 1500N
1000mm
RAV RAD
RAH
A
B C
D
45°
PCB
1500
PDC
1

PDC = 2121.32 N (Compression)
PCB = 1500 (TENSION)
ConsideringFBD AT point B
∑Fx = 0
-1500 COS 0 + 1500 COS 90 –PAB COS 225 –PDB COS 270 = 0
-1500 + 1
√ 2 PAB = 0
PAB = 1221.32N (Compression)
Again
∑FY = 0
-1500 SIN0 - 1500SIN 90–PAB SIN 270 = 0
-1500 + 1
√ 2 PAB + PDB = 0
-1500 + 1500 + PDB = 0
PDB = 0 N
Considering FBD AT point D
45°
PBC = -1500
1500
PAB
PDB
45°
(-2121.31 N)
PAD
PRD

By Lame’s Theorm
−2121.32
sin 90 = Rd
sin 135 = Pad
sin 135
RD = PAD = -1500 N
RD = 1500 N(TENSION)
Considering point A
Considering FBD at point D
∑Fx = 0
-1500 COS 0 + 2121.32COS 45 –PAH COS 180 –RAV COS 270 = 0
-1500 + 1500 + RAH = 0
RAH = 0
-1500 SIN 0 + 2121.32 SIN 45 –RAH SIN 180 –RAV SIN 270 = 0
1500 + RAV = 0
RAV= -1500N
i.e. RAV= 1500N (Tension)
if all the value changes to 30°
Considering FBD AT point C
45°
PAD = -1500
(PAB =-2121.31 N
N)
RAH
RAV
30°
PCB
1500 N
PDC

∑Fx = 0
-PCB + √ 3
2 POC = 0
PDC = 2
√ 3 PCB
Again
∑FY = 0
-1500 + Pdc
2 = 0
PDC = 3000 N (com)
PCB = 2598.076 N (Tension)
Considering FBD AT point B
∑Fx = 0
-2598.076 COS 0 – 1500 COS 90 –PAB COS 240 –PDB COS 270 = 0
-2598.076 + P ab
2 = 0
PAB = 5196.12 N (Compression)
∑Fy = 0
-2598.076 SIN 0 – 1500 SIN 90 –PAB COS 240 –PDB SIN 270 = 0
– 1500 –4500 –PDB = 0
1
30°
(-2598.076)
1500 N
PAB
PDB

PDB = 3000(Compression)
Considering FBD AT point D
∑Fx = 0
-3000 COS 30 + 3000 COS 90 –PDA COS 180 –RDCOS 270 = 0
2598.076 – PDA = 0
PDA = 2598.076(Tension)
∑Fy = 0
3000 SIN 30 + 3000 SIN 90 –PDA SIN 180 –RD Sin 270 = 0
RD = 4500 N (Tension)
Considering FBD AT point A
∑Fx = 0
2598.076COS 0 + 5196.152 COS 30 –RAH COS 180 –RAVCOS 270 = 0
7098.075 – RAH = 0
RAH = -7098.075 N
RAH = 7098.075 N (Tension)
30°
3000N
3000 N
PDA
RD
30°
5196.152N
RAH
RAV
2598.076

∑FY = 0
2598.076 SIN 0 + 5196.152 SIN 30 –RAH SIN 180 –RAV SIN 270 = 0
2598.076 + RAV = 0
RAV= -2598.076 N (Tension)
Que- 2
N = 2500 rpm
Power of Motor = 1 hp = 746W
Torque between Motor and the Gearbox
Power Tω Where T = Torque
T = T X 2π X 2500/60
T = 746 x 60/(2π X 2500)
= 2.8495 Nm
Since no Gear ratio is mention there for we consider torque is uniform through out the shaft
T/J = τ/R
Where J = polar M. o. I. of the Shaft
τ= TR/J
τ= (16 X 2.8495)/(π X (25 x 10^-3 )^3
τ = 0.9288 MPa
A B C
Motor Gearbox Blower

Que- 3
From Question 1 force in link AB will be PAB = 2121.32 N
Diameter of Link AB, d = 55mm
Stress at link AB = PAB / AAB
= 2121.31/(π/4 X 55^2)
БAB = 0.8929 Mpa