An aluminum bar 600mm long, with diameter 40mm long has a hole drilled in the center of the bar.The hole is 30mm in diameter and
1 answer:
Answer:
Total contraction on the Bar = 1.22786 mm
Explanation:
Given that:
Total Length for aluminum bar = 600 mm
Diameter for aluminum bar = 40 mm
Hole diameter = 30 mm
Hole length = 100 mm
elasticity for the aluminum is 85GN/m² = 85 × 10³ N/mm²
compressive load P = 180 KN = 180 × 10³ N
Calculate the total contraction on the bar = ???
The relation used in calculating the contraction on the bar is:
The relation used in calculating the total contraction on the bar can be expressed as :
Total contraction in the Bar = (contraction in part of bar without hole + contraction in part of bar with hole)
i.e
Total contraction on the Bar =
Let's find the area of cross section without the hole and with the hole
Area of cross section without the hole is :
Using A = πd²/4
A = π (40)²/4
A = 1256.64 mm²
Area of cross section with the hole is :
A = π (40²-30²)/4
A = 549.78 mm²
Total contraction on the Bar =
Total contraction on the Bar =
Total contraction on the Bar = 2.117( 0.398 + 0.182)
Total contraction on the Bar = 2.117*(0.58)
Total contraction on the Bar = 1.22786 mm
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Answer:
r₂=0.1 m
Explanation:
Given that
r₁= 10 m , β₁ = 20 dB
At r₂ ,β₂= 60 dB
As we know that intensity level of sound given as
10² x 10⁻¹² = I₁
I₁=10⁻¹⁰ W/m²
10⁶ x 10⁻¹² = I₂
I₂ = 10⁻⁶ W/m²
I₁=10⁻¹⁰ W/m²
P = I A
P=Power ,I =Intensity ,A=Area
r₂=0.1 m