Answer
given,
l₀ = 50 mm
d₀ = 13.1 mm
l₁ = 53.7 mm
d₁ = 12.6 mm
Area of cross section
A = 134. 782 mm²
The strain is
= 0.074
the tensile modulus of A356 aluminium is
E = 72.4 GPa
The stress is
σ = εE
= 72.4 × 10⁹ × 0.074
= 5357.6 × 10⁶ Pa
σ = 5357.6 MPa
Answer:
The maximum allowable distance = 5 m
Explanation:
Data:
There will be three forces on the jib. Let the forces be denoted as:
,
and the force on pole AB
To find the angle AB makes with the horizontal beam:
The load has a mass of 2 000 kg then, the force will be:
F = mg, where g = 9.81 m/s²
=
Breaking AB into its x and y coordinates:
Then,
∑ = 0
∑
∑
so the components of the forces will be 30 656.2 N and - 3 372.18 N
Answer:
22.90 × 10⁸ kg
Explanation:
Given:
Diameter, d = 0.02 m
ωₙ = 0.95 rad/sec
Time period, T = 0.35 sec
Now, we know
T=
where, L is the length of the steel cable
g is the acceleration due to gravity
0.35=
or
L = 0.0304 m
Now,
The stiffness, K is given as:
K =
Where, A is the area
E is the elastic modulus of the steel = 2 × 10¹¹ N/m²
or
K =
or
K = 20.66 × 10⁸ N
Also,
Natural frequency, ωₙ =
or
mass, m =
or
mass, m =
mass, m = 22.90 × 10⁸ kg