Answer:
<em>minimum required diameter of the steel linkage is 3.57 mm</em>
<em></em>
Explanation:
original length of linkage l = 10 m
force to be transmitted f = 2 kN = 2000 N
extension e = 5 mm= 0.005 m
maximum stress σ = 200 N/mm^2 =
maximum stress allowed on material σ = force/area
imputing values,
200 = 2000/area
area = 2000/() =
m^2
recall that area =
=
=
=
=
m = 3.57 mm
<em>maximum diameter of the steel linkage d = 3.57 mm</em>
Answer:
The answer is "".
Explanation:
Please find the complete question in the attachment file.
As the original prices are 24 euros, must be multiplied by 24.
Therefore the new price value will be
Answer:
For [1 1 0] and [1 0 1] plane, σₓ = 6.05 MPa
For [0 1 1] plane, σ = 0; slip will not occur
Explanation:
compute the resolved shear stress in [111] direction on each of the [110], [011] and on the [101] plane.
Given;
Stress direction: [1 0 0] ⇒ A
Slip direction: [1 1 1]
Normal to slip direction: [1 1 1] ⇒ B
∅ is the angle between A & B
Step 1: cos∅ = A·B/|A| |B| = ⇒ cos∅ = 1/
σₓ = τ/cos ∅·cosλ
where τ is the critical resolved shear stress given as 2.47MPa
Step 2: Solve for the slip along each plane
(a) [1 1 0]
cosλ = [1 1 0]·[1 0 0]/(·
)
note: cosλ = slip D·stress D/|slip D||stress D|
cosλ = 1/
∵ σₓ = τ/ ·
=
* 2.47MPa = 6.05MPa
Hence, stress necessary to cause slip on [1 1 0] is 6.05MPa
(b) [0 1 1]
cosλ = [0 1 1]·[1 0 0]/(·
) = 0
∵ σₓ = 2.47MPa/0, which is not defined
Hence, for stress along [1 0 0], slip will not occur along [0 1 1]
(c) [1 0 1]
cosλ = [0 1 1]·[1 0 0]/(·
)
cosλ = 1/
∵ σₓ = τ/ ·
=
* 2.47MPa = 6.05MPa
See attachment for the space diagram
