Question

ПОМОГИТЕ РЕШИТЬ ПЛИЗПод действием силы направленной уздовж оси стержня в нём возникает напряжение 150 мПа. Диаметр стержня 4мм

Answer 1

d =4 мм = 0.004 м

R = d/2 =0.002 м

S =пR²

Под действием силы направленной ВДОЛЬ оси стержня cos<a = 1

напряжение p = F/S *c0s<a

сила F = p*S/cos<a = p*пR²/cos<a= 150 мПа *п*0.002² / 1 = 1.9*10^-6 Н = 1.9 мкН ~ 2 мкН

 

Answer 2

Calculation of Stress

To calculate the stress (σ) in the rod, we can use the formula:

σ = F/A

Where: - σ = stress - F = force applied - A = cross-sectional area of the rod

Given: - Force (F) = force applied along the axis of the rod - Diameter (d) = 4 mm

First, we need to calculate the cross-sectional area (A) of the rod using the diameter.

Calculation of Cross-Sectional Area

The formula to calculate the cross-sectional area (A) of a rod is:

A = π * (d/2)^2

Where: - A = cross-sectional area - π (pi) = 3.14159 - d = diameter of the rod

Let's calculate the cross-sectional area using the given diameter.

Cross-Sectional Area Calculation

Using the given diameter (d = 4 mm), we can calculate the cross-sectional area (A) of the rod:

A = π * (4/2)^2

A = 3.14159 * (2)^2

A = 3.14159 * 4

A = 12.56636 mm^2

Now that we have the cross-sectional area, we can proceed to calculate the stress in the rod.

Calculation of Stress

Given: - Stress (σ) = 150 MPa - Cross-sectional area (A) = 12.56636 mm^2

Using the formula σ = F/A, we can rearrange it to solve for the force (F):

F = σ * A

Substituting the given values: F = 150 * 12.56636

F ≈ 1884.954 MPa * mm^2

Therefore, the force applied along the axis of the rod is approximately 1884.954 MPa * mm^2.