Asked by ADIVHAHO
A square cast-iron bar is placed under 50 kN pressure. 6.3.1 Calculate the dimensions of the bar if the stress in the bar is not to exceed 300 kN.
Answers
Answered by
GPT 3.5
Given data:
Pressure on the bar = 50 kN = 50000 N
Maximum allowable stress = 300 kN = 300000 N
Let the dimensions of the square cast-iron bar be:
Length = L
Width = W
Thickness = T
The formula for stress is:
Stress = Force / Area
Since the bar is square, its cross-sectional area can be calculated as:
Area = W * T
Given that Pressure on the bar = Force, we can write:
50000 / (W * T) <= 300000
Solving for W, we get:
W >= 50000 / (300000 / T)
W >= 1/6 T
Similarly, solving for T:
T >= 50000 / (300000 / W)
T >= 1/5 W
Since the bar is square, L = W. Therefore, substituting for W, we get:
L >= 1/6 T
This means that the dimensions of the square cast-iron bar should be as follows:
Length (L) >= 1/6 x Thickness (T)
Width (W) >= 1/6 x Thickness (T)
Thickness (T) >= 1/5 x Width (W)
Therefore, the dimensions of the bar can be designed such that the stress in the bar does not exceed 300 kN.
Pressure on the bar = 50 kN = 50000 N
Maximum allowable stress = 300 kN = 300000 N
Let the dimensions of the square cast-iron bar be:
Length = L
Width = W
Thickness = T
The formula for stress is:
Stress = Force / Area
Since the bar is square, its cross-sectional area can be calculated as:
Area = W * T
Given that Pressure on the bar = Force, we can write:
50000 / (W * T) <= 300000
Solving for W, we get:
W >= 50000 / (300000 / T)
W >= 1/6 T
Similarly, solving for T:
T >= 50000 / (300000 / W)
T >= 1/5 W
Since the bar is square, L = W. Therefore, substituting for W, we get:
L >= 1/6 T
This means that the dimensions of the square cast-iron bar should be as follows:
Length (L) >= 1/6 x Thickness (T)
Width (W) >= 1/6 x Thickness (T)
Thickness (T) >= 1/5 x Width (W)
Therefore, the dimensions of the bar can be designed such that the stress in the bar does not exceed 300 kN.