Civil Engineering - Steel Structure Design - Discussion

@All,

If we consider stress-strain curve of mild steel, Young's Modulus (longitudinal stress/longitudinal strain) is slope of stress-strain curve (up-to proportionality point) and the region in left side of elastic point (nearly equal to proportionality point) represents elastic range and right side of curve represents plastic range.

Given:
Axial tension = 2300 kg/cm^2 (axial stress),
Axial strain = 0.004,
Therefore, Axial stress/Axial strain = 2300/0.004 = 5.75 X 10^6 kg/cm^2,
which is lesser than Young's modulus of steel (2.1 x 10^6 kg/cm^2).
So, the bar is in the elastic range.

Can any one give me different types of ranges with their suitable values.

@Nilraj you are wrong. 2300 is not load check units duffer.

Here Area A=πD^2/4=3.14*20*20/4=315 mm sq.=3.14 cm sq.

Load P=2300 Kg.
Stress=P/A=2300/3.14=732.48kg/cm sq.

Now,

The stress/strain= 732.48/0.004=1.83 * 10^5 Kg/cm sq.....which is less than E=2.1*10^6 kg per cm sq.

So the bar is in elastic range.

There should be no units to strain. So t is assumed that the length is in meter to Strain= 0.004 / 100 = 0.00004 (m/m).

E=Stress/Strain so Strain= Stress/E = 2300/(2.1x10^6)=0.001. Since Strain here is 00004< 0.001, the member is in Elastic Range.

Load = 2300 kg/cm^2 = 23 kg/mm^2.
A = pi() x 20 x 20/4 = 314 sq.mm.

Axial tensile stress = P/A = 23/314 = 0.073 kg/sq.mm = 0.73 N/sq.mm.
Allowable tensile stress till yield point = 0.6 fy = 0.6 x 250 = 150 N/sq.mm.
0.73 < 150, Hence in elastic range.

How to find the answer? Please explain it.

Explain?

Why strain have unit?

How to check? Can I get any reference?