Solution to Problem 203 Stress-strain Diagram

Problem 203
The following data were recorded during the tensile test of a 14-mm-diameter mild steel rod. The gage length was 50 mm.
 

Load (N) Elongation (mm) Load (N) Elongation (mm)
0 0 46 200 1.25
6 310 0.010 52 400 2.50
12 600 0.020 58 500 4.50
18 800 0.030 68 000 7.50
25 100 0.040 59 000 12.5
31 300 0.050 67 800 15.5
37 900 0.060 65 000 20.0
40 100 0.163 65 500 Fracture
41 600 0.433

 

Plot the stress-strain diagram and determine the following mechanical properties: (a) proportional limits; (b) modulus of elasticity; (c) yield point; (d) ultimate strength; and (e) rupture strength.
 

Axial Deformation

In the linear portion of the stress-strain diagram, the tress is proportional to strain and is given by
 

$\sigma = E \varepsilon$

 

since $\sigma = P / A$ and $\varepsilon = \delta / L$, then $\dfrac{P}{A} = E \dfrac{\delta}{L}$
 

$\delta = \dfrac{PL}{AE} = \dfrac{\sigma L}{E}$

To use this formula, the load must be axial, the bar must have a uniform cross-sectional area, and the stress must not exceed the proportional limit.
 

Stress-strain Diagram

Suppose that a metal specimen be placed in tension-compression-testing machine. As the axial load is gradually increased in increments, the total elongation over the gauge length is measured at each increment of the load and this is continued until failure of the specimen takes place. Knowing the original cross-sectional area and length of the specimen, the normal stress σ and the strain ε can be obtained. The graph of these quantities with the stress σ along the y-axis and the strain ε along the x-axis is called the stress-strain diagram. The stress-strain diagram differs in form for various materials. The diagram shown below is that for a medium-carbon structural steel.
 

Simple Strain

Also known as unit deformation, strain is the ratio of the change in length caused by the applied force, to the original length.
 

Simple strain

 

$\varepsilon = \dfrac{\delta}{L}$

where δ is the deformation and L is the original length, thus ε is dimensionless.
 

Solution to Problem 142 Pressure Vessel

Problem 142
A pipe carrying steam at 3.5 MPa has an outside diameter of 450 mm and a wall thickness of 10 mm. A gasket is inserted between the flange at one end of the pipe and a flat plate used to cap the end. How many 40-mm-diameter bolts must be used to hold the cap on if the allowable stress in the bolts is 80 MPa, of which 55 MPa is the initial stress? What circumferential stress is developed in the pipe? Why is it necessary to tighten the bolt initially, and what will happen if the steam pressure should cause the stress in the bolts to be twice the value of the initial stress?