Solution to Problem 206 Axial Deformation

Problem 206
A steel rod having a cross-sectional area of 300 mm2 and a length of 150 m is suspended vertically from one end. It supports a tensile load of 20 kN at the lower end. If the unit mass of steel is 7850 kg/m3 and E = 200 × 103 MN/m2, find the total elongation of the rod.
 

Solution to Problem 205 Axial Deformation

Problem 205
A uniform bar of length L, cross-sectional area A, and unit mass ρ is suspended vertically from one end. Show that its total elongation is δ = ρgL2/2E. If the total mass of the bar is M, show also that δ = MgL/2AE.
 

Solution to Problem 204 Stress-strain Diagram

Problem 204
The following data were obtained during a tension test of an aluminum alloy. The initial diameter of the test specimen was 0.505 in. and the gage length was 2.0 in.
 

Load (lb) Elongation (in.) Load (lb) Elongation (in.)
0 0 14 000 0.020
2 310 0.00220 14 400 0.025
4 640 0.00440 14 500 0.060
6 950 0.00660 14 600 0.080
9 290 0.00880 14 800 0.100
11 600 0.0110 14 600 0.120
12 600 0.0150 13 600 Fracture

 

Plot the stress-strain diagram and determine the following mechanical properties: (a) proportional limit; (b) modulus of elasticity; (c) yield point; (d) yield strength at 0.2% offset; (e) ultimate strength; and (f) rupture strength.
 

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.
 

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?
 

Solution to Problem 141 Pressure Vessel

Problem 141
The tank shown in Fig. P-141 is fabricated from 1/8-in steel plate. Calculate the maximum longitudinal and circumferential stress caused by an internal pressure of 125 psi.
 

141-flat-oval-tank.gif

 

Solution to Problem 140 Pressure Vessel

Problem 140
At what angular velocity will the stress of the rotating steel ring equal 150 MPa if its mean radius is 220 mm? The density of steel 7.85 Mg/m3.
 

Solution 140
140-fbd-rotating-ring.gif$CF = M \omega^2 \bar x$
 

Where:
$M = \rho V = \rho A \pi R$

$x = 2R / \pi$

 

Thus,
$CF = \rho A \pi R \omega^2 (2R / \pi)$

Solution to Problem 139 Pressure Vessel

Problem 139
Find the limiting peripheral velocity of a rotating steel ring if the allowable stress is 20 ksi and steel weighs 490 lb/ft3. At what revolutions per minute (rpm) will the stress reach 30 ksi if the mean radius is 10 in.?
 

Solution to Problem 138 Pressure Vessel

Problem 138
The strength of longitudinal joint in Fig. 1-17 is 33 kips/ft, whereas for the girth is 16 kips/ft. Calculate the maximum diameter of the cylinder tank if the internal pressure is 150 psi.
 

Solution to Problem 137 Pressure Vessel

Problem 137
A water tank, 22 ft in diameter, is made from steel plates that are 1/2 in. thick. Find the maximum height to which the tank may be filled if the circumferential stress is limited to 6000 psi. The specific weight of water is 62.4 lb/ft3.
 

Pages

Subscribe to MATHalino RSS