November 2008

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 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 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 207 Axial Deformation

Problem 207
A steel wire 30 ft long, hanging vertically, supports a load of 500 lb. Neglecting the weight of the wire, determine the required diameter if the stress is not to exceed 20 ksi and the total elongation is not to exceed 0.20 in. Assume E = 29 × 106 psi.
 

Solution to Problem 208 Axial Deformation

Problem 208
A steel tire, 10 mm thick, 80 mm wide, and 1500.0 mm inside diameter, is heated and shrunk onto a steel wheel 1500.5 mm in diameter. If the coefficient of static friction is 0.30, what torque is required to twist the tire relative to the wheel? Neglect the deformation of the wheel. Use E = 200 GPa.
 

Solution to Problem 209 Axial Deformation

Problem 209
An aluminum bar having a cross-sectional area of 0.5 in2 carries the axial loads applied at the positions shown in Fig. P-209. Compute the total change in length of the bar if E = 10 × 106 psi. Assume the bar is suitably braced to prevent lateral buckling.
 

Aluminum bar loaded as indicated

 

Solution to Problem 210 Axial Deformation

Problem 210
Solve Prob. 209 if the points of application of the 6000-lb and the 4000-lb forces are interchanged.
 

Solution 210
P1 = 4000 lb compression
P2 = 11000 lb compression
P3 = 6000 lb compression
 

Solution to Problem 211 Axial Deformation

Problem 211
A bronze bar is fastened between a steel bar and an aluminum bar as shown in Fig. p-211. Axial loads are applied at the positions indicated. Find the largest value of P that will not exceed an overall deformation of 3.0 mm, or the following stresses: 140 MPa in the steel, 120 MPa in the bronze, and 80 MPa in the aluminum. Assume that the assembly is suitably braced to prevent buckling. Use Est = 200 GPa, Eal = 70 GPa, and Ebr = 83 GPa.
 

Figure P-211

 

Solution to Problem 212 Axial Deformation

Problem 212
The rigid bar ABC shown in Fig. P-212 is hinged at A and supported by a steel rod at B. Determine the largest load P that can be applied at C if the stress in the steel rod is limited to 30 ksi and the vertical movement of end C must not exceed 0.10 in.
 

Figure P-212

 

Solution to Problem 213 Axial Deformation

Problem 213
The rigid bar AB, attached to two vertical rods as shown in Fig. P-213, is horizontal before the load P is applied. Determine the vertical movement of P if its magnitude is 50 kN.
 

Figure P-213

 

Solution to Problem 214 Axial Deformation

Problem 214
The rigid bars AB and CD shown in Fig. P-214 are supported by pins at A and C and the two rods. Determine the maximum force P that can be applied as shown if its vertical movement is limited to 5 mm. Neglect the weights of all members.
 

Figure P-214

 

Solution to Problem 215 Axial Deformation

Problem 215
A uniform concrete slab of total weight W is to be attached, as shown in Fig. P-215, to two rods whose lower ends are on the same level. Determine the ratio of the areas of the rods so that the slab will remain level.
 

Figure P-215

 

Solution 215

Solution to Problem 216 Axial Deformation

Problem 216
As shown in Fig. P-216, two aluminum rods AB and BC, hinged to rigid supports, are pinned together at B to carry a vertical load P = 6000 lb. If each rod has a cross-sectional area of 0.60 in.2 and E = 10 × 106 psi, compute the elongation of each rod and the horizontal and vertical displacements of point B. Assume α = 30° and θ = 30°.
 

Figure P-216 and P-217

 

Solution to Problem 217 Axial Deformation

Problem 217
Solve Prob. 216 if rod AB is of steel, with E = 29 × 106 psi. Assume α = 45° and θ = 30°; all other data remain unchanged.
 

Solution to Problem 218 Axial Deformation

Problem 218
A uniform slender rod of length L and cross sectional area A is rotating in a horizontal plane about a vertical axis through one end. If the unit mass of the rod is ρ, and it is rotating at a constant angular velocity of ω rad/sec, show that the total elongation of the rod is ρω2 L3/3E.
 

Solution to Problem 219 Axial Deformation

Problem 219
A round bar of length L, which tapers uniformly from a diameter D at one end to a smaller diameter d at the other, is suspended vertically from the large end. If w is the weight per unit volume, find the elongation of ω the rod caused by its own weight. Use this result to determine the elongation of a cone suspended from its base.
 

Shearing Deformation

Shearing Deformation
Shearing forces cause shearing deformation. An element subject to shear does not change in length but undergoes a change in shape.
 

shearing-deformation.jpg

 

The change in angle at the corner of an original rectangular element is called the shear strain and is expressed as
 

$\gamma = \dfrac{\delta_s}{L}$

 

The ratio of the shear stress τ and the shear strain γ is called the modulus of elasticity in shear or modulus of rigidity and is denoted as G, in MPa.
 

$G = \dfrac{\tau}{\gamma}$

 

The relationship between the shearing deformation and the applied shearing force is
 

$\delta_s = \dfrac{VL}{A_s G} = \dfrac{\tau L}{G}$

 

where V is the shearing force acting over an area As.
 

Solution to Problem 222 Poisson's Ratio

Problem 222
A solid cylinder of diameter d carries an axial load P. Show that its change in diameter is 4Pν / πEd.
 

Solution to Problem 224 Triaxial Deformation

Problem 224
For the block loaded triaxially as described in Prob. 223, find the uniformly distributed load that must be added in the x direction to produce no deformation in the z direction.
 

Solution to Problem 225 Biaxial Deformation

Problem 225
A welded steel cylindrical drum made of a 10-mm plate has an internal diameter of 1.20 m. Compute the change in diameter that would be caused by an internal pressure of 1.5 MPa. Assume that Poisson's ratio is 0.30 and E = 200 GPa.
 

Solution to Problem 226 Biaxial Deformation

Problem 226
A 2-in.-diameter steel tube with a wall thickness of 0.05 inch just fits in a rigid hole. Find the tangential stress if an axial compressive load of 3140 lb is applied. Assume ν = 0.30 and neglect the possibility of buckling.
 

Solution to Problem 227 Biaxial Deformation

Problem 227
A 150-mm-long bronze tube, closed at its ends, is 80 mm in diameter and has a wall thickness of 3 mm. It fits without clearance in an 80-mm hole in a rigid block. The tube is then subjected to an internal pressure of 4.00 MPa. Assuming ν = 1/3 and E = 83 GPa, determine the tangential stress in the tube.
 

Solution to Problem 228 Biaxial Deformation

Problem 228
A 6-in.-long bronze tube, with closed ends, is 3 in. in diameter with a wall thickness of 0.10 in. With no internal pressure, the tube just fits between two rigid end walls. Calculate the longitudinal and tangential stresses for an internal pressure of 6000 psi. Assume ν = 1/3 and E = 12 × 106 psi.
 

Solution 228
$\varepsilon = \dfrac{\sigma_x}{E} - \nu \dfrac{\sigma_y}{E} = 0$

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