compressive stress

Solution to Problem 270 Thermal Stress

270 Bronze sleeve slipped over a steel bolt

Problem 270

A bronze sleeve is slipped over a steel bolt and held in place by a nut that is turned to produce an initial stress of 2000 psi in the bronze. For the steel bolt, A = 0.75 in2, E = 29 × 106 psi, and α = 6.5 × 10–6 in/(in·°F). For the bronze sleeve, A = 1.5 in2, E = 12 × 106 psi and α = 10.5 × 10–6 in/(in·°F). After a temperature rise of 100°F, find the final stress in each material.

Solution to Problem 265 Thermal Stress

265 Bronze bar between rigid walls

Problem 265

A bronze bar 3 m long with a cross sectional area of 320 mm2 is placed between two rigid walls as shown in Fig. P-265. At a temperature of −20°C, the gap Δ = 25 mm. Find the temperature at which the compressive stress in the bar will be 35 MPa. Use α = 18.0 × 10−6 m/(m·°C) and E = 80 GPa.

Solution to Problem 250 Statically Indeterminate

250 Bolt and tube assembly

Problem 250

In the assembly of the bronze tube and steel bolt shown in Fig. P-250, the pitch of the bolt thread is p = 1/32 in.; the cross-sectional area of the bronze tube is 1.5 in.2 and of steel bolt is 3/4 in.2 The nut is turned until there is a compressive stress of 4000 psi in the bronze tube. Find the stresses if the nut is given one additional turn. How many turns of the nut will reduce these stresses to zero? Use Ebr = 12 × 106 psi and Est = 29 × 106 psi.

Solution to Problem 233 Statically Indeterminate

233 Combined steel and cast iron bar

Problem 233

A steel bar 50 mm in diameter and 2 m long is surrounded by a shell of a cast iron 5 mm thick. Compute the load that will compress the combined bar a total of 0.8 mm in the length of 2 m. For steel, E = 200 GPa, and for cast iron, E = 100 GPa.

Solution to Problem 131 Bearing Stress

Problem 131
Repeat Problem 130 if the rivet diameter is 22 mm and all other data remain unchanged.

Solution to Problem 130 Bearing Stress

130 Trusss Joints

Strength of Materials 4th Edition by Pytel and Singer
Problem 130 page 22

Given:
Allowable shear stress = 70 MPa
Allowable bearing stress = 140 MPa
Diameter of rivets = 19 mm
The truss below:
Figure P-130 and P-131

Required:
Number of rivets to fasten member BC to the gusset plate
Number of rivets to fasten member BE to the gusset plate
Largest average tensile or compressive stress in members BC and BE

Solution to Problem 123 Shear Stress

123 Forces parallel and perpendicular to wood grain

Strength of Materials 4th Edition by Pytel and Singer
Problem 123 page 18

Given:
Cross-section of wood = 50 mm by 100 mm
Maximum allowable compressive stress in wood = 20 MN/m2
Maximum allowable shear stress parallel to the grain in wood = 5 MN/m2
Inclination of the grain from the horizontal = 20 degree

Required: The axial force P that can be safely applied to the block

Solution to Problem 116 Shear Stress

Strength of Materials 4th Edition by Pytel and Singer
Problem 116 page 16

Given:
Shear strength of plate = 40 ksi
Allowable compressive stress of punch = 50 ksi
The figure below:
Figure 1-11c

Required:

  1. Maximum thickness of plate to punch a 2.5 inches diameter hole
  2. Diameter of smallest hole if the plate is 0.25 inch thick

Solution to Problem 113 Normal Stress

113 Section of a truss

Strength of Materials 4th Edition by Pytel and Singer
Problem 113 page 15

Given:
Cross sectional area of each member = 1600 mm2.

Required: Stresses in members BC, BD, and CF

Solution to Problem 112 Normal Stress

112 Free Body Diagram of Truss

Strength of Materials 4th Edition by Pytel and Singer
Problem 112 page 14

Given:
Maximum allowable stress in tension = 20 ksi
Maximum allowable stress in compression = 14 ksi

Required: Cross-sectional areas of members AG, BC, and CE

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