Solution to Problem 270 Thermal Stress

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 250 Statically Indeterminate

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.
 

Figure P-250

 

Solution to Problem 123 Shear Stress

Problem 123
A rectangular piece of wood, 50 mm by 100 mm in cross section, is used as a compression block shown in Fig. P-123. Determine the axial force P that can be safely applied to the block if the compressive stress in wood is limited to 20 MN/m2 and the shearing stress parallel to the grain is limited to 5MN/m2. The grain makes an angle of 20° with the horizontal, as shown. (Hint: Use the results in Problem 122.)
 

Normal Stresses

Stress is defined as the strength of a material per unit area or unit strength. It is the force on a member divided by area, which carries the force, formerly express in psi, now in N/mm2 or MPa.
 

$\sigma = \dfrac{P}{A}$

 

where P is the applied normal load in Newton and A is the area in mm2. The maximum stress in tension or compression occurs over a section normal to the load.