October 2008

Solution to Problem 104 Normal Stress

Problem 104
A hollow steel tube with an inside diameter of 100 mm must carry a tensile load of 400 kN. Determine the outside diameter of the tube if the stress is limited to 120 MN/m2.
 

Simple Stresses

Simple Stresses

  1. Normal Stress
  2. Shear Stress
  3. Bearing Stress
  4. Thin-walled Pressure Vessel

Solution to Problem 105 | Normal Stress

Problem 105
A homogeneous 800 kg bar AB is supported at either end by a cable as shown in Fig. P-105. Calculate the smallest area of each cable if the stress is not to exceed 90 MPa in bronze and 120 MPa in steel.
 

Solution to Problem 106 Normal Stress

Problem 106
The homogeneous bar shown in Fig. P-106 is supported by a smooth pin at C and a cable that runs from A to B around the smooth peg at D. Find the stress in the cable if its diameter is 0.6 inch and the bar weighs 6000 lb.
 

106-beam-with-chords.jpg

 

Solution to Problem 107 Normal Stress

Problem 107
A rod is composed of an aluminum section rigidly attached between steel and bronze sections, as shown in Fig. P-107. Axial loads are applied at the positions indicated. If P = 3000 lb and the cross sectional area of the rod is 0.5 in2, determine the stress in each section.
 

107-composite-bar-two-forces_0.gif

 

Solution to Problem 108 Normal Stress

Problem 108
An aluminum rod is rigidly attached between a steel rod and a bronze rod as shown in Fig. P-108. Axial loads are applied at the positions indicated. Find the maximum value of P that will not exceed a stress in steel of 140 MPa, in aluminum of 90 MPa, or in bronze of 100 MPa.
 

Solution to Problem 109 Normal Stress

Problem 109
Determine the largest weight W that can be supported by two wires shown in Fig. P-109. The stress in either wire is not to exceed 30 ksi. The cross-sectional areas of wires AB and AC are 0.4 in2 and 0.5 in2, respectively.
 

109-weight-cable-system.gif

 

Solution to Problem 110 Normal Stress

Problem 110
110-footing-wooden-post.gifA 12-inches square steel bearing plate lies between an 8-inches diameter wooden post and a concrete footing as shown in Fig. P-110. Determine the maximum value of the load P if the stress in wood is limited to 1800 psi and that in concrete to 650 psi.
 

Solution to Problem 111 Normal Stress

Problem 111
For the truss shown in Fig. P-111, calculate the stresses in members CE, DE, and DF. The cross-sectional area of each member is 1.8 in2. Indicate tension (T) or compression (C).
 

Solution to Problem 112 Normal Stress

Problem 112
Determine the cross-sectional areas of members AG, BC, and CE for the truss shown in Fig. P-112. The stresses are not to exceed 20 ksi in tension and 14 ksi in compression. A reduced stress in compression is specified to reduce the danger of buckling.
 

112-inclined-truss.gif

 

Solution to Problem 113 Normal Stress

Problem 113
Find the stresses in members BC, BD, and CF for the truss shown in Fig. P-113. Indicate the tension or compression. The cross sectional area of each member is 1600 mm2.
 

Solution to Problem 114 Normal Stress

Problem 114
The homogeneous bar ABCD shown in Fig. P-114 is supported by a cable that runs from A to B around the smooth peg at E, a vertical cable at C, and a smooth inclined surface at D. Determine the mass of the heaviest bar that can be supported if the stress in each cable is limited to 100 MPa. The area of the cable AB is 250 mm2 and that of the cable at C is 300 mm2.
 

Strength of Materials

Reviewer in Strength of Materials

This page is the portal of the Reviewer in Strength of Materials. You can find here some basic theories and principles. Most of the content however for this online reviewer is solution to problems. You can find here a compiled step-by-step solution to problems in Strength of Materials. Feel free to explore the pages by selecting the topics tabulated below or browse it by chapters given as links below the tabulated data.
 

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.
 

Shear Stress

Forces parallel to the area resisting the force cause shearing stress. It differs to tensile and compressive stresses, which are caused by forces perpendicular to the area on which they act. Shearing stress is also known as tangential stress.
 

$\tau = \dfrac{V}{A}$

 

where V is the resultant shearing force which passes through the centroid of the area A being sheared.
 

Solution to Problem 115 Shear Stress

Problem 115
What force is required to punch a 20-mm-diameter hole in a plate that is 25 mm thick? The shear strength is 350 MN/m2.
 

Solution to Problem 116 Shear Stress

Problem 116
As in Fig. 1-11c, a hole is to be punched out of a plate having a shearing strength of 40 ksi. The compressive stress in the punch is limited to 50 ksi. (a) Compute the maximum thickness of plate in which a hole 2.5 inches in diameter can be punched. (b) If the plate is 0.25 inch thick, determine the diameter of the smallest hole that can be punched.
 

116-hole-puncher.gif

 

Solution to Problem 117 Shear Stress

Problem 117
Find the smallest diameter bolt that can be used in the clevis shown in Fig. 1-11b if P = 400 kN. The shearing strength of the bolt is 300 MPa.
 

117-clevis.gif

 

Solution to Problem 118 Shear Stress

Problem 118
A 200-mm-diameter pulley is prevented from rotating relative to 60-mm-diameter shaft by a 70-mm-long key, as shown in Fig. P-118. If a torque T = 2.2 kN·m is applied to the shaft, determine the width b if the allowable shearing stress in the key is 60 MPa.
 

Solution to Problem 119 Shear Stress

Problem 119
Compute the shearing stress in the pin at B for the member supported as shown in Fig. P-119. The pin diameter is 20 mm.
 

Solution to Problem 120 Shear Stress

Problem 120
The members of the structure in Fig. P-120 weigh 200 lb/ft. Determine the smallest diameter pin that can be used at A if the shearing stress is limited to 5000 psi. Assume single shear.
 

Solution to Problem 121 Shear Stress

Problem 121
Referring to Fig. P-121, compute the maximum force P that can be applied by the machine operator, if the shearing stress in the pin at B and the axial stress in the control rod at C are limited to 4000 psi and 5000 psi, respectively. The diameters are 0.25 inch for the pin, and 0.5 inch for the control rod. Assume single shear for the pin at B.
 

Solution to Problem 122 Shear Stress

Problem 122
Two blocks of wood, width w and thickness t, are glued together along the joint inclined at the angle θ as shown in Fig. P-122. Using the free-body diagram concept in Fig. 1-4a, show that the shearing stress on the glued joint is τ = P sin 2θ / 2A, where A is the cross-sectional area.
 

122-block-of-wood.gif

 

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.)
 

Pages