net Simple Stresses Simple stresses are expressed as the ratio of the applied force divided by the resisting area or σ = Force / Area. It is the expression of force per unit area to structural members that are subjected to external forces and/or induced forces. Stress is the lead to accurately describe and predict the elastic deformation of a body. Simple stress can be classified as normal stress, shear stress, and bearing stress.
Normal stress develops when a force is applied perpendicular to the cross-sectional area of the material. If the force is going to pull the material, the stress is said to be www.net tensile stress and compressive stress develops when the material is being compressed by two opposing forces. Shear stress is developed if the applied force is parallel to the resisting area. Example is the bolt that holds the tension rod in its anchor.
Another condition of shearing is when we twist a bar along its longitudinal axis. This type of shearing is called torsion and covered in Chapter 3. Another type of simple stress is the bearing stress, it is the contact pressure between two bodies. Suspension bridges are good example of structures that carry these stresses.
The weight of the vehicle is carried by the bridge deck and passes the force to the stringers (vertical cables), which in turn, supported by the main suspension cables. The suspension cables then transferred the force into bridge towers.net Normal Stress Stress Stress is the expression of force applied to a unit area of surface. It is measured in psi (English unit) or in MPa (SI unit). Another unit of stress which is not commonly used is the dynes (cgs unit).
Stress is the ratio of force over area. stress = force / area Simple Stresses There are three types of simple stress namely; normal stress, shearing stress, and bearing stress. Normal Stress www.net The resisting area is perpendicular to the applied force, thus normal. There are two types of normal stresses; tensile stress and compressive stress.
Tensile stress applied to bar tends the bar to elongate while compressive stress tend to shorten the bar. 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. SOLVED PROBLEMS IN 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.net Solution 104 Problem 105 www.net A homogeneous 800 kg bar AB is supported at either end by a cable as shown in Fig. Calculate the smallest area of each cable if the stress is not to exceed 90 MPa in bronze and 120 MPa in steel. Figure P-105 Solution 105 www.net 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.net Problem 107 A rod is composed of an aluminum section rigidly attached between steel and bronze sections, as shown in Fig. 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.net Problem 108 An aluminum rod is rigidly attached between a steel rod and a bronze rod as shown in Fig. 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. Figure P-108 Solution 108 www.net Problem 109 Determine the largest weight W that can be supported by two wires shown in Fig. The stress in either wire is not to exceed 30 ksi. The cross-sectional areas of wires AB and AC are 0.net Solution 109 www.net Problem 110 A 12-inches square steel bearing plate lies between an 8-inches diameter wooden post and a concrete footing as shown in Fig.
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.net Problem 111 For the truss shown in Fig. P-111, calculate the stresses in members CE, DE, and DF. The crosssectional area of each member is 1.net Problem 112 Determine the crosssectional areas of members AG, BC, and CE for the truss shown in Fig. 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.net Solution 112 www.net Problem 113 Find the stresses in members BC, BD, and CF for the truss shown in Fig. Indicate the tension or compression. The cross sectional area of each member is 1600 mm2.net Solution 113 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.net Solution 114 www.net Shearing 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.
where V is the resultant shearing force which passes which passes through the centroid of the area A being sheared.net SOLVED PROBLEMS IN SHEARING 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.net 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.net Problem 117 Find the smallest diameter bolt that can be used in the clevis shown in Fig. The shearing strength of the bolt is 300 MPa. Solution 117 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.2 kN·m is applied to the shaft, determine the width b if the allowable shearing stress in the key is 60 MPa.net Solution 118 www.net Problem 119 Compute the shearing stress in the pin at B for the member supported as shown in Fig. The pin diameter is 20 mm.net 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.net 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.net Solution 121 www.net 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. 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.net Solution 122 Problem 123 A rectangular piece of wood, 50 mm by 100 mm in cross section, is used as a compression block shown in Fig.
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 5 MN/m2. The grain makes an angle of 20° with the horizontal, as shown. (Hint: Use the results in Problem 122.net Solution 123 www.net Bearing Stress Bearing stress is the contact pressure between the separate bodies. It differs from compressive stress, as it is an internal stress caused by compressive forces.net SOLVED PROBLEMS IN BEARING STRESS Problem 125 In Fig.
1-12, assume that a 20-mm-diameter rivet joins the plates that are each 110 mm wide. The allowable stresses are 120 MPa for bearing in the plate material and 60 MPa for shearing of rivet. Determine (a) the minimum thickness of each plate; and (b) the largest average tensile stress in the plates.net Problem 126 The lap joint shown in Fig. P-126 is fastened by four ¾-in.
Calculate the maximum safe load P that can be applied if the shearing stress in the rivets is limited to 14 ksi and the bearing stress in the plates is limited to 18 ksi. Assume the applied load is uniformly distributed among the four rivets.net Problem 127 In the clevis shown in Fig. 1-11b, find the minimum bolt diameter and the minimum thickness of each yoke that will support a load P = 14 kips without exceeding a shearing stress of 12 ksi and a bearing stress of 20 ksi.net Solution 127 Problem 128 A W18 × 86 beam is riveted to a W24 × 117 girder by a connection similar to that in Fig. The diameter of the rivets is 7/8 in., and the angles are each 4 × 31/2 × 3/8 in.
For each rivet, assume that the allowable stresses are τ = 15 ksi and σb = 32 ksi.net Find the allowable load on the connection.net Solution 128 Note: Textbook is Strength of Materials 4th edition by Pytel and Singer www.net Problem 129 A 7/8-in.-diameter bolt, having a diameter at the root of the threads of 0., is used to fasten two timbers together as shown in Fig. The nut is tightened to cause a tensile stress of 18 ksi in the bolt. Compute the shearing stress in the head of the bolt and in the threads. Also, determine the outside diameter of the washers if their inside diameter is 9/8 in.
and the bearing stress is limited to 800 psi.net Solution 129 www.net Problem 130 Figure P-130 shows a roof truss and the detail of the riveted connection at joint B. Using allowable stresses of τ = 70 MPa and σb= 140 MPa, how many 19-mm diameter rivets are required to fasten member BC to the gusset plate? Member BE? What is the largest average tensile or compressive stress in BC and BE? www.net Solution 130 www.net Problem 131 Repeat Problem 130 if the rivet diameter is 22 mm and all other data remain unchanged.net Thin-Walled Pressure Vessels A tank or pipe carrying a fluid or gas under a pressure is subjected to tensile forces, which resist bursting, developed across longitudinal and transverse sections. TANGENTIAL STRESS (Circumferential Stress) Consider the tank shown being subjected to an internal pressure p. The length of the tank is L and the wall thickness is t.
Isolating the right half of the tank: www.net If there exist an external pressure po and an internal pressure pi, the formula may be expressed as: LONGITUDINAL STRESS, σL Consider the free body diagram in the transverse section of the tank: The total force acting at the rear of the tank F must equal to the total longitudinal stress on the wall PT = σL Awall.
