Problems strenght materials by f. singer

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 normalstress, 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 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 resistingarea. 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 weightof 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.
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 ofstress 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
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 400kN. Determine the outside diameter of the tube if the stress is limited to 120 MN/m2.
Solution 104
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.
Figure P-105
Solution 105
Problem 106 The homogeneous bar shown in Fig. P-106 issupported 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.
Solution 106
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 crosssectional area of the rod is 0.5 in2, determine the stress in each section.
Solution 107
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.
Figure P-108Solution 108
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.
Solution 109
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....