EJERCICIOS RESUELTO ESTATICA

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 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.

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

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.

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 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.

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 cross sectional 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-108

Solution 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. 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 110

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 1.8 in2. Indicate tension (T) or compression (C).

Solution 111

Problem 112

Determine the crosssectional areas of members AG, BC, and CE for the truss shown in Fig. P-112 above. 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.

Solution 112

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 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.

Solution 114

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.

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.

Solution 115

Problem 116

As in Fig. 1-11c, a hole is to be punched

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