1.1 - Introduction - Simple Stresses

MEC 103 MECHANICS OF DEFORMABLE BODIES

INTRODUCTION ENGR. ROGELIO FRETTEN C. DELA CRUZ, CE INSTRUCTOR

ENGINEERING MECHANICS: 1. Statics 2. Dynamics We considered only the external effect of forces acting on a body. The bodies are assumed perfectly rigid (no deformation).

STRENGTH OF MATERIALS: Internal effects of the forces on the body will be considered. Deformations will be of great importance.

The difference between rigid-body mechanics and mechanics of materials can be appreciated if we consider the bar shown in Fig. 1.1.

In mechanics of materials, the statics solution is extended to include analysis of the forces acting inside the bar to be certain that the bar will neither break nor deform excessively

ANALYSIS OF INTERNAL FORCES

It is convenient to represent both R and CR in terms of two components: one perpendicular to the cross-section and the other lying in the cross-section. These components are given physically meaningful names. P - the component of the resultant force that is perpendicular to the cross-section, tending to elongate or shorten the bar. It is called the normal force or axial force.

V - the component of the resultant force lying in the plane of the cross-section, tending to shear (slide) one segment of the bar relative to the other segment. It is called the shear force.

T - the component of the resultant couple that tends to twist (rotate) the bar. It is called the twisting moment or torque.

M - the component of the resultant couple that tends to bend the bar. It is called the bending moment.

SIMPLE STRESSES Stress is known as the intensity of load per unit area. Stress is also a measure of the unit strength of a material.

SIMPLE STRESSES Three types of simple stress: 1. Normal Stress 2. Shearing Stress 3. Bearing Stress.

NORMAL STRESS The resisting area is perpendicular to the applied force, thus normal.

NORMAL STRESS Two types of normal stress: 1. Tensile stress 2. Compressive stress

NORMAL STRESS The normal stress acting at any point on a cross-section is given by the formula:

Where: σ = Normal Stress P = Axial force A = Cross-sectional Area

NORMAL STRESS The normal stress acting at any point on a cross-section is given by the formula:

Units of stress: N N 1 2  1 Pa ; 1  1 MPa 2 m mm lb kip psi  2 ; ksi  2 in in

Acceleration due to influence of gravity: g = 32.2 ft/sec2 g = 9.81 m/sec2

For water at 4°C = 62.4 lb/ft3 = 9.81 kN/m3 = 1000kg/m3

Illustrative Problems

The compound bar ABCD consists of three segments, each of a different material with different dimensions. Compute the stress in each segment when the axial loads are applied.

Determine the largest weight W that can be supported by the two wires AB and AC. The working stresses are 100 MPa for AB and 150 MPa for AC. The cross-sectional areas of AB and AC are 400 mm2 and 200 mm2, respectively.

The 1000-kg uniform bar AB is suspended from two cables AC and BD each with cross-sectional area 400 mm2. Find the magnitude P and location x of the largest additional vertical force that can be applied to the bar. The stresses in AC and BD are limited to 100 MPa and 50 MPa, respectively.

Determine the largest weight W that can be supported safely by the structure shown in the figure. The working stresses are 16,000 psi for the steel cable AB and 720 psi for the wood strut BC. Neglect the weight of the structure.

The homogeneous 120-N sign is suspended from a ball and socket joint at O and cables AD and BC. Determine the tensile stresses in the cables if each cable has a cross-sectional area of 10 mm2.

The wood pole is supported by two cables of 1/4-in. diameter. The turn buckles in the cables are tightened until the stress in the cables reaches 60,000 psi. If the working compressive stress for wood is 200 psi, determine the smallest permissible diameter of the pole.

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