1.1 - Introduction - Simple Stresses

7/24/2019 1.1 - Introduction - Simple Stresses

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MEC 103MECHANICS OF DEFORMABLE BODIES

INTRODUCTION

ENGR. ROGELIO FRETTEN C. DELA CRUZ, CEINSTRUCTOR


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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 beconsidered.

Deformations will be of great importance.


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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


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ANALYSIS OF INTERNAL FORCES


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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 elongateor shorten the bar. It is called the

normal force or axial force.


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V - the component of the

resultant force lying in theplane of the cross-section,

tending to shear (slide) one

segment of the bar relative to

the other segment. It is calledthe shear force.


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T - the component of theresultant couple that tends

to twist (rotate) the bar. It is

called the twistingmoment

ortorque.


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M - the component of theresultant couple that

tends to bend the bar. It

is called the bending

moment.


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SIMPLE STRESSES

Stress is known as the intensity ofload per unit area.

Stress is also a measure of the unit

strength of a material.


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Three types of simple stress:

1. Normal Stress

2. Shear ing Stress3. Bearing Stress.

SIMPLE STRESSES


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NORMAL STRESS

The resisting area is perpendicular

to the applied force, thus normal.


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NORMAL STRESS

Two types of normal stress:

1. Tensile stress2. Compressive stress


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NORMAL STRESS

The normal stress acting at any point

on a cross-section is given by theformula:

Where:

= Normal Stress

P = Axial force

A = Cross-sectional Area


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NORMAL STRESS

The normal stress acting at any point

on a cross-section is given by theformula:

Units of stress:

22

22

in

kipksi;

in

lbpsi

MPa1mm

N1;Pa1

m

N1


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For water at 4C

= 62.4 lb/ft3

= 9.81 kN/m3

= 1000kg/m3

Acceleration due to influence of gravity:

g= 32.2 ft/sec2

g = 9.81 m/sec2


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Illustrative Problems


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The compound bar ABCD consis ts of three segments,

each of a dif ferent mater ia l w ith dif ferent d imension s.

Compu te the stress in each segment when the axia lloads are app lied.


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Determine the largest weight W that can be

supported by the two wires AB and AC. The

working stresses are 100 MPa forABand 150MPa for AC. The cross-sectional areas of AB

and AC are 400 mm2 and 200 mm2,

respectively.


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The 1000-kg uniform barABis suspended from two cables

ACand BDeach 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.


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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 thesteel cable AB and 720 psi for the wood strut BC.

Neglect the weight of the structure.


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The homogeneous 120-N sign is suspended from

a ball and socket joint at Oand cablesADand

BC

. Determine the tensile stresses in the cables ifeach cable has a cross-sectional area of 10 mm2.


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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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END

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1.1 - Introduction - Simple Stresses