2 handouts-lecture-7

14

Click here to load reader

description

STATICS

Transcript of 2 handouts-lecture-7

Page 1: 2  handouts-lecture-7

COURSE:COURSE: CECE 201201 (STATICS)(STATICS)

LECTURELECTURE NO.:NO.: 0707

FACULTY:FACULTY: DR.DR. SHAMSHADSHAMSHAD

AHMADAHMAD DEPARTMENT:DEPARTMENT: CIVILCIVIL

ENGINEERINGENGINEERING

UNIVERSITY:UNIVERSITY: KINGKING FAHDFAHD UNIVERSITYUNIVERSITY OFOF PETROLEUMPETROLEUM

&& MINERALS,MINERALS, DHAHRAN,DHAHRAN, SAUDISAUDI ARABIAARABIA

Page 2: 2  handouts-lecture-7

TEXTTEXT BOOK:BOOK: ENGINEERINGENGINEERING MECHANICMECHANIC S-S- STATICSSTATICSbyby R.C.R.C. HIBBELER,HIBBELER, PRENTICEPRENTICE HALLHALL

Page 3: 2  handouts-lecture-7

LECTURELECTURE NO.NO. 77

EQUILIBRIUMEQUILIBRIUM OFOF AA PARTICLEPARTICLE ANDAND FREFREEE -BODY-BODY DIAGRAMDIAGRAM

Objectives:Objectives:

►► ToTo definedefine thethe equil ibriumequil ibrium ofof aa particleparticle

►► ToTo explainexplain thethe condit ioncondit ion forfor equil ibriumequil ibrium ofof aa particleparticle

►► To explainexplain thethe conceptconcept ofof free-bodyfree-body diagramdiagram forfor

aa particle

Page 4: 2  handouts-lecture-7

DEFINITIONDEFINITION OFOF EQUILIBRIUMEQUILIBRIUM OFOF AA PARTICLEPARTICLE

A particle is said to be in equilibrium:1. provided it is at rest if originally at rest, or2. particle has constant velocity if originally in

motion

Page 5: 2  handouts-lecture-7

CONDITIONSCONDITIONS FORFOR EQUILIBRIUMEQUILIBRIUM OFOF AA PARTICLEPARTICLE

According to the Newton’s First Law, whichstates that “a particle originally at rest will remain in rest condition if resultant force acting on the particle is zero”, the condition for equilibrium can be obtained by summing up all the forces algebraically and equating the sum tozero, as follows: ΣF = 0

Page 6: 2  handouts-lecture-7

CONNECTIONSCONNECTIONS OFTENOFTEN ENCOUNTEREDENCOUNTERED ININ PARTICLEPARTICLE EQUILIBRIUMEQUILIBRIUM PROBLEMSPROBLEMS

(i) Springs

When a force (F) is applied to a linear elastic spring having stiffness k (also called spring constant), there will be change in the length of spring (s) in direct proportion to the applied force, as follows: F = ks

For example, the force required to stretch or compress the spring, as shown in the following figure, is given as: F = ks = 500 × 0.2 = 100 N

Page 7: 2  handouts-lecture-7

CONNECTIONSCONNECTIONS OFTENOFTEN ENCOUNTEREDENCOUNTERED ININ PARTICLEPARTICLE EQUILIBRIUMEQUILIBRIUM PROBLEMSPROBLEMS

(ii) Cables and Pulleys

Let us consider a cable subjected to tension T and passing over a pulley, as shown in the following figure:For the cable and pulley system, as shown in the above figure, the following points may be noted:

• Cable is assumed to have negligible weight and they cannot stretch

• A cable may support only a tensile force, T, and this force always acts in the direction of the cable

• For any θ value, the cable is subjected to aconstant tensile force T throughout its length.

Page 8: 2  handouts-lecture-7

PROCEDUREPROCEDURE FORFOR DRAWINGDRAWING AA FREFRE EE -BODY-BODY

DIAGRAMDIAGRAM

Free-body diagram of a given system is required to apply the equil ibrium condit ions for analysis of the system.

Following steps are adopted for drawing free-body

diagrams:

• Draw outl ined shape of the given system

• Isolate or cut “free” the particle from its surroundings and the sketch of the particle

Show all known and unknown forces

• Indicat e

all the forces acting on the particle, on the sketch

each force

Page 9: 2  handouts-lecture-7

• Label the magnitudes and the directions of all the known and unknown forces acting on the particle

Page 10: 2  handouts-lecture-7

PROBLEMPROBLEM SOLVING:SOLVING: ExampleExample ## 11

Page 11: 2  handouts-lecture-7

PROBLEMPROBLEM SOLVING:SOLVING: ExampleExample ## 22

Page 12: 2  handouts-lecture-7

PROBLEMPROBLEM SOLVING:SOLVING: ExampleExample ## 33

Page 13: 2  handouts-lecture-7

Multiple Choice Problems

1. A 0.7 m long spring having stiffness k = 400 N/m is subjected to tensile force of 200 N. The length of the spring after applying load will be

(a) 0.2 m (b) 0.5 m(c) 0.7 m (d) 1.2 m

Ans: (d) Feedback:

For a spring, we have:F = ks

⇒ s = F

k = 200

400= 0.5 m

The length of the spring after applying load= 0.7 + 0.5 = 1.2 m

Page 14: 2  handouts-lecture-7

Multiple Choice Problems2. If angle θ is 30° for the cable passing over a pulley,

as shown in the figure below, the tension T in the cable will be

(a) 0 (b) 600 N (c) 693 N (d) 1200N

Ans: (b)

Feedback:

Since the same cable is passing over a pulley, the tension at any location of the cable will remain same irrespective of the orientation of the cable.