lecture note for solid mechanicscontents.kocw.net/document/solid mechanics - essential... · 2014....

Advanced Materials & Smart Structures Lab.금오공대기계공학과 윤성호교수

Lecture Note for Solid Mechanics- Essential Concept and Statics Review -

Prof. Sung Ho YoonDepartment of Mechanical EngineeringKumoh National Institute of Technology


Text book : Mechanics of Materials, 6th ed.,

W.F. Riley, L.D. Sturges, and D.H. Morris, 2007.

Prerequisite : Knowledge of Statics, Basic Physics, Mathematics, etc.


Objective of mechanics of materials

Relationships between loads applied to a nonrigid body and internal forces and deformations induced in the body

- Determination of load carrying capability of structural members

- Development of mathematical and experimental methods of analysis

Subject matter of this course

Given a certain function to perform, of what materials should be constructed, and what sizes and proportions of various elements? (Designer’s task)

Given the completed design, does it perform the function economically and without excessive deformation? (Checkers’ problem)

Given a completed structure, what is its actual load carrying capability? (Rating problem)

Essential Concepts and Statics Review


Classification of forces

Forces : action of one body on another(equal magnitude, opposite direction pairs)

Direct and indirect contact between two bodies Surface forces (direct contact)

- Concentrated force and distributed force (uniform or non-uniform)

Body forces (indirect contact)- Gravitational attraction force and electromagnetic force

unit : force/length (N/m)



Applied forces and reactions Applied forces

- Concentrated and/or distributed forces designed to carry Reactions

- Forces at supports to prevent movements

External forces and internal forces

Static forces and dynamic forces Static forces

- Forces which do not vary with time Dynamic forces

- Forces which vary with time such as impact load and repeated load



Equilibrium of a rigid body Rigid body

- Body that does not deform under the action of applied forces

Equilibrium condition- Resultant of the system of forces acting on the body is zero- ):( bodytheofforonoMandF o 00

00 oMandF

000

000

zyx

zyx

MMM

FFF

000 zyx MFF

00 yx FF

Free body diagram Free body : Body of interest separated from all other interacting bodies

Free body diagram : Free body with all external forces Use for determining the unknown forces acting on rigid bodies

(3-Dim)

(2-Dim)

(2-Dim)



(Example) 900kg mass is supported by a roller that can move along the beam. Determine (a) reactions at A and B neglecting the mass of beam and (b) reactions at A and B if the mass of the beam is 8.5kg/m.

(Sol)

000 zyx MFF(a) Using

03882980

088290

00

)()(BM

BAF

AF

A

yy

xx

NBNAA yx 331055200

000 zyx MFF(b) Using

04398383882980

03983888290

00

)().()()(

).(

BM

BAF

AF

A

yy

xx

NBNAA yx 364058500

(83.39 N/m x 8m)



(Example) Calculate the support reactions and the forces in the BD, DE, and CE. Assume that the truss supports one side of bridge and floor beams carry vehicle loads (W=3400 lb) to the truss joints.

(Sol)

00 MandF

Since the truss is a two-force member, it is subjected to either tension or compression.

lbElbAAEM

EAF

AF

yx

A

yy

xx

425127500)32(850)64(0

08508500

00

Half of the car’s weight (W=1700 lb) is carried by the truss and the other half is carried by the truss on the other side of the bridge.

Using



Free body diagram of pin E

lbTlbT CEDE 7.5048.659

0sin425sin0

0cos0

DEDEy

DECEx

TTEF

TTF

Since = tan -1 (16/19) = 40.10 O

Summing moments about point C

)(850850

0)32(425)16(0

ClblbT

TM

BD

BDC

26 ft



(Example) The weight of books on a shelf bracket is 75 lb. Determine all forces on all three members of this frame.

(Sol) 00 MandF

Using

lbClbAlbA

AF

CAF

CM

yx

yy

xx

A

0.1200.750.120

0750

00

0)75(850

lbBlbT

BM

TM

AC

A

ACB

0.1300.50

0)75(8135120

0)75(1212)75(40

Dismember the bracket and draw free body diagrams of each member



(Example) Determine reactions at support A neglecting the weights of the pipe and the wrench.

(Sol)

00 MandF(a) Using

00

01802000

0230035002000

02000

00

00

zz

yy

xx

zz

yy

xx

MM

MM

MM

AF

AF

AF

).(

)..(

NANANA zyx 20000

mNMmNMmNM zyx 00360116 ..



(Example) Determine the forces at A and B on rigid structure AB subjected to a 5000-lb force.

(Sol)

00 MandF

Using

0)4(5000)4)(45sin(0

0500045sin0

045cos0

TM

TAF

TAF

z

yy

xx

lbTlblbAlbA yx 7071004795.05000



Equilibrium of a deformable body

Interaction between loads acting on a body, deformations of the body, and the geometry of the free body diagram makes the solution of the deformable body problems more complex

Solution requires either a trial-and-error solution or a numerical solution or an iterative solution method

(Example) 5000 lb weight is to be supported by AB (rigid bar) and BC (nonrigid wire). Using TBC=k and =Lf-Li, determine the tension in the wire after the load is applied for k=5000 lb/in.



0cos4)2/45sin(40)cos4()sin4()2/45cos(

)cos4()2/45sin(0

0)2/45sin(0

0)2/45cos(0

WTWT

TM

WTAF

TAF

A

yy

xx

Using the trigonometric relation, BABABA sincoscossin)sin(

Force-deformation relation is introduced

inLL

inL

L

inftL

if

f

f

i

118867

886718867

18867

90484824848

8867657544222

22

sin.

.sin.

sin.

)cos(

..

Thus, cos)/sin(sin. Wk 245118867

If k is 5000 lb/in, =2.465deg, T=7221 lb, Ax= 4995 lb, Ay = -215 lb

(Example) Determine the tension in the wire if k=5000 lb/in.

(Sol)



Internal forces

- Forces that exist throughout the interior of a body

C : centroidR : resultant forceC : resultant couple

Internal forces : R & C

0:0

0:0

21

21

CMMM

RFFF

C

where M1 and M2 are the moments of F1 and F2 about the centroid.

We find the resultant of the internal force system using the equilibrium equations. However, we can not find the exact distribution of the internal forces without information of the deformation of the body.



000

000

zyx

zyx

MMM

FFF

Resolve the resultants R and C into components along and perpendicular to the section

P : Normal forceV : Shear force

T : Twisting momentM : Bending moment

Components of the internal force system can be found using the equilibrium equations



(Example) Determine (a) the support reactions and (b) the internal forces on a section 4m to the right of the support A.

(Sol)

00 MandF

(a) Using

03552100

05100

00

)().(AA

yy

xx

MM

AF

AF

mkNMkNAkNA Ayx 40150

(b)

0)1(5)2(8)4(15400

058150

00

zz

yy

x

MM

VF

PF

mkNMkNVkNP zy 120



(Homework)

(1-8), (1-9), (1-22), (1-34), (1-56), (1-61)

lecture note for solid mechanicscontents.kocw.net/document/solid mechanics - essential... · 2014....

Documents