1

Mechanics of Solids I Mechanics of Solids I

Introduction & Concept of stress

MechanicsMechanics

Concerned with the state of rest or motion of bodies subjected to the action of forcesbodies subjected to the action of forces

Rigid-Body Mechanics

Statics

- Equilibrium of bodies

Mechanics Deformable-Body Mechanics

Fluid Mechanics

Dynamics

- Accelerated motion

2

MechanicsMechanicsThe main objectives:

Provide the future engineer with the means of Provide the future engineer with the means of analyzing and designing various machines and load bearing structures

Involve the determination of stresses and deformations

Mechanics of SolidsMechanics of SolidsBased upon understanding of the equilibrium of rigid bodies under the action of forces.

Concerned with the relationships between external loads (forces and moments) and internal loadsInternal loads and stressand strainStrain and deformations or displacement

3

Review of Static EquilibriumReview of Static Equilibrium

Types of loads: External loads and Internal loadsE t l l d d t f f d b d fExternal loads are due to surface forces and body forcesInternal loads can be considered as forces of interaction between the constituent material particles of the body

The equations of static equlibrium require:∑F 0 ∑F 0 d ∑F 0

External loads are balanced by internal forces

∑Fx = 0; ∑Fy = 0, and ∑Fz = 0∑Mx = 0; ∑My = 0, and ∑Mz = 0For 2D, ∑Fx = 0; ∑Fy = 0, and ∑MA = 0

Free-Body Diagram (FBD)

Support reactions - 2D problems

ReviewReview

4

ExampleExample

The structure is designed to support a 30 kN loadsupport a 30 kN loadThe structure consists of a boom and rod joined by pins (zero moment connections) at the junctions and supportsPerform a static analysis to determine the internal force in each structural member and the reaction forces at the supportsreaction forces at the supports

Structure FreeStructure Free--Body DiagramBody Diagram

Structure is detached from supports and the loads and reaction forces are indicatedand reaction forces are indicated

AAx =Cx =

5

Component FreeComponent Free--Body DiagramBody Diagram

Ay =Cy =

Method of Joints Method of Joints –– internal forceinternal force

The boom and rod are two-force members i e the members are members, i.e., the members are subjected to only two forces which are applied at member ends.

For equilibrium, the forces must be parallel to to an axis between the force application points, equal in magnitude, and in opposite directions.

FAB =FBC =

6

Stress AnalysisStress AnalysisCan the structure safely support the 30-kN load?

MPa159m10314N1050

26-

3=

×

×==

AP

BCσ

At any section through member BC, the internal force is 50 kN with a force intensity or stressstress of

dBC = 20 mm

• Conclusion: the strength of member BC is adequate.

MPa165all =σ

From the material properties for steel, the allowable stress is

DesignDesign• Selection of appropriate materials and component

dimensions to meet performance requirementsdimensions to meet performance requirements.

• To construct the rod from aluminum (σall = 100 MPa). What is an appropriate choice for the rod diameter?

m10500Pa10100N1050

2

266

3×=

×

×=== −

π

σσ

dA

PAAP

allall

( ) mm2.25m1052.2m1050044

4

226

=×=×

==

=

−−

ππ

π

Ad

A

• An aluminum rod 26 mm or more in diameter is adequate.

7

Basic Principles of AnalysisBasic Principles of Analysis

Equilibrium ConditionsTh ti f t ti ilib i f f t b The equations of static equilibrium of forces must be satisfied throughout the member

Material BehaviorThe stress–strain or force–deformation relations (Hooke’s law) must apply to the behavior of the material of which the member is made of

Geometry of DeformationThe conditions of compatibility of deformations must be satisfied: that is, each deformed portion of the member must fit together with adjacent portions

Boundary conditions are used in the analysis

o The resultant of the internal forces for an axially loaded member is normal to a section cut perpendicular to the

Axial Loading: Normal stressAxial Loading: Normal stress

member is normal to a section cut perpendicular to the member axis.

Stress, σ = load /force intensity on that section = load per unit area

0lim aveA

F PA A

σ σΔ →

Δ= =

Δ

8

o A uniform distribution of stress in a section infers

Centric & Eccentric loadingCentric & Eccentric loading

that the line of action for the resultant of the internal forces passes through the centroid of the section = centric loading

o If a two-force member is eccentrically loaded, then the resultant of the stress distribution in a section must yield an axial force and a moment.

Shearing StressShearing Stress

o Forces P and P’ are applied transverselyo o ces a d a e app ed a s e se yto the member AB.

AP

=aveτo The average shear stress is,

o Corresponding internal forces act in the plane of section C and are called shearing forces.

o The shear stress distribution cannot be assumed to be uniform.

o Shear stress distribution varies from zero at the member surfaces to maximum values that may be much larger than the average value.

9

ShearingShearing StressStress ExamplesExamplesSingle Shear Double Shear

aveP FA A

τ = = ave 2P FA A

τ = =

Bearing Stress in ConnectionsBearing Stress in Connectionso Bolts, rivets, and pins create

stresses on the points of contact or stresses on the points of contact or bearing surfaces of the members they connect.

o The resultant of the force distribution on the surface is equal and opposite to the force exerted on the pin.

bP PA td

σ = =

o Corresponding average force intensity is called the bearing stress,

10

• Would like to determine the stresses in the members

Stress AnalysisStress Analysis & Design Example& Design Example

stresses in the members and connections of the structure shown.

• From a statics analysis:FAB = 40 kN (compression) FBC = 50 kN (tension)

• Must consider maximum normal stresses in AB and BC, and the shearing stress and bearing stress at each pinned connection

Rod & Boom Normal StressesRod & Boom Normal Stresses

• At the rod center, the average normal stress σBC = +159 MPa

( )( )

MPa167m10300

1050

m10300mm25mm40mm20

26

3,

26

=×

×==

×=−=

−

−

NAP

A

endBCσ

• At the flattened rod ends, the smallest cross-sectional area occurs at the pin centerline,

+159 MPa.

• The boom is in compression with an axial force of 40 kN and average normal stress of –26.7 MPa.

• The minimum area sections at the boom ends are unstressed since the boom is in compression.

11

Pin Shearing StressesPin Shearing Stresses

• The force on the pin at C is equal to

MPa102m10491N1050

26

3, =

×

×== −A

PaveCτ

the force exerted by the rod BC,

• The pin at A is in double shear with a total force equal to the force exerted by the boom AB,

MPa7.40m10491

kN2026, =

×== −A

PaveAτ

Pin Bearing StressesPin Bearing Stresses

• To determine the bearing stress at A in the boom AB, we have t = 30 mm and d = 25 mm,

( ) ( )σ = = =

40kN 53.3MPa30mm 25mmb

Ptd

• To determine the bearing stress at A in the bracket, we have t = 2(25 mm) = 50 mm and d( )= 25 mm,

( ) ( )σ = = =

40kN 32.0MPa50mm 25mmb

Ptd

12

Stress in TwoStress in Two--force membersforce memberso Axial forces on a two force member

result in only normal stresses on a plane result in only normal stresses on a plane cut perpendicular to the member axis.

o Will show that either axial or transverse forces may produce both normal and shear stresses with respect to a plane other than one cut perpendicular to the member axis.

o Transverse forces on bolts and pins result in only shear stresses on the plane perpendicular to bolt or pin axis.

Stress on an Oblique PlaneStress on an Oblique Planeo Pass a section through the member forming

an angle q with the normal plane.g q p

cos sinF P V Pθ θ= =

o Resolve P into components normal and tangential to the oblique section,

o From equilibrium conditions, the distributed forces (stresses) on the plane must be equivalent to the force P.

2

0 0

0 0

coscos

cossin

sin cos

cos

F P PAA A

V P PAA A

θ

θ

θσ θθ

θτ θ θθ

= = =

= = =

o The average normal and shear stresses on the oblique plane are

13

Stress under General LoadingsStress under General Loadingso A member subjected to a general combination of

loads is cut into two segments by a plane passing g y p p gthrough Q

0lim

lim lim

x

xA

x xy z

FA

V V

σ

τ τ

Δ →

Δ=

Δ

Δ Δ= =

o The distribution of internal stress components may be defined as,

o For equilibrium, an equal and opposite internal force and stress distribution must be exerted on the other segment of the member.

0 0lim limxy xzA AA A

τ τΔ → Δ →

= =Δ Δ

State of StressState of Stress

o Stress components are defined for the planes cut parallel to the x, y and z axes.

o For equilibrium, equal and opposite stresses are exerted on the hidden planes.

14

State of StressState of Stresso The combination of forces generated by the

stresses must satisfy the conditions for equilibrium:stresses must satisfy the conditions for equilibrium:0

0x y z

x y z

F F F

M M M

= = =∑ ∑ ∑

= = =∑ ∑ ∑

( ) ( )0z xy yx

xy yx

M A a A aτ τ

τ τ

= = Δ − Δ∑

=

i il l d

o Consider the moments about the z axis:

o It follows that only 6 components of stress are required to define the complete state of stress

τ τ τ τ= =similarly, andyz zy yz zy

Factor of SafetyFactor of SafetyStructural members or machines must be designed such that the working stresses are less than the ultimate strength of the

σ= Factor of safety

ultimate stressFS

the working stresses are less than the ultimate strength of the material.

Factor of safety considerations:• uncertainty in material properties • uncertainty of loadings• uncertainty of analyses• number of loading cycles• types of failureσ

σ= =u

all

ultimate stressallowable stress

FSyp

• maintenance requirements and deterioration effects

• importance of member to integrity of whole structure

• risk to life and property• influence on machine function