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CHAPTER - 3

LOAD AND STRESS ANALYSIS

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Equilibrium

Equilibrium is the condition of a system in which competinginfluences are balanced.

The two major classification of equilibrium in a structuralanalysis are,

1. Static Equilibrium

2. Mechanical Equilibrium

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Classification

STATIC EQUILIBRIUM:

A system of particles is in static equilibrium when allthe particles of the system are at rest and the total force on

each particle is permanently zero

MECHANICAL EQUILIBRIUM:

The necessary and sufficient conditions for a particleto be in mechanical equilibrium is that the net force actingupon the particle is zero.

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Free body diagrams A diagram of a body (or a part of it) which shows all theforces and couples applied on it, and which has all the forcesand couples labeled for use in the solution of the problem iscalled a free-body diagram.

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Purpose of free body diagram

The diagram establishes the directions of reference axis,provides a place to record the dimensions of the subsystemand the magnitudes and directions of the known forces andhelps in assuming the directions of unknown forces.

The diagram simplifies our thinking because it provides aplace to store one thought while proceeding to the next.

The diagram provides a means of communicating ourthoughts clearly and unambiguously to other people.

.

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Purpose of free body diagramCareful and complete construction of the diagram clarifiesfuzzy thinking by bringing out various points that are notalways apparent in the statement or in the geometry of thetotal problem. Thus, the diagram aids in understanding allfacets of the problem.

The diagram helps in the planning of a logical attack on theproblem and in setting up the mathematical relations

The diagram helps in recording progress in the solution andin illustrating the methods used.

The diagram allows others to follow our reasoning, showing

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Steps for free body diagramFollow these steps to draw a free-body diagram.

1. Select the body (or part of a body) that we wantto analyze, and draw it.

2. Identify all the forces and couples that are appliedonto the body and draw them on the body. Place each forceand couple at the point that it is applied.

3. Label all the forces and couples with uniquelabels for use during the solution process.

4. Add any relevant dimensions onto our picture.

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Forces and couples 1. Identify all the forces which come from the interaction of one body with

another. Remember that for each way in which a support restricts thefree motion of the body, a force or a moment must be applied to the bodyto impose the restriction on the motion.

2. Apply the weight of the body to its center of gravity (if it is uniform, thenapply it to the centroid).

3. Remember that strings and cables can only pull on an object.

4. Remember that internal loads cancel out and should not be put on thefree-body diagram.

5. Remember that if we have selected the direction of forces or couples of

interaction on one body, then Newtons 3rd

law states that you must applythe forces or couples in the opposite direction on the other body.

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Solving for unknowns 1. In 2-D problems the equilibrium equationsresult in three independent scalar equations (two

components of force and one component ofmoment). Therefore, we can only solve for threescalar unknowns.

2. While taking moment , select a point suchthat the line of action of at least one unknownforce passes through that point. This will eliminateone unknown from our moment equation and willresult in simpler equations to work with.

3. We can sometimes take moments about twoor three different points in a problem. Select each

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Solving for unknowns 4. Remember that the additional equations we generate in theabove way are not independent of the original equations and

therefore we will still have only three independent equations in 2-Dproblems per free-body diagram and you can only solve for threeunknowns per free-body diagram.

5. For a composite body, if we have drawn a free-body diagramand written the equilibrium equations for each of its subsections,we will gain no additional information if we draw the free-bodydiagram of the entire composite body and write its equilibriumequations.

6. In 3-D problems the equilibrium equations result in sixindependent scalar equations (three components of force andthree components of moment). Therefore, we can solve for up tosix scalar unknowns per free-body diagram.

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Example - beamThe figure shows below is a beam supported by reactions F 1 , F 2 and F 3. If the beam is cut at some section located at x = x 1 and the left-handportion is removed as a free body, an internal shear force V and bendingmoment M must act on the cut surface to ensure equilibrium.

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stressSTRESS ( ) :

Internal resistance force offered by abody against deformation.

1. NORMAL STRESS ( n) : The internal resistance force acting in the

NORMAL direction.

2. SHEAR STRESS (

):The internal resistance force acting in theTANGENTIAL direction.

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ONE DIMENSIONAL STRESSEngineering stress / Nominal stress

The simplest definition of stress, = F / A, where A is theinitial cross-sectional area prior to the application of theload

True stressTrue stress is an alternative definition in which the initialarea is replaced by the current area

eetrue )1(

Relation between Engineering & Nominal stress

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TYPES OF STRESSES

TENSILE

BENDING

COMPRESSIVE

SHEAR

TORSION

http://www.allstar.fiu.edu/aero/pic1-2e.htmhttp://www.allstar.fiu.edu/aero/pic1-2d.htmhttp://www.allstar.fiu.edu/aero/pic1-2c.htmhttp://www.allstar.fiu.edu/aero/pic1-2b.htmhttp://www.allstar.fiu.edu/aero/pic1-2a.htm

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normal Stress componentsCartesian co- ordinate system :

1. In the Cartesian co-ordinates system, we make useof the axes, X, Y and Z

2. Let us consider the small element of the materialand show the various normal stresses acting the faces

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Normal Stress componentsCylindrical co-ordinate system:

In the Cylindrical - co-ordinate system we make use ofco-ordinates r, q and Z.

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SHEAR STRESS

TORSION

12

12

B Az z

dz

dx

z d

z d y

z

d z

d y

xdxdy

xdxdy

dz dxdy dx dzdy xz

D

C

xz

Taking moment about CD, We get

This implies that if there is a shear in one plane then there will be a shear inthe plane perpendicular to that

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stress componentsShear Stresses :

With shear stress components, the single subscript notation isnot practical, because such stresses are in direction parallel to thesurfaces on which they act. We therefore have two directions to specify,that of normal to the surface and the stress itself.. To do this, we attachtwo subscripts to the symbol ' t' , for shear stresses.

For Cartesian co-ordinates,

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stress componentsFor cylindrical co-ordinates

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Stress components in 3-d

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Elastic strain When a material is placed in tension, there exists not only an axialstrain, but also negative strain (contraction) perpendicular to the axialstrain.

Assuming a linear, homogeneous, isotropic material, this lateral strain isproportional to the axial strain.

If the axial direction is x, then the lateral strains are y = z = v x . Theconstant of proportionality v is called Poissons ratio, which is about 0.3

for most structural metals.

If the axial stress is in the x direction,

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For a stress element undergoing x , y , and z simultaneously, thenormal strains are given by,

Shear strain is the change in a right angle of a stress element when

subjected to pure shear stress, and Hookes law for shear is given by

where the constant G is the shear modulus of elasticity or modulus of

rigidity

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STRESS-STRAIN CURVE

Mild steel

Thermoplastic

Copper

http://en.wikipedia.org/wiki/Image:Stress_v_strain_A36_2.png