classification of forces

8/3/2019 Classification of Forces

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Classification of forces

What is force?

It is an action of 1 body to another.Exist in equal magnitude, opposite directionpairs

Type of force Definition

External All force acting on FBD

W Applied force/load(dimension of force/length)Concentrated/distributed

Concentrated , F Area of the contact is smallcompared to the size of body(act at a point)

Reaction Supporting forcesStatic load (dead weight) Do not vary with time,

pressure/temperature

Repeat load Load that applied/remove manythousand time

Impact load Transfer large amount energy in

short time


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Equilibrium of rigid body

Rigid body-body which is not deformunder applied load.

3 dimension but always count on 2dimension

Fy

Fx

F=0 (force)M=0 (momentum)


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Example Problem1

Force at B & C are equal

Determine the force A and B on ABHow many unknown ?For moment choose the right sidePoint A was chosen as a main.

Fx=0 +

Fy=0 +

M=0 +(Anti clockwise)


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Example Problem 2

A 900 kg mass is supported by a roller thatcan move along a beam. The beam issupported by a pin at A and roller at B.a)Neglect the mass of the beam anddetermine the reaction at A & Bb)If the mass of the beam is 8.5 kg/m,

determine the reaction at A and B


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Internal forces

In mechanical analysis problem, is to be able todetermine the resultant force and moment actingwithin a body, which are necessary to hold thebody together when the body is subjected toexternal loads

Normal force, NShear force, VTorsional moment, TBending moment, MFR & MR


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Example Problem 3

a) Determine the support reaction

b) The internal force system on asection 6 ft to the right of a supportat A.


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Example Problem 4

The cantilever beam is subjected to both concentrated

and distributed loads. Determine:-a) the support reactionsb) The internal forces on a section 4 m to the right of

the support at A


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Normal stress under axial loading

Body must be able to withstand the intensity ofinternal force

Area

ForceStress

stressnormalecompressivfor

stressnormaltensilefor

A

Favg


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Shearing stress in connection

Loads apply generally transmitted to the individual

through rivets, bolts, pin, nails. It resultant to shear force V equal to applied load P

areaboltA

forceshearV

A

Vavg

Shear force on each cross section equalto half of the applied load


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Bearing stress

Known as compressive normal stress

It occur on the surface of contact between the headof the bolt and the top plate and between the nutand the bottom plate

Since metal can sustain stress of several thousand

pound per square inch, unit of ksi per suare inchalways been used (1 ksi =1000psi)

areaprojectedA

surfacecontacttheacross

dtransmitteforceV

A

Fb

PamNstressforunitSI /2


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The flat steel bar has axial loads applied at point A,

B, C and D as shown. Of the bar has crosssectional area of 3 inch2, determine the normalstress in the bar

a) on a cross section 20 inch, to the right of point A

b) on a cross section 20 inch, to the right of point B

c)on cross section 20 inch, to the right of point C

Example Problem 5


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A brass tube with an outside diameter of 2 inch and wallthickness of 0.375 inch is connected to a steel tube with aninside diameter of 2 inch and wall thickness of 0.25 inch byusing 0.75 inch diameter pin. Determine:-

a)The shearing stress in the pin when the joint is carrying an

axial load P of 10 kip b) The length of joint required if the pin is replace by a glued

joint and the shearing stress in the glue must be limited to250 psi

Example Problem 6


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Stresses on inclined plane in axiallyloaded member

A

Favg

A

Vavg

For inclined :-

2sin2cossincos

sin

)2cos1(2

cos

cos

cos 2

A

P

A

P

A

P

A

V

A

P

A

P

A

P

A

N

nn

n

n

Why it is cosand the other is sin?N and V depend on the angle ,

and with respect to incline plane with respect ofapplied load.


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Stresses on inclined plane in axially loadedmember

Graph below show the magnitudes of andas a function of .

max when is 0O or 180O

max when is 45O or 135O

Therefore, the maximum normal and shearingstress for axial tensile or compressive loading are

n

n

n

n

A

P

A

P

2

,maxmax


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Example Problem 7

A plastic bar with circular cross section will be used

to support an axial load of 1000 lb. the allowablenormal and shearing stress in the adhesive jointused to connect the two parts of the bar are 675 psiand 350 psi respectively. Determine the required

diameter, d of the bar. Draw the FBD.


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Stress transformationequations

This equation relates normal and shearing stress on oneplane (whose normal is oriented at an angle wrt a referencex-axis), through a point and known stress x ,y , andxy =yxon the reference plane that can be develope using FBD.

All stress are positive

Dotted line A-A represent any plane through the point

Counter clockwise angle is positive, where is measured

from the positive x-axis to the positive n-axis.

dA cos for the vertical and dA sin for the horizontal face.

N-axis is perpendicular to the incline face. The t-axis parallelto the inclined face.

The xy and nt axes shown are positive.

Stress are multiplied by the areas over which they act.

S f


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Summation of force in n-direction

2sin2cos22

:

cossin2sincos

0sin)cos(cos)sin(sin)sin(cos)cos(

22

xy

yxyx

n

xyyxn

xyyx

xyyxyxnn

angledoubleoftermsin

knowing

dAdAdAdAdAF

St t f ti


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Summation of force in t-direction


2cos2sin2

:

)sin(coscossin)(

0cos)cos(sin)sin(cos)sin(sin)cos(

22

xy

yx

nt

xyyxnt

xyyx

xyyxyxntt

angledoubleoftermsin

knowing

dAdAdAdAdAF

Tensile normal stress +ve, while compressive normal stressve

Sign of normal stress independent of the coordinate being used

Angle measured anti-clockwise from positive x-axis

All force in figure are positive

Sign of shearing stress pointing opposite direction would beve

The sign of shearing stress depend on the coordinate being used

If its positive at first coordinate, the value will remain


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At a point on structural member subjected

to plane stress there are normal andshearing stresses on horizontal andvertical planes through a point. Usestress transformation equation todetermine:

a) Normal and shearing stress on planeAB

Example Problem 7

P i i l d i


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Principal stresses and maximum

shearing stress-plane stress

For design purpose, critical stresses at the point usually themaximum tensile stress and the maximum shearing stress

Maximum and minimum values of n occur at value of forwhich d n/d =0, the differentiation yields :-

Equal to zero and solving gives

Orientation of the element

2cos2sin2

@ xyyx

yxnt

2sin2cos22@ xy

yxyx

xn

2cos22sin)(

n

xyyxd

d

yx

xy

p

22tan


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Principal stresses and maximumshearing stress-plane stress

Values from above give the orientation of twoprincipal plane. The third principal plane for theplane stress state has an outward normal to the z-direction

This give principal stress

The orientation maximum in plane shear stress

2

2

2,122

xy

yxyx

xy

yx

2

)(2tan

P i i l d i


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Principal stresses and maximum

shearing stress-plane stress

2

yx

avg

2

2

max2

xy

yx

planein

Maximum in plane shear stress

Average plane shear stress

2

minmaxmax

2

2

2,1

22xy

yxyx


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Example Problem 7

Calculate the orientation of element,maximum in plane shear stress andaverage normal stress.

M h Ci l Pl


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Mohrs Circle-Planestress Set x-y reference

Identify stresses x, y, xy=yx

Draw set of - coordinate with and positive to the rightand upward.

Plot (x, -xy) and lable it in point V(vertical plane)

Plot (x,

yx) and lable it in point H(horizontal plane)

Draw line between V and H. Establish the center C and theradius R of the circle.

Draw the circle

For center, y=0, so value for is for x. (,0)

2

yx

avg

M h Ci l Pl


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Mohrs Circle-Planestress

Point D provides the principal stress P1Point E provides the principal stress P2Point A provides the maximum in-planeshearing stressp and accompanyingnormal stress average that act on plane.

E ample Problem


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Example Problem

8 example 9.72Determine and show on sketch :a) The principal and maximum shearingstress at the pointb) The normal stress on plane

classification of forces

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