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Statics and Mechanics of Materials

Internal force, normal and shearing Stress

Chapter 4-1

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OutlinesOutlines

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Internal Forces -

cutting plane

Result of mutual attraction (or repulsion) between molecules on both sides of the cutting plane

These result is distributed over the entire surface of the cutting plane

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Internal Forces -

cutting plane

Each part of the body satisfies the equilibrium equation

The resultant of the internal forces R must be in equilibrium with other applied forces in the body part

Stress is the intensity of the R

So either body part can be used to determine the internal forces

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Internal Force –

cutting plane

If the cutting plane is perpendicular to the bar axis the internal forces, internal stress, and the resultant will be perpendicular in normal direction

If the cutting plane is not

perpendicular the resultant will still be perpendicular, but it has normal and tangential

components

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Normal stress –

axial loading

Axial loading = the loading/force is collinear with the axis of the bar

Stress = intensity of the internal force

Generally speaking,

Or symbolically,

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Normal stress –

some notesGenerally, the stress is not uniformly

distributed over the areaFor many applications, it’d be assumed that it

is uniformly distributedCross area changes under loadingEngineering stress

uses initial

cross sectional

areaTrue stress

uses the deformed area

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Shearing stresses in connectionsLoads are transmitted to individual members

through connections that use rivets, bolts, pins, nails, or welds

Single shear Double shearPunching shearBearing stress

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Single shear Double shear

Single shear and double shear

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Punching shearExample: Shear stress developed due to action

of punch in forming a rivet hole

sAP

PP

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Bearing stress

Bearing stress = compressive normal stress

While the amount of the force = compression load, the area depends on the mode of the contact

Examples: –

Between the head of the bolt and the top plate (a)

–

Between the surfaces of the shanks and hole which they pass (b)

bb A

F

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Units of Stress

Dimension = FL-2

USCS; –

psi

(pounds per square inch),

–

ksi

(kilo pounds per square inch), –

ksi

= 1000 psi

SI; –

Pa (Pascal = N/m2),

–

kPa

(kilo Pa) = 1000 Pa, or –

MPa

(mega Pa) = 106

Pa

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Example Problem 4-1

The cross-sectional area = 3 in2.

Determine the axial stress in the bar on a cross section;–

20” to the right of A

–

20” to the right of B–

20” to the right of C

First thing to do; to determine the internal force on the section

use cutting plane

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Example Problem 4-4

The column experiences compression

Determine the bearing stress on the surface between the bearing plate and the column

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Example Problem 4-4FBD of the bearing plate

Compression developed in the column

Compression developed in the timber beam

Cross section of the column

do

di

22

4 iobb

b ddAAF

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Problem 4-12

Average punching shear stress in the collar

Average bearing stress between the collar and the plate

Punching shear stressBearing stress

Plate

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Please read and practice example problems 4-3, 4-4 and 4-5

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Maximum and minimum stresses

Maximum normal stress when = 0 (or 180)

Maximum shearing stress when = 45 (or 135) (opposite directions)

Minimum stress = 0, when = 90

Note: maximum stresses don’t appear on the same angle

AP

max

AP

2max

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Example Problem 4-7

Given:─

A = 200x100 mm2

─

AB

= 12.00 MPa─

= 36o

Questions:–

P?

–

AB

=?–

Max normal and shearing stresses

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Example Problem 4-7: AnswerFollow the solution in

the book or use

And the maximum are

)2cos1(2

A

Pn

kNAP n 7.694)2cos1(

2

MPaA

P 52.162sin2

MPaA

P

MPaAP

37.172

7.34

max

max