Mechanics of Materials - Home | University of Pittsburghqiw4/Academic/ENGR0135/Chapter4-1.pdf ·...
Transcript of Mechanics of Materials - Home | University of Pittsburghqiw4/Academic/ENGR0135/Chapter4-1.pdf ·...
Department of Mechanical Engineering
Statics and Mechanics of Materials
Internal force, normal and shearing Stress
Chapter 4-1
Department of Mechanical Engineering
OutlinesOutlines
Department of Mechanical Engineering
Department of Mechanical Engineering
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
Department of Mechanical Engineering
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
Department of Mechanical Engineering
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
Department of Mechanical Engineering
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,
Department of Mechanical Engineering
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
Department of Mechanical Engineering
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
Department of Mechanical Engineering
Single shear Double shear
Single shear and double shear
Department of Mechanical Engineering
Punching shearExample: Shear stress developed due to action
of punch in forming a rivet hole
sAP
PP
Department of Mechanical Engineering
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
Department of Mechanical Engineering
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
Department of Mechanical Engineering
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
Department of Mechanical Engineering
Example Problem 4-4
The column experiences compression
Determine the bearing stress on the surface between the bearing plate and the column
Department of Mechanical Engineering
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
Department of Mechanical Engineering
Problem 4-12
Average punching shear stress in the collar
Average bearing stress between the collar and the plate
Punching shear stressBearing stress
Plate
Department of Mechanical Engineering
Please read and practice example problems 4-3, 4-4 and 4-5
Department of Mechanical Engineering
Department of Mechanical Engineering
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
Department of Mechanical Engineering
Example Problem 4-7
Given:─
A = 200x100 mm2
─
AB
= 12.00 MPa─
= 36o
Questions:–
P?
–
AB
=?–
Max normal and shearing stresses
Department of Mechanical Engineering
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