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Chapter 1:
Simple Stress
Chapter 2:
Simple Strain
Chapter 3:
Torsion
Strength of Materials deals with the relationship between externally applied
loads and their internal effects on bodies.
bodies are no longer assumed rigid.
Objective:
The ability of structures (structural members) to safely
resist the maximum internal effects produced by any
loading combination should be ensured.
Internal Forces Axial Forces – any force acting along the centroidal axis of a body.
Normal Force – any force acting normal or perpendicular to a
plane or cross-sectional area.
Shear Force – any force acting parallel to a plane or cross-
sectional area.
Torque – any couple acting about the longitudinal or
centroidal axis
Bending Moment – any couple acting along the longitudinal
or centroidal axis.
Concept of Deformation
Any solid, non-rigid body when acted upon by an
external force will experience a change from its natural
shape or form. The body is said to have deformed.
Deformation may occur in different kinds
depending on external force.
Elasticity and Plasticity
Elasticity is the tendency of a body to return to its
original state after deformation.
Plasticity is the propensity of a body to remain
deformed even after the load is removed
Plastic conditions occur when elastic boundaries
are exceeded.
Concept of Actual and Allowable Actual forces, stresses, moments, etc., are derived from
the effects of externally applied loads acting on the body beinganalyzed.
Allowable or working forces, stresses, moments, etc.,are computed from the structural properties of the member inquestion. These allowable quantities speak of the prescribedcapacity of the member.
Ultimately, a body should be capable of resisting the appliedload and the effects of which should be within the safe andacceptable or tolerable range.
Put simply, the ALLOWABLE should be greater
than the ACTUAL.
Multiplication Factor
Prefix Symbol
x109 Giga G
x106 Mega M
x103 Kilo k
x10-3 Milli m
x10-6 Micro µ
x10-9 Nano η
x10-12 Pico p
1 MPa = 1N/mm2
1 Pa = 1N/m2
Stress
Stress is a force intensity per unit area.
The strength of material is quantified through its stress
capacity – how large of a force a certain area can
withstand.
In other words, the larger the force a unit area can
resist the higher the material’s strength.
Normal or Simple Stress, , is caused by a force
that acts perpendicular to the area resisting
the force.
Centroid
Area, A
Force, F
Normal Stress,
= dF / dA
Ave = F / A
Non-Uniform Stress Distributions occur at sections near
the point load application and at varied cross-sections.
By Saint Vernant’s Principle, the stress due to a point
load and the stress due to an equivalent pressure
causes similar stress distributions at a certain distance
from the point of application
Normal Stress, ,may also be termed AXIAL STRESS if
the force acts along the longitudinal axis of the member
(as in truss members).
By character, members may experience either TENSILE
STRESS or COMPRESSIVE STRESS.
Compressive Stress“towards the body”
Tensile Stress“away from the body”
Simple Shearing Stress, ,is caused by a force that is
parallel to the area resisting the force. Also called
tangential stress, occurs whenever a load causes one
body to slide past its adjacent section.
Force, F
Force, F
Some Common Types of
Simple Shear in Stressed
Bodies.
Single Shear – Resisting Area on one plane.
Rivets
Force, F
Force, F
Area, A
Double Shear – Resisting Area on two planes
Rivets
Force, F
Area, A
Punching Shear – Resisting Area is non-planar.
Force, F
Area, A(Cylindrical Surface Area)
Bearing Stress – b ,occurs as contact pressure
between separate bodies, compressive in nature.
Force, F
Body 1Body 2
Bearing Stress
Example: For the truss shown a reduced stress in
compression is specified to avoid the danger ofbuckling. Determine the cross-sectional
area of bars CF, BE and BF so that the
stresses will not exceed
100 MN/m2 in tension
or 80 MN/m2 in
compression.
B
AC G
D
E
F6m
8m
3m 3m
40 kN50 kN
Example The bars of the pin connected frame are each
30mm x 60mm in section. Determine the maximumload P that can be applied so that the stresses of bar
AB, BC and AC will not exceed 100 MN/m2 in
tension or 80 MN/m2 in compression.
B
A C
P
6m8m
10m
Example : A cast-iron column supports an axial
compressive load of 250 kN. Determine the outsidediameter of the column (a) if its inside diameter is
200 mm and (b) if its thickness
is 0.1 D and the limiting
stress is 50 MPa.
Example : The homogeneous bar is supported by a smooth
pin at C and a cable that runs at A to B around the
smooth peg at D. The bar weighs 6 kN. Find the
stress in the cable if its diameter is 15 mm.
A B
D
C
3m
5m 5m
Example : A steel tube is rigidly attached between an
aluminum rod and a bronze rod as shown. Axialloads are applied at the positions indicated. Find themaximum value of P that will not exceed a stress of
80 MPa in Aluminum, 150 MPa in steel or 100 MPain bronze.
P3P 2P
AluminumA = 200 mm2
SteelA = 400 mm2
BronzeA = 500 mm2
1m 2m 2.5m
Example : A rod is composed of an aluminum section
rigidly attached between steel and bronze section.Axial loads are applied at the positions indicated. If
P = 3000 lb and the cross sectional area of the rod is
0.5 in2. Determine the stresses in steel, bronze andaluminum.
Steel Aluminum Bronze
2ft 3ft 2.5ft
P4P
Example : The end chord off a timber truss is framed into
the bottom chord as shown in the figure. Neglectingfriction, (a) Compute
dimension “b” if the
allowable shearing
stress is 900 kPa
and (b) determine
dimension “c” so that
bearing stress does
not exceed 7 MPa.
Example : Two block of wood, 50-mm wide and 20-mm
thick are glued together as shown. (a) Determine theshear load and the shearing stress on the glued jointif P= 6000N (b) generalize the procedure to showthat the shearing stress on a plane inclined at an
angle to a transverse section of area A is
= Psin2 /2A.
θ
θ
P50 mm
20
600
Example : A rectangular piece of wood, 50 mm by 100 mm
in cross-section, is used as a compression block asshown in the figure. The grain makes an angle of 20o
with the horizontal. Determine the maximum
axial load P which can safely applied to
the block (a) if the compressive stress
in the wood is limited to 20 MN/m2
and (b) if the shearing stress parallel
to the grain is limited to 5 MN/m2.
P
200
100 mm
Example : The bell crank shown is in equilibrium. (a)
Determine the required diameter of the connectingrod AB if its axial stress is limited to 100 MPa. (b)Determine the shearing stress in the pin at D if itsdiameter is 20 mm.
240 mm200 mm
A B
DC
30 kN
600
DH
DV
P
Example: Two 130-mm wide plates are fastened by three
20 mm diameter rivets. Assuming that P = 50kN,determine (a) the shearing stress in each rivets; (b)the bearing stress in each plate and (c) themaximum average tensile stress in each plateAssume that the applied load P is distributedequally among the rivets.
Force, P
25 mm
Example: Two 130-mm wide plates are fastened by three
20 mm diameter rivets. Determine the maximumsafe load P which may be applied (a) if the shearingstress in the rivets is limited to 60 MPa, (b) if thebearing stress of the plate is limited to 110 MPa, and(c) average tensile stress of the plate is limited to140 MPa.
Force, P
25 mm
Example : The figure shows a W460 x 97 beam riveted to a
W610 x 125 girder by two 100 x 90 x 10 mm angleswith 19 mm diameter rivets. The web of the girder is11.9 mm thick and the web of the beam is 11.4 mmthick. Determine the allowable end reaction.
- Shop-Driven rivets (Angles to Beam) = 80 MPa,
b = 170 MPa, Field-driven rivets (Angles to Girder)
= 70 MPa and = 140 MPa.
W610 x 125 Girder
W460 x 97 Beam
Example : The figure shows a roof truss and the detail of
the riveted connection at joint B. Using allowableshearing stress of 70 MPa and bearing stress of 140MPa, Area of 75 x 75 x 6 = 864 mm2, Area of 75 x 75 x13 = 1780 mm2. (a) How many 19 mm diameter rivetsare required to fasten members BC and BE to gussetplate? (b) Determine the largest averagetensile/compressive stresses in members BC and BE.
PBC
PBE
14 mm Gusset Plate
75 x 75 x 13 mm
75 x 75 x 6 mm
Joint B
Example : A truss joint shown consists of a bottom chord
C made up of two angles, and web members A and B
each carrying the given loads. Using A 502-1 rivets
with an allowable shearing stress of 120 MPa and
bearing stress of 600 MPa and AISC specifications.Determine the required number of 18mm diameterrivets to develop fully the truss joint for members A,B and bottom chord C.
70○ 38○
9.5 mm Gusset Plate 2 angles
2 angles
2 angles AB
C
220 kN150 kN
400 kN624.67 kN
70○ 38○
9.5 mm Gusset Plate 2 angles
2 angles
2 angles AB
C
220 kN150 kN
400 kN624.67 kN
Answer:
THIN-WALLED PRESSURE VESSELS
A vessel is said to be thin-walled when the ratio ofthe thickness to the radius of the vessel is small suchthat the internal stress in the material is constantthroughout the thickness of the vessel.
1 = circumferential or hoop stress
2 = longitudinal stress
t = thickness
ro = outside radius 1
ri= inside radius 2
p = internal pressure
Example :
A cylindrical pressure vessel is fabricated fromsteel plates which have a thickness of 20mm. Thediameter of the pressure vessel is 500mm and itslength is 3m. Determine the maximum internalpressure which can be applied if the stress in thesteel is limited to 140MPa
Example :
A water tank is 8m in diameter and 12m high.If the tank is to be completely filled, (a) Determinethe tangential force on the tank and the (b)minimum thickness of the tank plating if thestress is limited to 40 MPa.
12 m
8 m
Example :
The strength per meter of the longitudinaljoint in the figure is 480 kN, where as for the girthjoin, it is 200kN. Determine the maximumdiameter of the cylindrical tank if the internalpressure is 1.5 MN/m2.
Girth joint
Longitudinal joint
Example :
The tank shown in the figure is fabricated from10mm steel plate. Determine the maximumlongitudinal and circumferential stresses caused by aninternal pressure of 1.2 MPa.
600 mm
400 mm
Example :
The tank shown in the figure is fabricated fromsteel plate. Determine the minimum thickness of platewhich may be used if the stress is limited to 40 MPaand internal pressure is 1.5 MPa.
600 mm
400 mm