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