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 crosssectional 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 depending on external force.

in

different

kinds

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 being analyzed. Allowable or working forces, stresses, moments, etc., are computed from the structural properties of the member in question. These allowable quantities speak of the prescribed capacity of the member. Ultimately, a body should be capable of resisting the applied load and the effects of which should be within the safe and acceptable 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).

Compressive Stress “towards the body”

Tensile Stress “away from the body”

By character, members may experience either TENSILE STRESS or COMPRESSIVE STRESS.

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. Force, F Rivets 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 2

Bearing Stress

Body 1

Example: For the truss shown a reduced stress in compression is specified to avoid the danger of buckling. Determine the cross-sectional B area of bars CF, BE and BF so that the stresses will not exceed E

100 MN/m2 in tension or 80 MN/m2 in compression.

8m A

6m

C 3m

F

G 3m

40 kN

D

50 kN

Example The bars of the pin connected frame are each 30mm x 60mm in section. Determine the maximum load 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. P B

8m

6m

A

C

10m

Example : A cast-iron column supports an axial compressive load of 250 kN. Determine the outside diameter 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. D 3m A

B 5m

C 5m

Example : A steel tube is rigidly attached between an aluminum rod and a bronze rod as shown. Axial loads are applied at the positions indicated. Find the maximum value of P that will not exceed a stress of 80 MPa in Aluminum, 150 MPa in steel or 100 MPa in bronze. 3P

P

Aluminum A = 200 mm2

Steel A = 400 mm2

1m

2m

2P Bronze A = 500 mm2

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 and aluminum. Steel

Aluminum

4P

Bronze

P 2ft

3ft

2.5ft

Example : The end chord off a timber truss is framed into the bottom chord as shown in the figure. Neglecting friction, (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 the shear load and the shearing stress on the glued joint if P= 6000N (b) generalize the procedure to show that the shearing stress on a plane inclined at an angle θ to a transverse section of area A is = Psin2 θ /2A. 20

P

50 mm 600

Example : A rectangular piece of wood, 50 mm by 100 mm in cross-section, is used as a compression block as shown in the figure. The grain makes an angle of 20o with the horizontal. Determine the maximum axial load P which can safely applied to P 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.

200

100 mm

Example : The bell crank shown is in equilibrium. (a) Determine the required diameter of the connecting rod AB if its axial stress is limited to 100 MPa. (b) Determine the shearing stress in the pin at D if its A diameter is 20 mm. B P

200 mm 240 mm D

C

DH 600 DV

30 kN

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) the maximum average tensile stress in each plate Assume that the applied load P is distributed equally among the rivets. Force, P 25 mm

Example: Two 130-mm wide plates are fastened by three 20 mm diameter rivets. Determine the maximum safe load P which may be applied (a) if the shearing stress in the rivets is limited to 60 MPa, (b) if the bearing stress of the plate is limited to 110 MPa, and (c) average tensile stress of the plate is limited to 140 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 angles with 19 mm diameter rivets. The web of the girder is 11.9 mm thick and the web of the beam is 11.4 mm thick. 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.

W460 x 97 Beam

W610 x 125 Girder

Example : The figure shows a roof truss and the detail of the riveted connection at joint B. Using allowable shearing stress of 70 MPa and bearing stress of 140 MPa, Area of 75 x 75 x 6 = 864 mm2, Area of 75 x 75 x 13 = 1780 mm2. (a) How many 19 mm diameter rivets are required to fasten members BC and BE to gusset plate? (b) Determine the largest average tensile/compressive stresses in members BC and BE.

14 mm Gusset Plate

Joint B

PBE 75 x 75 x 13 mm

75 x 75 x 6 mm

PBC

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 diameter rivets to develop fully the truss joint for members A, B and bottom chord C.

9.5 mm Gusset Plate

150 kN

2 angles

624.67 kN

2 angles

A

220 kN

B

70○

38○

2 angles

400 kN

C

Answer: 9.5 mm Gusset Plate

150 kN

2 angles

624.67 kN

2 angles

A

220 kN

B

70○

38○

2 angles

400 kN

C

THIN-WALLED PRESSURE VESSELS A vessel is said to be thin-walled when the ratio of the thickness to the radius of the vessel is small such that the internal stress in the material is constant throughout 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 from steel plates which have a thickness of 20mm. The diameter of the pressure vessel is 500mm and its length is 3m. Determine the maximum internal pressure which can be applied if the stress in the steel is limited to 140MPa

Example : A water tank is 8m in diameter and 12m high. If the tank is to be completely filled, (a) Determine the tangential force on the tank and the (b) minimum thickness of the tank plating if the stress is limited to 40 MPa.

12 m

8m

Example : The strength per meter of the longitudinal joint in the figure is 480 kN, where as for the girth join, it is 200kN. Determine the maximum diameter of the cylindrical tank if the internal pressure is 1.5 MN/m2. Girth joint

Longitudinal joint

Example : The tank shown in the figure is fabricated from 10mm steel plate. Determine the maximum longitudinal and circumferential stresses caused by an internal pressure of 1.2 MPa.

400 mm

600 mm

Example : The tank shown in the figure is fabricated from steel plate. Determine the minimum thickness of plate which may be used if the stress is limited to 40 MPa and internal pressure is 1.5 MPa.

400 mm

600 mm