Lecture 1 1

Chapter 1

Objective

Strengthen the knowledge: important principle in static

Introduction of the normal and shear stress

Stresses: Normal and Shear

Normal Stress

Lecture 1 2

Normal Stress: Units

Lecture 1 3

PaNmmA

NPave 2

2)(

)(

Basic units of stress is N/m2 or Pa (Pascal)

If force is N and A in mm2

MPaNmmA

NPave 26

210

)(

)(

It is more common to have unit in MPa, if the properties of the material is in MPa, preferably use force in N and any length mmwhich can result in MPa

4

A

Pave P: force that is perpendicular to plane

( the force is acted on the center of designated area)

A: area of the plane

Calculate the average normal stress:

Structure (Assume P = 100 kN )Wide Flange Structure: W310 x 74Angles L203 x 203 x 19.0

d = 200 mm

Allowable Normal Stress

The allowable stress is the maximum stress allowed due to the properties of the material.

Lecture 1 5

aveall Considered as SAFE

aveall Considered as FAIL

ave

allSF

.

Safety Factor

Safe F.S >= 1

Fail F.S < 1

Link BC is 6 mm thick

and is made of a steel

with a 450 MPa yield

strength in tension. What

should be its width w if

the structure shown is

being designed to support

a 20 kN load P with a

factor of safety of 3?

EXAMPLE 1

REVIEW

Lecture 1 8

A load W = 1000 kg is supported by a 3‐cable system as shown in figure 1. The

cable run along 2 pulleys F and D which is attached to the solid wall by a round beam EF and CD both of which are 30mm diameter and 300mm long. Ignore the mass of cable and beam. All cables are the same size. The sizes of the pulleys at D & F can be ignored. Gravity = 9.81m/s2

Answer the following questions:Calculate the tension of cable AB, BD, and BFCalculate the diameter of the cable is the allowable stress is 140 MPa, assume all the cables have the same diameter.

Find the internal stress at the mid point of round beam FE

Solve it

Lecture 1 11

Solve it

Lecture 1 12

Solve it

Lecture 1 13

Solve it

Lecture 1 14

Direct Shear Stress

Lecture 1 15

V: shear load( the force is acted on the center of designated area)

A: area of the plane

FF

V = F

A

V

Double Shear Stress

Lecture 1 16

V: shear load( the force is acted on the center of designated area)

A: area of the plane

V

A

F

A

F

A

V

2

2/

V

Application

Lecture 1 17

Single Shear

Double Shear

Multiple Shear (complex)

50 kN

40 kN

Pin C

Lecture 1 18

MPa

Pax

dA

V

8.101

108.101

4/)025.0(

000,50

4/

000,50

6

2

2

*N/m2 = Pa

Or

N/mm2 = MPa

Direct Shear Stress on pin at C

FBD at pin C

Pin A

Lecture 1 19

MPa

dA

V

7.40

4/)25(

000,20

4/

2/000,40

2

2

Direct Shear Stress on pin at A

FBD at pin AFBD at pin A

V

Pin B

Lecture 1 20

50 kN

40 kN

20 kN

15 kN 15 kN

50 kN

Draw the FBD of BE and DG portion of the pin?

What is the highest shear load on the pin B?

21

Shear load due to torsion

Shear Stress due to torque

T

VV

D

TV

VDT

T

0

0

Equilibrium equation:

22

Shear Stress: torque

T= 10 Nm ccwBolt d = 10mmPosition of the bolt R = 100 mm

Calculate the V and A.Sketch the FBD.

Allowable Shear Stress

The allowable stress is the maximum shear stress allowed due to the properties of the material.

Lecture 1 23

aveall Considered as SAFE

aveall Considered as FAIL

ave

allSF

.

Safety Factor

Safe F.S >= 1

Fail F.S < 1

Q 1-75 pp 55

Lecture 1 24

The joint is fastened together using two blots. Determine the requiredDiameter of the bolts if the failure shear stress of the bolt is 350 Mpa. Use factor of safety of F.S = 2.5.

Lecture 1 25

V for each bolt

kNV 000,204/000,80

Shear stress resulted from V

mmd

mmd

all

14

5.13

4/

000,202d

Allowable shear stress

MPaSF

all 1405.2

350

.

Therefore:

EXAMPLE 2Member B is subject to a

compressive force of 3600 N.

If A and B are both made of

wood and are 10 mm thick,

determine the nearest 5 mm

the smallest dimension h of

the support so that the

average shear stress doest not

exceed τallow = 2000 kPa

h

hN = 3323.08 N

V = 1384.62 N

3600 Nτallow = 2000 x 103

= 1384.62/10h

h = 69.23 mm

= 70 mm

The two members are pinned together at B as shown. Topview of the pin connections at A and B are also given in thefigure. If the pins have an allowable shear stress of τallow =90 MPa and the allowable tensile stress of rod CB is (σt)allow = 115 MPa, determine:

a. to nearest mm the smallest diameter of pin A and B,

b. the diameter of rod CB necessary to support the load.

EXAMPLE 3

EXAMPLE 4

Determine the required cross-

sectional area of member BC

and the diameter of the pins at

A and B if the allowable

normal stress is σallow = 20

Mpa and the allowable shear

stress is τallow = 30 MPa.

In the structure shown, an 8 mm diameter pin is used at A, and 12 mm

diameter pins are used at B and D. Knowing that the ultimate shearing stress

is 100 MPa at all connections and that the ultimate normal stress is 250 MPa

in each of the two links joining B and D, determine the allowable load P if

an overall factor of safety of 3.0 is desired.

EXAMPLE 5

Solve it

Lecture 1 36