mm222 lec 10-12

Hafiz Kabeer Raza Research Associate

Faculty of Materials Science and Engineering, GIK Institute Contact: Office G13, Faculty Lobby

[email protected], [email protected], 03344025392

MM222

Strength of Materials

Lecture 10

Spring 2015

Spring 2015 By Hafiz Kabeer Raza MM222 Strength of Materials

Factor of safety Self Reading

Article 1.12

Article 1.13

Ultimate strength (normal and shear) u

Allowable stress all Factor of safety = u/all


Sample problem 1.3

dAB = ? dC = ? t = ? Determine P and C (Cx, Cy) Determine (, )all Find the relevant dimension


Sample Problem 1.4

Maximum allowable force upward at C = ? Find C in terms of B and D Find C from all the allowable stresses

Control rod AB Bolt at B Bolt at D Bolt at C

The Lowest value is the safe for all


Stress on an Oblique Plane Pass a section through the member forming

an angle q with the normal plane.

qq

q

q

q

q

q

q

q

cossin

cos

sin

cos

cos

cos

00

2

00

A

P

A

P

A

V

A

P

A

P

A

F

The average normal and shear stresses on

the oblique plane are

qq sincos PVPF

Resolve P into components normal and

tangential to the oblique section,

From equilibrium conditions, the

distributed forces (stresses) on the plane

must be equivalent to the force P.




MM222


Lecture 11

Spring 2015


Problem 1.29 & 1.30

Problem 1.29 Simply put the values of

parameters in the formula = 70.0 psi, = 40.4 psi

Problem 1.30

= cos2 , =

(a) 1.5 kips, (b) 43.3 psi


Problem 1.60

Method of joints Also a force triangle can be

made AB = 7.32 kips (T), BC =

8.966 kips (C) Max. area AB = 0.5(1.8-0.8) AB max = 14.64 ksi Area BC = 0.5*1.8 BC max = 9.96 ksi


Problem 1.61

(a) 8.92 ksi

(b) 22.415 ksi

(c) 11.207 ksi


Problem 1.63


Home work

Home work 1.33, 1.37, 1.67, 1.18

For interest only 1.39, 1.41, 1.42, 1.51


Self reading 2.1 to 2.5

2.6 and 2.7 will not be discussed at this point

Next 2.8




MM222


Lecture 12

Spring 2015


Deformations Under Axial Loading

AE

P

EE

From Hookes Law:

From the definition of strain:

L

Equating and solving for the deformation,

AE

PL

With variations in loading, cross-section or

material properties,

i ii

ii

EA

LP

L

AE

dxP

0


Example 2.01

Determine the deformation of

the steel rod shown under the

given loads.

in. 618.0 in. 07.1

psi1029 6

dD

E

SOLUTION:

Divide the rod into components at

the load application points.

Apply a free-body analysis on each

component to determine the

internal force

Evaluate the total of the component

deflections.


Example 2.01 SOLUTION:

Divide the rod into three

components:

221

21

in 9.0

in. 12

AA

LL

23

3

in 3.0

in. 16

A

L

Apply free-body analysis to each

component to determine internal forces,

lb1030

lb1015

lb1060

33

32

31

P

P

P

Evaluate total deflection,

in.109.75

3.0

161030

9.0

121015

9.0

121060

1029

1

1

3

333

6

3

33

2

22

1

11

A

LP

A

LP

A

LP

EEA

LP

i ii

ii

in. 109.75 3


Sample Problem 2.1

The rigid bar BDE is supported by two

links AB and CD.

Link AB is made of aluminum (E = 70

GPa) and has a cross-sectional area of 500

mm2. Link CD is made of steel (E = 200

GPa) and has a cross-sectional area of (600

mm2).

For the 30-kN force shown, determine the

deflection a) of B, b) of D, and c) of E.

SOLUTION:

Apply a free-body analysis to the bar

BDE to find the forces exerted by

links AB and DC.

Evaluate the deformation of links AB

and DC or the displacements of B

and D.

Work out the geometry to find the

deflection at E given the deflections

at B and D.


Sample Problem 2.1

2 - 18

Free body: Bar BDE

ncompressioF

F

tensionF

F

M

AB

AB

CD

CD

B

kN60

m2.0m4.0kN300

0M

kN90

m2.0m6.0kN300

0

D

SOLUTION:

Displacement of B:

m10514

Pa1070m10500

m3.0N1060

6

926-

3

AE

PLB

mm 514.0B

Displacement of D:

m10300

Pa10200m10600

m4.0N1090

6

926-

3

AE

PLD

mm 300.0D


Sample Problem 2.1 Suppose the new positions of B, D

and E are B, D and E respectively. So that the deformations point are BB, DD and EE.

Since BDE is a rigid member, the line joining B, D and E or B, D and E will be straight line and will follow all the rules of straight line.

Suppose coordinate points B(0,B), D(200,-D) and E(600,-E); where only E is unknown.

It can be computed by equating the slopes of BD and DE.

E = 1.928 mm


Home work Problems

2.1

2.2

2.3

2.14


Relative end displacement The previous examples had

one end attached to a fixed support. In each case, therefore, the deformation of the rod was equal to the displacement of its free end

What will be the of the bar of both ends are moving?

/ = =

Example: compute, how much point B will move downward? If the applied load P was 10 kN. For L = 1.0 m, A = 2.5 mm2, and E = 200 GPa B/A = 20 mm A = 10 mm So B = 30 mm


Quizzes Started from the last week (it is 4th week now)

Academic

Week # Quiz # Date

(Thursday) Remarks

3 1 25-02-2015 Done with date change

5 2 12-03-2015 Date and time updated

7 3 26-03-2015 -do-

9 4 16-04-2015 -do-

11 5 30-04-2015 -do-

13 6 14-05-2015 -do-

15 7 28-05-2015 -do-


1st submission of note books Deadline

Monday, March 16, 2015, 01:00 PM.

mm222 lec 10-12

Documents

ksi spring

psi spring

uall spring

ultimate strength normal

psi problem

relevant dimension spring

kips c

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