Hafiz Kabeer Raza Research Associate

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

raza@giki.edu.pk, hkabeerraza@gmail.com, 03344025392

MM222

Strength of Materials

Lecture 22

Spring 2015

Spring 2015 By Hafiz Kabeer Raza MM222 Strength of Materials

Section moment of inertia General Formula

Where = section moment of inertia of area component A = area of area component = vertical distance (along the plane of moment) of

centroid of area component from a reference point

= vertical distance (along the plane of moment) of overall centroid

d =

2dAII x

A

AyY

Section moment of inertia, also called area moment of inertia

Distance of centroid from a reference point

Spring 2015 By Hafiz Kabeer Raza MM222 Strength of Materials

Exercise

b h A d Ad2 +Ad2

1 30 40 1200 20 24000 18 388800 160000

2 90 20 1800 50 90000 12 259200 60000

= 3000 = 114000

38 =

General Formula Where = section moment of inertia of area component A = area of area component = vertical distance (along the plane of moment) of centroid of area component from a reference point = vertical distance (along the plane of moment) of overall centroid d =

Spring 2015 By Hafiz Kabeer Raza MM222 Strength of Materials

Sample Problem 4.2

A cast-iron machine part is acted upon

by a 3 kN-m couple. Knowing E = 165

GPa and neglecting the effects of

fillets, determine (a) the maximum

tensile and compressive stresses, (b)

the radius of curvature.

SOLUTION:

Based on the cross section geometry,

calculate the location of the section

centroid and moment of inertia.

2dAIIA

AyY x

Apply the elastic flexural formula to

find the maximum tensile and

compressive stresses.

I

Mcm

Calculate the curvature

EI

M

1

Spring 2015 By Hafiz Kabeer Raza MM222 Strength of Materials

Sample Problem 4.2 SOLUTION:

Based on the cross section geometry, calculate

the location of the section centroid and

moment of inertia.

mm 383000

10114 3

A

AyY

3

3

3

32

101143000

104220120030402

109050180090201

mm ,mm ,mm Area,

AyA

Ayy

49-3

23

12123

121

23

1212

m10868 mm10868

18120040301218002090

I

dAbhdAIIx

Spring 2015 By Hafiz Kabeer Raza MM222 Strength of Materials

Sample Problem 4.2 Apply the elastic flexural formula to find the

maximum tensile and compressive stresses.

49

49

mm10868

m038.0mkN 3

mm10868

m022.0mkN 3

I

cM

I

cM

I

Mc

BB

AA

m

MPa 0.76A

MPa 3.131B

Calculate the curvature

49- m10868GPa 165mkN 3

1

EI

M

m 7.47

m1095.201 1-3

Spring 2015 By Hafiz Kabeer Raza MM222 Strength of Materials

=

1

=

We can also compute the radius of curvature by first calculating maximum strain

Spring 2015 By Hafiz Kabeer Raza MM222 Strength of Materials

Home work Problems 3.56, 3.69, 3.73, 3.74

Problems 4.2, 4.5, 4.9, 4.10, 4.18

Spring 2015 By Hafiz Kabeer Raza MM222 Strength of Materials

Problem 4.3