mechanics of materials lec02

LECTURE 2

NOOR MAZNI ISMAIL

FACULTY OF MANUFACTURING ENGINEERING

Example 02: Determine the internal loads

Q2: Determine the resultant internal loadings acting on the cross section at B of the

pipe. The pipe has a mass of 2 kg/m and is subjected to both a vertical force of 50 N and

a couple moment of 70 N·m at its end A. It is fixed to the wall at C.

Solution:Free-Body Diagram

( )( )( )( )( )( ) N 525.2481.925.12

N 81.981.95.02

==

==

AD

BD

W

W

Calculating the weight of each segment of pipe,

Applying the six scalar equations of equilibrium,

( )( )( )( ) (Ans) N 3.84

050525.2481.9 ;0

(Ans) 0 ;0

(Ans) 0 ;0

=

=−−−=

==

==

∑∑∑

zB

zBz

yBy

xBx

F

FF

FF

FF

( ) ( ) ( ) ( ) ( )( )

( ) ( ) ( ) ( )( )

( ) ( ) (Ans) 0 ;0

(Ans) mN8.77

025.150625.0525.24 ;0

(Ans) mN3.30

025.081.95.0525.245.05070 ;0

==

⋅−=

=++=

⋅−=

=−−−+=

∑

∑

∑

zBzB

yB

yByB

xB

xBxB

MM

M

MM

M

MM

Stress1. NORMAL STRESS

2. SHEAR STRESS

3. ALLOWABLE STRESS

Stress� Distribution of internal loading is important in mechanics of

materials.

� The intensity of internal force at a point is called

stress, also known as force per unit area.

� SI Unit: N/m2

Types of StressNormal Stress, σ (sigma)

� Force per unit area acting normal to ΔA. Since ΔFz is normal to

the area, then

Shear Stress, τ (tau)

� Force per unit area acting tangent to ΔA. The shaer stress

components,

A

Fz

Az ∆

∆=

→∆ 0limσ

A

F

A

F

y

Azy

x

Azx

∆

∆=

∆

∆=

→∆

→∆

0

0

lim

lim

τ

τ

Normal force acting “pull” on ΔA, tensile stress.

If acting “pushes” on ΔA, compressive stress.

Note:

the subscript notation z specifies the orientation of the area ΔA, and x and y indicate the axes along which shear stress acts.

1.0: Average Normal Stress in an Axially Loaded Bar

� When a cross-sectional area bar is subjected to axial

force through the centroid, it is only subjected to normal

stress.

� Stress is assumed to be averaged over the area.

Average Normal Stress Distribution

� When a bar is subjected to a

constant deformation,

Normal Stress Equilibrium

� 2 normal stress components

that are equal in magnitude

but opposite in direction.

A

P

AP

dAdFA

=

=

= ∫∫

σ

σ

σ

σ = average normal stressP = resultant normal forceA = cross sectional area of bar

Example 01The bar has a constant width of 35 mm and a thickness of 10 mm. Determine the maximum average normal stress in the bar when it is subjected to the loading shown.

Solution:By inspection, different sections have different internal forces.

Graphically, the normal force diagram is as shown.

Solution:

By inspection, the largest loading is in region BC,

kN 30=BCP

Since the cross-sectional area of the bar is constant, the largest average normal stress is

( )( )( )

(Ans) MPa 7.8501.0035.0

10303

===A

PBCBCσ

1 N/m2 = 1 Pa

3kN/m 80=stγ

Example 02The casting is made of steel that has a specific weight of

. Determine the average compressive stress acting at points A and B.

Solution:By drawing a free-body diagram of the top segment,the internal axial force P at the section is

( )( ) ( )kN 042.8

02.08.080

0 ;0

2

=

=−

=−=↑+ ∑

P

P

WPF stz

π

The average compressive stress becomes

( )(Ans) kN/m 0.64

2.0

042.8 2

2===

πσ

A

P

2.0: Average Shear Stress� The average shear stress distributed over each sectioned

area that develops a shear force.

� 2 different types of shear:

A

Vavg =τ

τ = average shear stressV = internal resultant shear forceA = area at that section

a) Single Shear b) Double Shear

� If two parts are joined together, the applied load may

cause shearing of of the material, it is generally assumed

that an average shear stress acts over the cross-sectional

area.

� Shear Stress Equalibrium

All shear stresses must have equal magnitude and be directed

either toward or away from each other at opposite edges of

the element.

Example 01The inclined member is subjected to a compressive force of 3000 N. Determine the

average compressive stress along the smooth areas of contact defined by AB and BC,

and the average shear stress along the horizontal plane defined by EDB.

The shear force acting on the sectioned horizontal plane EDB is

Solution:

N 1800 ;0 ==→+ ∑ VFx

Average compressive stresses along the AB and BC planes are

( )( )

( )( )(Ans) N/mm 20.1

4050

2400

(Ans) N/mm 80.14025

1800

2

2

==

==

BC

AB

σ

σ

( )( )(Ans) N/mm 60.0

4075

1800 2==avgτ

Average shear stress acting on the BD plane is

The compressive forces acting on the areas of contact (inclined member) are

( )( ) N 240003000 ;0

N 180003000 ;0

5

4

5

3

=⇒=−=↑+

=⇒=−=→+

∑∑

BCBCy

ABABx

FFF

FFF

AVERAGE STRESS:

INTERNAL LOADING:

3.0: Allowable Stress & Factor of Safety

� Proper design is critical to ensure the

structure is safe.

� Many unknown factors that influence

the actual stress in a member.

� A factor of safety is needed to

obtained allowable load. Restrict the

applied load.

� The factor of safety (F.S.) is a ratio of

the failure load divided by the

allowable loadallow

fail

allow

fail

allow

fail

SF

SF

F

FSF

τ

τ

σ

σ

=

=

=

.

.

.

Example 01

The control arm is subjected to the loading. Determine to the nearest 5 mm the required diameter of the steel pin at C if the allowable shear stress for the steel is . Note in the figure that the pin is subjected to double shear.

Solution:For equilibrium we have

MPa 55=allowableτ

( ) ( ) ( )( )( )( ) kN 3002515 ;0

kN 502515 ;0

kN 150125.025075.0152.0 ;0

5

3

5

4

5

3

=⇒=−−=+↑

=⇒=+−−=+→

=⇒=−==+

∑∑∑

yyy

xxx

ABABC

CCF

CCF

FFM

Solution:

The pin at C resists the resultant force at C. Therefore,

( ) ( ) kN 41.3030522 =−=CF

mm 8.18

mm 45.2462

m 1045.2761055

205.15

2

26

3

2

=

=

×=×

== −

d

d

VA

allowable

π

τ

The pin is subjected to double shear, a shear force of 15.205 kN acts over its cross-sectional area between the arm and each supporting leaf for the pin.

The required area is

Use a pin with a diameter of d = 20 mm. (Ans)

25 October 2011 19

END OF LECTURE 02Thank you!

25 October 2011 20

ASSIGNMENT 1 !