normal stress1 normal stress (1.1-1.5) mae 314 – solid mechanics yun jing

Normal Stress 1

Normal Stress (1.1-1.5)

MAE 314 – Solid Mechanics

Yun Jing

2

Pretest

This is a structure which was designed to support a 30kN load, it consists of a boom AB and of a rod BC. The boom and rod are connected by a pin at B and are supported by pins and brackets at A and C. (1) Is there a reaction moment at A? why? (2) What is the reaction force in the vertical direction at A? (3) What is the internal force in AB? (4) What is the internal force in BC?

Normal Stress 3

Statics Review

Pins = no rxn moment

Normal Stress 4

Statics Review

• Ay and Cy can not be determined from these equations.

kN30

0kN300

kN40

0

kN40

m8.0kN30m6.00

yy

yyy

xx

xxx

x

xC

CA

CAF

AC

CAF

A

AM

• Solve for reactions at A & C:

Normal Stress 5

Statics Review

• Results: kN30kN40kN40 yx CCA

0

m8.00

y

yB

A

AM

• Consider a free-body diagram for the boom:

kN30yC

substitute into the structure equilibrium equation

See section 1.2 in text for complete static analysis and review of method of joints.

Bx

Normal Stress 6

Normal Stress 7

Introduction to Normal Stress Methods of statics allow us to determine forces and moments in a structure, but how do we determine if a load can be safely supported? Factors: material, size, etc. Need a new concept….Stress

Normal Stress 8

Introduction to Normal Stress

Stress = Force per unit areaA

F

Normal Stress 9

If stress varies over a cross-section, the resultant of the internal forces for an axially loaded member is normal to a section cut perpendicular to the axis. Thus, we can write the stress at a point as We assume the force F is evenly distributedover the cross-section of the bar. In realityF = resultant force over the end of the bar.


A

FdA

A

F

A

lim0

Sign convention Tensile (member is in tension)Compressive (member is in compression)Units (force/area)English: lb/in2 = psi kip/in2 = ksiSI: N/m2 = Pa (Pascal) kN/m2 = kPa MPa, GPa, etc.

Normal Stress 10


00

Tensile

Compressive

Normal Stress 11

Homogenous: material is the same throughout the bar Cross-section: section perpendicular to longitudinal axis of bar

Prismatic: cross-section does not change along axis of bar

Definitions and Assumptions

F’ F

A

Prismatic Non-Prismatic

Normal Stress 12

Uniaxial bar: a bar with only one axis Normal Stress (σ): stress acting perpendicular to the cross-section. Deformation of the bar is uniform throughout. (Uniform Stress State) Stress is measured far from the point of application. Loads must act through the centroid of the cross-section.


Normal Stress 13

The uniform stress state does not apply near the ends of the bar. Assume the distribution of normal stresses in an axially loadedmember is uniform, except in theimmediate vicinity of the points ofapplication of the loads (Saint-Venant’s Principle).


“Uniform” Stress

Normal Stress 14

How do we know all loads must act through the centroid of the cross-section? Let us represent P, the resultant force, by a uniform stress over the cross-section (so that they are statically equivalent).


Normal Stress 15

Moments due to σ:

Set Mx = Mx and My = My


A

y

A

x

dAxM

dAyM

AA

AA

xdAA

dAxP

x

ydAA

dAyP

y

11

11

Equations for the centroidMM xx

MM yy

Normal Stress 16

Example Problem Can the structure we used for our statics review safely support a30 kN load? (Assume the entire structure is made of steel with a maximum allowable stress σall=165 MPa.)

Cross-section 30 mm x 50 mm

Normal Stress 17

Example Problem Two cylindrical rods are welded together and loaded as shown. Find the normal stress at the midsection of each rod.

mmd

mmd

30

50

2

1

Shearing and Bearing Stress 18

Shearing and Bearing Stress (1.6-1.8, 1.12)


Yun Jing


What is Shearing Stress?

We learned about normal stress (σ), which acts perpendicular to the cross-section. Shear stress (τ) acts tangential to the surface of a material element.

Normal stress results in a volume change.

Shear stress results in a shape change.


Where Do Shearing Stresses Occur?

Shearing stresses are commonly found in bolts, pins, and rivets.

Free Body Diagram of Bolt

Bolt is in “single” shear

Force P results in shearing stressForce F results in bearing stress (will discuss later)


We do not assume τ is uniform over the cross-section, because this is not the case. τ is the average shear stress.

The maximum value of τ may be considerably greater than τave, which is important for design purposes.

Shear Stress Defined

A

F

A

Pave


Double Shear

Free Body Diagram of Bolt Free Body Diagram of Center of Bolt

A

F

A

F

A

Pave 2

2

Bolt is in “double” shear


Bearing Stress

Bearing stress is a normal stress, not a shearing stress.

Thus,

where Ab = projected area where bearing pressure is

appliedP = bearing force

td

P

A

P

bb

Single shear case

Read section 1.8 in text for a detailed stress analysis of a structure.

Would like to determine the stresses in the members and connections of the structure shown.

Must consider maximum normal stresses in AB and BC, and the shearing stress and bearing stress at each pinned connection

From a statics analysis:FAB = 40 kN

(compression) FBC = 50 kN (tension)

Rod & Boom Normal StressesThe rod is in tension with an axial force of 50 kN.

The boom is in compression with an axial force of 40 kN and average normal stress of –26.7 MPa.

The minimum area sections at the boom ends are unstressed since the boom is in compression.

MPa167m10300

1050

m10300mm25mm40mm20

26

3

,

26

N

A

P

A

endBC

At the flattened rod ends, the smallest cross-sectional area occurs at the pin centerline,

At the rod center, the average normal stress in the circular cross-section (A = 314x10-6m2) is BC = +159 MPa.

Pin Shearing StressesThe cross-sectional area for pins at A, B,

and C,

262

2 m104912

mm25

rA

MPa102m10491

N105026

3

,

A

PaveC

The force on the pin at C is equal to the force exerted by the rod BC,

The pin at A is in double shear with a total force equal to the force exerted by the boom AB,

MPa7.40m10491

kN2026,

A

PaveA

Divide the pin at B into sections to determine the section with the largest shear force,

(largest) kN25

kN15

G

E

P

P

MPa9.50m10491

kN2526,

A

PGaveB

Evaluate the corresponding average shearing stress,

Pin Shearing Stresses

Pin Bearing Stresses

To determine the bearing stress at A in the boom AB, we have t = 30 mm and d = 25 mm,

MPa3.53mm25mm30

kN40

td

Pb

To determine the bearing stress at A in the bracket, we have t = 2(25 mm) = 50 mm and d = 25 mm,

MPa0.32mm25mm50

kN40

td

Pb


Equilibrium of Shear Stresses Consider an infinitesimal element of material. Apply a single shear stress, τ1. Total shear force on surface is (τ1)bc. For equilibrium in the y-direction, applyτ1 on (-) surface. For moment equilibrium about the z-axis,apply τ2 on top and bottom surfaces. Moment equilibrium equation about z-axis: Thus, a shear stress must be balanced by three other stresses for the element to be in equilibrium.

1

2

2

1

bacabc )()( 21

21


Equilibrium of Shear Stresses

1

2

2

1

Face DirectionShear Stress

+ + +

+ - -

- - +

- + -

What does this tell us? Shear stresses on opposite (parallel) faces of an element are equal in magnitude and opposite in direction. Shear stress on adjacent (perpendicular) faces of an element are equal in magnitude and both point towards or away from each other.

Sign convention for shear stresses Positive face – normal is in (+) x, y, or z direction Negative face - normal is in (-) x, y, or z direction +

+

-

-


σx = stress in x-direction applied in the plane normal to x-axis σ y = stress in y-direction applied in the plane normal to y-axis σ z = stress in z-direction applied in the plane normal to z-axis τxy = stress in y-direction applied in the plane normal to x-axis τ xz = stress in z-direction applied in the plane normal to x-axis τ zy = stress in y-direction applied in the plane normal to z-axis And so on…

Define General State of Stressy

zx


There are 9 components of stress: σx, σ y, σ z, τxy, τ xz, τ yx, τ yz, τ zx, τ zy As shown previously, in order to maintain equilibrium:τ xy= τ yx, τ xz = τ zx, τ yz = τ zy There are only 6 independent stress components.

Define General State of Stressy

zx


Example Problem

A load P = 10 kips is applied to a rod supported as shown by a plate with a 0.6 in. diameter hole. Determine the shear stress in the rod and the plate.


Example Problem Link AB is used to support the end of a horizontal beam. If link AB is subject to a 10 kips compressive force determine the normal and bearing stress in the link and the shear stress in each of the pins.

ind

int

inb

1

41

2

Oblique Planes and Design Considerations 35

Oblique Planes and Design Considerations (1.11, 1.13)


Yun Jing


Stress on an Oblique Plane What have we learned so far?

Axial forces in a two-force member cause normal stresses.

Transverse forces exerted on bolts and pins cause shearing stresses.


Stress on an Oblique Plane Axial forces cause both normal and shearing stresses on planes which are not perpendicular to the axis. Consider an inclined section of a uniaxial bar. The resultant force in the axial directionmust equal P to satisfy equilibrium. The force can be resolved into components perpendicular to the section, F, and parallel to the section, V. The area of the section iscosPF sinPV

cos/cos 00 AAAA


Stress on an Oblique Plane We can formulate the average normal stress on the section as The average shear stress on the section is

Thus, a normal force applied to a bar on an inclined section produces a combination of shear and normal stresses.

2

00

coscos/

cos

A

P

A

P

A

F

cossincos/

sin

00 A

P

A

P

A

V


Stress on an Oblique Plane Since σ and τ are functions of sine and cosine, we know the maximum and minimum values will occur at θ = 00, 450, and 900.

At θ=±900 σ=0

At θ=±450 σ=P/2A0

At θ=00 σ=P/A0 (max)

At θ=±900 τ=0

At θ=±450 τ=P/2A0 (max)

At θ=00 τ=0

2

0

cosA

P

cossin0A

P

Oblique Planes and Design Considerations

Stress on an Oblique Plane What does this mean in reality?


Design Considerations From a design perspective, it is important to know the largest load which a material can hold before failing. This load is called the ultimate load, Pu. Ultimate normal stress is denoted as σu and ultimate shear stress is denoted as τu.

A

Puu

A

Puu


Design Considerations Often the allowable load is considerably smaller than the

ultimate load.

It is a common design practice to use factor of safety.

all

u

P

P

loadallowable

loadultimateSF ..

all

u

stressallowable

stressultimateSF

..all

u

stressallowable

stressultimateSF

..


Example Problem Two wooden members are spliced as shown. If the maximum allowable tensile stress in the splice is 75 psi, determine the largest load that can be safely supported and the shearing stress in the splice.


Example Problem A load is supported by a steel pin inserted into a hanging wooden piece. Given the information below, determine the load P if an overall factor of safety of 3.2 is desired.

mmd

mmc

mmb

MPa

MPa

tensioninMPa

steelu

woodu

woodu

12

55

40

145

5.7

)(60

_

_

_

normal stress1 normal stress (1.1-1.5) mae 314 – solid mechanics yun jing

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