Topic1 Staticstatic

8/19/2019 Topic1 Staticstatic

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Static load design 1

Fx

FY

FZ

Design for Static Load

ObjectiveStrengthen the knowledge: static and strength of material.Know how to do analysis on component under static loadFind the relationship between these topicsKnow how to use Table in appendix A

Notation

Axis: Right Hand Rule To maintain the position of X, Y and Z axis

To show the rotation direction of Moment and Torsion

Force: arrow with single headEg: Fx = 10 kN, Fy = 20 kN and Fz = 30 kN

F = (10i + 20j + 30k) kN


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Moment and torsion: arrow with two head on direction of rotation.Eg: Mx = 10 Nm, My = 20 Nm and Mz = 30 Nm

M = (10i + 20j + 30k) Nm

Are torsion and moment same and why they are named differently?

`

Mx

My

Mz

Mx

My

Mz

Mx

My

Mz


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Static:

000 z y x F F F

000 z y x M M M

Discuss the following problem


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Based on T, which one has higher force value (F1 or F2). If pulleydiameter is D, find F1 in function of D and T.

Find F1 in function of D (sprocket diameter) and d (pulley diameter)


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Pulley

Chain

Sprocket

Ball Bearing

V-Belt

F=600 N

F1=1.75F2

F2Shaft rotated at

350 rpm (ccw)

F=600 N

Isometric View

1. 3-80 pp 140

2. 3-84 pp 140


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3. 3-68 pp 137

FBD Figure 3-72, 3-73


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INDETERMINATE PROBLEM


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STRESS ANALYSIS

Axial Load (P)

Axial Stress A

P

A: cross sectional area

Stresses on element

Shear load (V) : can be neglected when the part is subjected tobending or torsion

Transverse shear stress A

V

A: cross sectional area

P

V


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Moment

Bending stress

Z

M

I

Mc

c: distance from the center axisI: second moment of area (last page, Table

A-6 to A-8 pp1008 - 1012)Z: section modulus (Table A-6 to A-8)

Stresses on element

M


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Torsion

T

Note:In design we are responsible our design is safe against failure.Therefore, parameters such A, I, Z and J play important role as wellas the loads themselves.

NoteWe should identify the most critical element in order to proceed withour analysis.

We should calculate the maximum stresses resulted from stress onthe element of the most the critical position. How?

Shear Stress J Tr

R: distance from center axisJ: second polar of momentinertia


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Compounded Load

P M Compounded

Which element is most critical and why?

M

M

P

P

A

B, D

C

A

B, D

C

A

B, D

C


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T M Compounded


M

M T

T

A

B, D

C

A

B, D

C

A

B, D

C


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M1 M2 Compounded


M1

M1

M2

M2

A

B, D

C

A

B, D

C

A

B, D

C


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Mohr CircleMohr Circle is used to represent graphically the state of stress on theentire structure.

Element

Refer to Eq 3-13, 3-14 pp 77, 78 to calculate the principal stressRead Mohr Circle background theory: Section 3-6 (pp 70 – 84)

τxy

sxy


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STATIC FAILURE THEORIES (Chap 5)

Ductile materials (yield criteria) Maximum shear stress (MSS): Sec 5-4 pp 211 Distortion energy (DE): Sec 5-5 pp 213 Ductile Coulomb-Mohr (DCM): Sec 5-6 pp 219

Brittle materials Maximum normal stress (MNS) Sec 5-8 Brittle Coulomb-Mohr (BCM) Sec 5-9

Modified Mohr (MM) Sec 5-9

Maximum shear stress Theory (MSS)

Yielding begins whenever the maximum shear stress in any elementequals or exceeds the maximum shear stress in a tension test

specimen of the same material when that specimen begins to yield.

LIMIT DUE YIELDING

LIMIT

DUET

O

SHEAR

SytSyc


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Therefore, any Mohrcircle generated fromthe most critical element

must lie inside this limit

Failure occurs when the Mohr

circle lies outside the limits

When the limits are plotagainst sa and sb.

Case1: when sa and sb are +ve

Case2: when sa is +ve and sb is –ve

Case3: when sa and sb are -ve

Syt Syc

SytSyc


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Case 1

Case 2

Case 3

QA

QB

Sy

Sy

Syc

Syc


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Distortion Energy Theory (DE)

The effective stress is called Von Misses

222

22

3'

'

xy y y x x

B B A A

If sy = 0

223' xy x

Therefore failure happens when s’ > Sy. To plot the graph sa against sb for DE Theory, s’ = Sy.

QA

QB

Sy

Sy

Syc

Syc


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Load line and its meaning on MSS and DE.Line when it is drawn from (0, 0) to another point and this line acrossthe (sa, sb) of the Mohr Circle of the element.Intersection between load line and the MSS and DE Curve: point of

yielding according the theories defn. Any stresses exceed theseintersecting point is considered fail due to yielding ( FS < 1 ).Before the intersection point ( FS > 1)

At the intersection point (FS = 1)Beyond the intersection point ( FS < 1 ) NO GO

Pure Shear LineWhen the element is subject to shear only. The intersection of this

line with MSS and DE curve will remark the relationship between yieldstress and shear yield stress. ThereMSS : Ssy = 0.5 SyDE: Ssy = 0.577 Sy

s A

sB

y

Sy

Syc

Syc

Pure Shear Ln

Load Line


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Static failure using mathematical approach

Three-dimensional Stresses (Section 3-7 pp 82)

Principal Stresses s3 < s2 < s1Maximum Shear Stress

When there is no pressure, sz is equal to zero, therefore one of

principal stress must equal to zero.

τ xy

s xy

s1s2s3


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What is the value of s3, s2, s1 and τmax for a, b and c?

MSST theory

Maximum shear 2

31

max

Allowable shear stress n

S S

y

sy2

Safety factor31

yS n

DET


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Total stress22

3' xy x

22' B B A A

Safety factor '

yS n

*NOTEPlease differentiate between

stresses on element xy x ,

2D Principal stresses B A ,

3D Principal stresses 321 ,,

Brittle Material


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Note:Please take note of the following appendix(All the tables on the appendices should be at your finger tips)

Table A-6 to A-8

Table A-9


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Table A-18

Table A-20 to A-22

Topic1 Staticstatic

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