1-1 stress analysis and design

7/31/2019 1-1 Stress Analysis and Design

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EM302/ EV202

Ms. Chu May Yen


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Review: The following list of topics covers the material we need to

have mastered (or be prepared to review) for this course:

Use of significant figures and appropriate units in the solution ofproblems.

Force & moment vectors. Reactions at supports in simple structures. Free body diagrams. The use of equilibrium equations. The idea of statically determinate and statically indeterminate

problems. Centroid of composite plane areas; moments of areas.

Area moments of inertia of composite plane areas.

* All of the above are usually covered in a course in statics.


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Briefly: in 2-Dimensional, the following equilibriumequations:

Fx = 0

Fy = 0

M = 0 These are 3 Linear equations we can solve for

a maximum of 3 unknowns.

* if there are more unknowns, then the number ofequations of statics, we called it:

A statically Indeterminate System


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Some examples to refresh your memories:

Example 1:

To solve these types of problems, using theanalysis we learn in engineering static,

Consider the joint C, Using Fx = 0 and Fy = 0.

And find the forces FAC and FBC


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A

B

C

FCB

FCA

CFCB

FCA

B C

A


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At joint C:

Fy = 0 FCA Sin 45 20 = 0

FCA = 20/ Sin 45= 28.2845 (Tension)

Fx = 0 FCB = 28.2845 x Cos 45

FCB = 20/ Cos 45= 20.0 (Compression)


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Find the reactions

B C

3m 5m

20kN(compression)

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20 x 4/5 = 16.0 kN

20 x 3/5 = 12.0 kN


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Solutions:

Fx = 0,Bx = 12.0 kN

MB = 0, 8Cy = 3 (16)

Cy = 6.0 kN

Fy = 0, By = 10.0 kN


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Consider example No. 1

We found that member BC was incompression under a force of 20kN & AC,was in tension, under the force of 28.3kN.

All this is good, and tells us the forces in themembers , it does not tell us whether

the given load may be .


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Will member BC under this load?

This depends on the cross-sectional area ofthe member & the material the member madeup of.

i.e


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Lets look at the following: The force , is actually distributed over the

entire area , of the cross section. It is theresultant of many elementary forces.


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The intensity of these distributed forces isequal to the force per unit area, F/A, in the

section. This force/unit area is called the in that

section, and is denoted by the Greek letter

Formula

sign, indicates sign, indicates

A

P=


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Units:

Force- N

Area- - m2

Stress- N/m2 = Pa

1kPa= 103 N/m2 = 103 Pa

1MPa= 106 N/m2 = 106 Pa

1GPa= 109 N/m2 = 109 Pa


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Considering member BC, we shall assumethat it is made of steel, and has a diameter of

25.4mm.

=20.0kN = 20 x 103N

= (25.4mm)2/4

= (0.0254m)2/4

= 5.067 x 10-4 m2

= 20 x 103N 5.067 x 10-4 m2= 39.47 x 106Pa

=


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*In order to determine whether member BC can

be used to support the given load, we mustcompare the value obtained for under thisloading with the maximum value of stresswhich may be safely applied to steel. i.e.


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2 Force member Acting along the axis called:

The forces act perpendicular to the plane ofthe section.

The stress produce is called:


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Example:

For the system loaded as shown, find theNormal stresses at the rods AB & BC,

d1 =75mm and d2 = 50mm.

160k

N

120kN 120kN1000mm

750mm

A

B

C


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For Rod AB:

= 81,487.3kN/m2

= 81.49 MPa (Tension)

A

F=

4)05.0(

1602

kN=

F=160

P=160


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For Rod BC,

At point B:

Fy = 0120 + 120 160 Cy = 0

Cy = 80kN

= 18108.3 kN/m2

= 18.11 MPa (Compression)

4)075.0(

802

kN=

C

B

A

160

80

80

120 120160


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Find stresses in members AB & BC.

Solution: Draw Free Body Diagram for the system.A

B

C

P=30 kN

D1=30mm

D2= 50mm

40k

N


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Find the length, L for the plate below

A = , P = ,

Plate area = 125mmx L

100 kN


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0.125 m x L = 0.0333

L= 0.0333/0.125= , or L =

A

P=


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1-1 stress analysis and design

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