ch 4 bending...

8/11/2019 Ch 4 Bending Stressesnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnn

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CHAPTER 4: BENDINGSTRESSES



Normal Stresses in Beams -Bending

2



Normal Stresses in Beams

the normal stress in the member , which occurs at a point on the cross-sectfarthest away from the neutral axis ( N/m 2 = Pa )

M the resultant internal moment , determined from the method of section

equations of equilibrium and calculated about the neutral axis of the cross sectiony the perpendicular distance from the neutral axis to a point farthest aneutral axis. This is where max acts ( m )

I the moment of inertia of the cross sectional area about the neutral axis ( m 4 )R radius of curvature ( m )

E Young Modulus ( N/m 2 = Pa )

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M = = EI y RFlexure formula :



Maximum Stresses at a Cross-Section

Relationships betweensigns of bendingmoments anddirections of normalstresses;(a) +ve bendingmoment.(b) -ve bendingmoment.

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Properties of Plane Areas5



Properties of Plane Areas.6



Combination Shape

IX/Y moment of inertia

A cross sectional areah length between neutral axis

I = (I G + Ah) 1+2+3+.

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Neutral Axis

UnsymmetricSymmetrical

=yAA

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c = y = 0.2/2 = 0.1 m

3kNm

100mm

200mm

I = bh 3/12 = 0.1 x 0.2 3 / 12 = 6.667x10 -5 m4

= McI

= (3kNm)(0.1m)/(6.667x10 -5 m 4)

= 4.4998 MPa (compression and tension)

12

Example 1: Cantilever Beam



EXAMPLE 213



EXAMPLE 314



Example 4: Simply Supported Bea

40mm

10mm

100mm

5kN/m

6m

15



Area A :

y A = 0.01/2 = 0.005 m;

A A = 0.04 x 0.01 = 0.0004 m 2;

Area B :

yB = 0.01 + 0.08/2 = 0.05 m;

AB= 0.01 x 0.08 = 0.0008 m 2

Area C :

yC = 0.01 + 0.08 + 0.01/2 = 0.095 m;

AC = 0.1 x 0.01 = 0.001 m 2

Step 1: Identify value c from neutral axi40mm

10mm

100mm

10m

10

80m

A

B

C

Neutral Axis

C1

C2

c1 = yA/ A

c1 = [y A A A + y B AB + y C AC ]/[A A + AB + AC]

c1 = [(0.005 x 0.0004) + (0.05 x 0.0008) + (0.095 x 0.001)] / [0.0004 + 0.0008 + 0.001]

c1 = 0.062 m16



Area A :

h A = 0.062-0.005=0.057m;

IA=3.33x10 -9m 4

Area B :

h B = 0.062-0.050=0.012m;

IB= 4.267x10 -7m 4

Area C :

h C = 0.095-0.062=0.033m;

IC=8.33x10 -9m 4

The shape for A,B and C are rectangular

Step 2: Determine moment of inertia

40mm

10mm

100mm

10mm

10mm

80mm

A

B

C

Neutral Axis

C1 = 0.062m

C2 = 0.038m

Total of combination shape; I = (IG + Ah)I = (I + Ah) A + (I + Ah) B + (I + Ah) C

= 1.30293x10 -6 m 4 + 5.419x10 -7 m 4 + 1.09733x10

= 2.942x10-6

m4

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Step 3: Determine maximum bending stresse5kN/m

6m

0 kNm

6m

+22.5kNm

+30kN

-30kN

Shear Force Diagram

Moment Diagram

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Step 4: Calculate normal stress

I = 2.94216x10 -6m 4

Mmax =22.5kNm (+ve) max = Mc 1 I

= (22.5kN)(0.062)/(2.94216x10 -6)

= 474.14x10 6N/m(compression)

= Mc 2 I

= (22.5kN)(0.038)/(2.94216

= 290.60x10 6N/m(tension

compress

tension

40mm

10mm

100mm

A

B

C

N

C1 = 0.062m

C2 = 0.038m

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Maximum bending stress20

L/2L/2

L L

L

Mmax =FL/4

Mmax =FL/8

(F unit N)

(F un it N/m)

F

F

F

L



Example 6

A cast-iron machine part is acted upon by a 3 kNm couple. Know= 165 GPa and neglecting the effects of fillets, determine (a) the

maximum tensile and compressive stresses, (b) the radius of

SOLUTION



Example 6: Continue

SOLUTION:

Based on the cross section geomcalculate the location of the sectiand moment of inertia.

I I A A y

Y x

Apply the elastic flexural formuthe maximum tensile and comprstresses.

I

Mcm

Calculate the curvature

1

SOLUTION B d th ti t l l t th l ti f th



SOLUTION: Based on the cross section geometry, calculate the location of thesection centroid and moment of inertia.

mm383000

10114 3

A A y

Y

2

300020120030402

50180090201

mm,mmArea,

A

y

49-43

12123

121

231212

m10868mm10868

301218002090

I

d Abhd A I I x

Example 6: Continue

ch 4 bending...

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