CHAPTER 4 (b)

BENDING STRESS

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Learning Outcomes:At the end of this lecture, student should be

able to;

Understand the bending equationCalculate the bending momentCalculate the bending stressDraw the stress distribution

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Bending Equation

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Bending Stress Guideline (Refer to BMD)

Sogging (Positive) Hogging (Negative)

Upper surface = Compression

Lower surface = Tension

Upper surface = Tension

Lower surface = compression

Base Base

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Example:

4 mx

120 mm

30 mmx

3 KN/m

30 mm

120 mm

BA

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The beam of the indicated section is loaded with a distributed load, 3KN/m along the beam. Based on the diagram shown earlier, determine the;

i) Centroid about neutral axis, yii) Moment inertia about neutral axis, Ixx

iii) Maximum bending moment, Mmax

iv) Maximum tensile stress and maximum compressive stress, σmax

v) Draw the stress distribution

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Answer

i) Centroid about neutral axis, y = 52.5 mm @ 0.0525 m

ii) Moment inertia about neutral axis Ixx = 14.71 x 106 mm4 @ 14.71 x 10-6 m4

Refer answer

at earlier exercise.

(Chapter 4a)

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iii) Maximum bending momentFormula;

Mmax = wL2 8

= (3 x 103) (42) 8 = 48 x 103

8 = 6 KNm #

This formula is only VALID when;

i)The beam is simply supported at both ends.ii) Carry uniformly distributed load along the beam.

* Other than this conditions, the Mmax is obtained by referring to Chapter 3.

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Based on the BMD, bending moment occurs in the positive range, therefore, the beam will be in sogging condition.

Upper surface = Compression

Lower surface = Tension

BaseBMD

SFD

Sogging (Positive)

0

0

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iv) Bending stress

We know that the beam is in sogging condition, thus;

PN

y = 52.5

150 mm 120 mm

30 mm

y lower = 52.5 mm (tensile)y upper = 150 – 52.5 = 97.5 mm (compress)

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iv) Bending stressIxx = 14.71 x 10-6 m4 yupper = 97.5 mm @ 0.0975 m

Mmax = 6 KNm ylower = 52.5 mm @ 0.0525 m

a) Max tensile stress

σt = Mylower

I = (6 x 103)(0.0525) 14.71 x 10-6

= 21.4 MPa #

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b) Max compressive stress

σc = Myupper

= (6 x 103)(0.0975) 14.71 x 10-6

= 39.8 MPa #

I

v) Stress distribution

Cross sectional Stress distribution

97.5 mm

52.5 mm

- 39.8 MPa

+ 21.4 MPa

PNσ = 0

150mm

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120 mm

Exercise120 mm

x

120 mm

40 mm

20 mm

x

40 mm

100 mm

30 KN/m

A B

5 m

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Based on the diagram, determine the;

i) Centroid about neutral axis, yii) Moment inertia about neutral axis, Ixx

iii) Maximum bending moment, Mmax

iv) Maximum tensile stress and maximum

compressive stress, σmax

v) Draw the stress distribution

Refer answer

at earlier exercise.

(Chapter 4a)

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