JJ310 STRENGTH OF MATERIAL Chapter 4(b)Bending Stress
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Transcript of JJ310 STRENGTH OF MATERIAL Chapter 4(b)Bending Stress
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 #
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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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