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INCLINED BENDING - PowerPoint PPT Presentation


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INCLINED BENDING. z. z. z. . y. y. y. . If bending moment does not coincide with any of principal central axis of cross-section inertia we have to deal with inclined bending. Direction of bending moment vector. Plane of load acting. +. z. z. z. z. z. . y. y. y. y. y.

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slide2

z

z

z

y

y

y

If bending moment does not coincide with any of principal central axis of cross-section inertia we have to deal with inclined bending.

Direction of bending moment vector

Plane of load acting

+

slide3

z

z

z

z

z

y

y

y

y

y

We use superposition principle

Jy>Jz

For neutral axis:

Jy<Jz

Neutral axis

slide4

z

y

is represented by the equation of the plane

Stress distribution:

This plane cuts thorough cross-section plane at the neutral axis

Tension zone

Compression zone

Neutral axis

slide6

z

Angle

80o

70o

60o

50o

40o

y

30o

20

10o

h/b=1,41

00o

Loading plane

Jy/Jz =2

Neutral axis

Bending moment vector

slide7

z

80o

70o

60o

50o

40o

y

30o

20

10o

h/b=1,41

00o

Jy/Jz =2

Loading plane

Neutral axis

Bending moment vector

slide8

z

80o

70o

60o

50o

40o

y

30o

20

10o

b/h=1,41

00o

Jy/Jz =1/2

Loading plane

Neutral axis

Bending moment vector

slide9

z

z

z

y

y

y

Zmax =a/2=0,5a

Zmax =a/2½=0,7a

h=b=a

Ideal shape: independent of plane of loading angle maximum stress is the same!

Jy/Jz =1

Płaszczyzna obciążenia

Oś obojętna

Wektor momentu

slide11

z

Angle

80o

70o

60o

50o

40o

y

30o

20

10o

h/b=1,41

00o

Loading plane

Jy/Jz =2

Neutral axis

Bending moment vector

slide12

z

Angle

80o

70o

60o

50o

40o

y

30o

20

10o

h/b=1,41

00o

Loading plane

Jy/Jz =2

Neutral axis

Bending moment vector

slide13

z

z

y

y

α=45°

Zmax =a/2=0,5a

Zmax =a/2½=0,7a

h=b=a

Ideal shape: independent of plane of loading angle maximum stress is the same!

Jy/Jz =1

Płaszczyzna obciążenia

Oś obojętna

Wektor momentu