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☻. 1.0 Axial Forces. 2 .0 Bending of Beams. Now we consider the elastic deformation of beams (bars) under bending loads. M. M. www.engineering.auckland.ac.nz/mechanical/MechEng242. Normal Force:. F n. F n. S.B. Shear Force:. F t. F t. Bending Moment:. M t. M t. K . J .

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Presentation Transcript
slide1

1.0 Axial Forces

2.0 Bending of Beams

Now we consider the elastic deformation of beams (bars) under bending loads.

M

M

www.engineering.auckland.ac.nz/mechanical/MechEng242

slide2

Normal Force:

Fn

Fn

S.B.

Shear Force:

Ft

Ft

Bending Moment:

Mt

Mt

K.J.

Torque or Twisting Moment:

Mn

Mn

Application to a Bar

slide3

Excavator

Atrium Structure

Yacht

Car Chassis

Examples of Devices under Bending Loading:

slide4

(Refer: B,C & A –Sec’s 6.3-6.6)

2.2 Stresses in Beams

sx

sx

P

x

Mxz

Mxz

2.3 Combined Bending and Axial Loading

(Refer: B,C & A –Sec’s 6.11, 6.12)

P1

P2

2.4Deflections in Beams

(Refer: B,C & A –Sec’s 7.1-7.4)

2.5Buckling

(Refer: B,C & A –Sec’s 10.1, 10.2)

2.0 Bending of Beams

2.1 Revision – Bending Moments

(Refer: B,C & A – Sec’s 6.1,6.2)

slide5

12 kN

Q

(SFD)

3m

3m

0

(BMD)

M

0

2.1Revision – Bending Moments

RECALL…

(Refer: B, C & A – Chapter 6)

Last year Jason Ingham introduced Shear Force and Bending Moment Diagrams.

slide6

y

x

Deflected Shape

B

A

Mxz

Mxz

RAy

RBy

Mxz

Mxz

What stresses are generated within, due to bending?

Consider the simply supported beam below:

(Refer: B, C&A – Sections 1.14, 1.15, 1.16, 6.1)

Radius of Curvature, R

P

slide7

P

A

B

W

u

Mxz

Mxz

RAy

RBy

Recall: Axial Deformation

Bending

Load (W)

Bending Moment (Mxz)

Axial Stiffness

Flexural Stiffness

Extension (u)

Curvature (1/R)

slide8

sx (Compression)

Mxz

Mxz

sx=0

sx (Tension)

sx is NOT UNIFORM through the section depth

sxDEPENDS ON:

Axial Stress Due to Bending:

Mxz=Bending Moment

y

x

Beam

Unlike stress generated by axial loads, due to bending:

(i) Bending Moment, Mxz

(ii) Geometry of Cross-section

slide9

Mxz

Mxz

-vesx

Qxy

Qxy

+vesx

+VE (POSITIVE)

“Happy” Beam is +VE

“Sad” Beam is -VE

Qxy=Shear Force

Sign Conventions:

y

Mxz=Bending Moment

x

slide10

Mxz=P.L

RAy=P

P.L

Mxz

P

Qxy

P

x

Mxz

Mxz

Mxz

Qxy

Qxy

Qxy

Q & M are POSITIVE

Example 1: Bending Moment Diagrams

P

y

x

A

B

L

slide11

P.L

Mxz

Qxy

x

P

P

+ve

Qxy

0

Mxz

0

-ve

-P.L

To find sx and deflections, need to know Mxz.

P

y

L

x

A

B

Shear Force Diagram (SFD)

Bending Moment Diagram (BMD)

slide12

a

b

x

P

a

Mxz

A

Qxy

Where can only be +VE or ZERO.

y

P

Example 2: Macaulay’s Notation

x

A

C

B

slide13

(i) When

0

(ii) When

Mxz

BMD:

Eq.

1

Eq.

2

+ve

0

A

C

B

y

P

a

b

x

A

C

B

x

1

2

slide14

Distributed Load w per unit length

x

RAy=wL

wL2

Mxz=

2

Mxz

wx

Qxy

wL

wL2

Mxz

2

Qxy

Example 3: Distributed Load

y

A

B

x

L

slide15

Mxz

L

BMD:

0

-ve

x

-wL2

2

slide16

Apart from the revision problems on Sheet 4, you might try these sources:

  • B, C & A

Worked Examples, pg 126-132

Problems, 6.1 to 6.8, pg 173

  • Jason Ingham’s problem sheets:

www.engineering.auckland.ac.nz/mechanical/EngGen121

Summary – Is anything Necessary for Revision

Generating Bending Moment Diagrams is a key skill you must revise. From these we will determine:

  • Stress Distributions within beams,
  • and the resulting Deflections