Bending In Beams: Professor Jonathan Q.s. Li

Lecture 7 Bending in Beams Professor Jonathan Q.S. Li

Dept. of Architecture and Civil Engineering City University of Hong Kong Tel: 34424677, E-mail: [email protected] Room: B6309

[CA3632: Mechanics of Structural and Materials]

1

Mechanics of Structures and Materials Other names:

Mechanics of Materials, Structural Mechanics, Engineering Mechanics, Applied Mechanics To prevent failure of structural or engineering components by fracture or deformation Strength and Rigidity of engineering components should be adequate to meet the loading conditions. Objective of the study of Mechanics of Materials To provide engineers with the means of analysis and design of various building components and load-bearing structures [Mechanics of Structures and Materials]

2

Recommended Text Books : Any Text Book of Mechanics of Materials or Structural Mechanics

Based on the same principles, use one of them. Mechanics of Materials F.P. Beer, E.R. Johnston, J.T. DeWolf Third Edition, McGraw-Hill, London, 2002.

Mechanics of Materials J. M. Gere, S.P. Timoshenko PWS Publishing Company, Fourth Edition, 1997.

Mechanics of Materials R.C. Hibbeler 3/E, Prentice Hall, New Jersey, 1997 [Mechanics of Structures and Materials]

3

Types of Structural Components

Tie

Strut

Cable [Mechanics of Structures and Materials]

Column

Beam 4

Types of Structural Components

Point Load Cantilever Distributed Load

Plate [Mechanics of Structures and Materials]

Shaft

Shell 5

Type of Supports and Connection for Structural Components Equivalent force system

A

A

Built-in Fixed Support

Fx

M

Fy Support force in any direction and moment. [Mechanics of Structures and Materials]

6

Type of Supports and Connection for Structural Components Equivalent force system

Fx

Pin Connection

Fy

Support force in any direction and but not moment.

[Mechanics of Structures and Materials]

7

Type of Supports and Connection for Structural Components Equivalent force system

Roller Support

Fy

Can support force only in the vertical direction [Mechanics of Structures and Materials]

8

Type of Supports and Connection for Structural Components Equivalent force system

Fx Frictionless Sliding Support

Can support force only in the horizontal direction [Mechanics of Structures and Materials]

9

Stress  Key Role in Mechanics of Materials Why ?

Because every material should be used below an Ultimate Stress (Ultimate Strength) which can be determined by materials testing.

How far below ? Allowable Stress = Ultimate Stress / Factor of Safety Factor of Safety > 1

Criteria for Design Working Stress  Allowable Stress []

[Mechanics of Structures and Materials]

10

Arrows

Must be satisfied!! In critical situations, you can consider i) changing the service conditions, by decreasing F ? ? ii) changing the size of the member, by increasing A ? ? iii) Using a stronger material , by increasing [] ? ? ii) solution for i) solution for iii) solution for

[Mechanics of Structures and Materials]

11

Use of Allowable Stress in Practice — Design of Cross Section for the known load and the selected material

A

F

 

— Calculation of the allowable load, for the known structural components

F  A 

— Check the Safety of structural components under known service conditions

[Mechanics of Structures and Materials]

12

Example: Shenzhen Universiade Center 世界大学生运动会(Global University Student Sports Game)

Simulation of Wind Effects on the Stadium by Computational Fluid Dynamics

13 13

View of mesh arrangement The computational region contains all major surrounding buildings and hills within a diameter of 10 km. The total number of grid is more than 10 millions

14

Mesh for hills

Mesh for buildings 15 15

Velocity distribution

Pressure distribution 16 16

Local distribution of velocity

17 17

Mean pressure coefficient on outer faces

WD

WD

RMS pressure coefficient on outer faces

Mean pressure coefficient on inner faces

WD

WD

RMS pressure 18 coefficient on inner faces 18

Topics in this Lecture Bending of Beams 

Pure Bending (symmetric and linearly elastic member) 

Normal Stress Distribution



Deformation



Neutral Axis and Second Moment of Area



Nonprismatic Beams



Shearing Stress in Beams



Deflection of Beams



Nonsymmetric Bending & Biaxial Bending

[Mechanics of Structures and Materials]

19

Bending of Beams (Animation)

[Mechanics of Structures and Materials]

20

Pure Bending F A a

F

C

D

•

On the studied section CD, there is no shear force Q. The bending on that section called pure bending, would only result in normal stress.

•

In contrast, nonuniform bending refers to flexure in the presence of shear forces. This is a general case.

B a

Q

A mx A

x B

Shear force

Pure bending

B

Bending moment [Mechanics of Structures and Materials]

x

21

Pure Bending Deformation o1o2     Radius of Curvature



m1m2 becomes shorter n1n2 becomes longer



o1o2 remains the same length

y

Neutral Axis y =0 m1

oo1 1 n1

oo

y

m2

oo2 2

n2

[Mechanics of Structures and Materials]

x

m1m2     y 



m1m2  o1o2 y  o1o2  22

Stress Distribution in Pure Bending y

 

Hooke’s Law

y

  E



Supposition of Elasticity y   E



z x



?

y x



[Mechanics of Structures and Materials]

23

Assumptions for bending 



Plane cross sections remain plane and perpendicular to its neutral axis after bending. Elongation occurs on one side and shortening on the opposite side.

[Mechanics of Structures and Materials]

24

Calculation of , Radius of Curvature y

  E

y

 dF  dA

 dA  0

z



A

  ydA  m x A

 ydA  0

A



E



2 y  dA  m x

A

x y

J z   y 2 dA A

dF

 dF   dA [Mechanics of Structures and Materials]

Second Moment of Area

E JZ mx

mx   y Jz 25

Stage Summary  Bending Moment yields Normal Stress in the Cross Section, which is proportional to the distance to the Neutral Axis and to the Bending Moment, but inversely proportional to the Second Moment of the y

Area.

 

M Jz

y

[Mechanics of Structures and Materials]

z x

26

Bending Moment and Normal Stress in Beams: Two Possibilities A M

A B

 

M Jz

Positive Moment

[Mechanics of Structures and Materials]

Normal Stress

y M

B

σ

Negative Moment

σ

Normal Stress 27

Sign for the shearing force Q R Q R Q R Q Q

Sign for the bending Moment M M

M

R positive clockwise

negative anticlockwise

[Mechanics of Structures and Materials]

positive

negative M

M

28

Stress and Deformation in Pure Bending y

1

M  Radius of Curvature, :  EJ z Normal Stress*:   

M Jz

z x

y

*The sign of this formula is dependent on

the definition of normal stress. In structural engineering, tension and compression stresses are defined as positive and negative, respectively. The direction of normal stress can be judged according to the bending moment direction and the location in a beam. [Mechanics of Structures and Materials]

29

Exercise 1(a simple question to warm everyone up)



A steel bar of 20mm x 65mm. rectangular cross section is subjected to two equal and opposite couples acting in the vertical plane of symmetry of the bar. Determine the value of the bending moment M that cause the bar to yield. Assume

 Y  248 N mm 2

[Mechanics of Structures and Materials]

 

M Jz

y

30

Exercise 1 1) Second Moment of Area 1 3 1 J z  bh   20  653  457708mm 4 12 12

2)Bending moment M

[Mechanics of Structures and Materials]

Jz 457708 y   248  3492664 N  mm ymax 32.5

31

Calculation of the Centroid 1   u udA c   The centroid of any Area AA  is at the point O(uc, vc), with   1  v vdA  c  AA  1 yc 

A A

v O(uc, vc)

u

y

ydA  0

means the Neutral Axis o1o2 should lie in the plane of centroid.

z

O

A

It is convenient to pose the origin O of the coordinate system on the centroid of the cross section [Mechanics of Structures and Materials]

32

Determination of the Centroid (Neutral Axish ) r

y s

h

1 1 s   rdA   r  bdr AA A0 (1) After the position of Centroid is determined, we place the origin of the coordinate system xyz on the Centroid; (2) The Neutral Axis should lie in the plane of centroid.

z O

b

s

w y

y = h s

(3) It becomes possible to calculate the Second Moment of Area;

mx   y Jz (4) Finally the Stress Distribution can be obtained.

1 1 rdA  r  bdr   AA A0

[Mechanics of Structures and Materials]

 y=s

33

Second Moment of Area Iz Different from I p in Torsion (Polar Moment of Inertia of Cross Section)

o

I p moment about a point I p   r dA 2

dA

A

r is distance to the rotating point O y

I z moment about an axis J z   y 2 dA A

y is distance to the neutral axis z

[Mechanics of Structures and Materials]

r

dA

y

z

34

Exercise 2 b y z

h

s B

A beam with trapezoid cross section is to be used. It is known for the material that the ratio of the allowable stress for tension and that for compression is       Select the optimal ratio for b/B, so that the material can be most efficiently used. (hints: calculate the location of neutral axis first)

Tips: The normal stress follows   

M y Jz

In the case of a trapezoid cross section the ratio of the maximal tensile

 max  s  stress to the maximal compressive stress is  max  h  s    If this ratio is equal to    , the tensile stress and the compressive

stress reach the allowable value and material can be most effectively used. [Mechanics of Structures and Materials]

35

Exercise 2 Solution:

s

r

h

1 1 rdA  r  2 wdr   AA A0

h B  b  2 B  r  br w  1    2  h  2h

b

A

h

2 B b   s r  rdr   B  h B  b  0  h  

2 1 B b 3 1 h    Bh 2  3 h h B  b   2 



2 h     B   B  b  3 Bb  

[Mechanics of Structures and Materials]

h

2w

w B

h 1 Bb s  1    2 3 Bb

36

Exercise 2  max   max 

1 1  s   2 6 hs 1  1 2 6

B  2b  2B  b



B b Bb   B b Bb b 2  1  B 2 

It suggest a triangular beam for   0.5

[Mechanics of Structures and Materials]

37

Centroid and Second Moment of Area of Common Geometric Shapes y h

s b

rectangular

sh 2

h

1 I z  bh 3 12

b

s

z

bh 3 Iz  36

triangular y

y

A  h2

h z

sh

s

h

y

A  bh

z

1 bh 2 1 s h 3 A

A

2

h4 Iz  12

cubic [Mechanics of Structures and Materials]

d s

z

4 d s 2 Iz 

circular



 64

d2

d4

38

Centroid and Second Moment of Area of Common Geometric Shapes y Bb A h 2 h B b s  1   2 3  Iz

b z

h B

 

y D d

A

s z

Hollow

s

 D 4



r2

Semicircular

2

d

2

y





 D 64

z s

2b

D 2

Iz 



I z  0.11r 4

Trapezoid

   3  B  b  

A

2 z 4 s  r s 3

r

s

2  B  b h 3  1  B  b    1

24

y

2a 4

[Mechanics of Structures and Materials]

d4



A   ab sb Iz 

 4

ab 3

Elliptical 39

Nonprismatic Beams 





The beam theory described above was derived for prismatic beams, that is, straight beams having the same cross section throughout their length. However, non-prismatic beams are commonly used to reduce weight and improve appearance. Such beams are found in automobiles, airplanes, machinery, bridges, building, tools and many other applications.

[Mechanics of Structures and Materials]

40

Exercise 3

A cantilever beam of length L is being designed to support a concentrated load P at the free end. The cross section of the beam is rectangular with constant width b. To assist them in designing this beam, the designers would like to know how the height of an idealized beam should vary in order that the maximum normal stress at every cross section will be equal to the allowable stress σ allow.( A beam meeting this condition is called a fully stressed beam or a beam of constant strength.). Considering only the bending stresses obtained from the flexure formula, determine the height of such a beam. [Mechanics of Structures and Materials]

41

Exercise 3 Solution: 1)Bending moment and section modulus at distance x from free end of the beam M=Px I=bhx3/12 Where hx is the height of the beam at distance x. 2)The allowable stress  allow  3) The height of beam hx 

My Px(hx / 2) 6 Px   2 3 I bhx / 12 bhx

6 Px b allow

4) Fixed end of the beam(x=L), the height hB is [Mechanics of Structures and Materials]

6 PL hB  b allow 42

Exercise 3 4) Therefore we can express the height hx in the following form

hx  hB

x L

Note: 1. At the loaded end of the beam (x=0), the theoretical height is zero, because there is no bending moment at that point. However, the beam of this kind is not practical because it is incapable of supporting the shear forces near the end of the beam. 2. Nevertheless, knowing the properties of a fully stressed beam can be an important aid to engineers when designing structures for constant maximum stress. [Mechanics of Structures and Materials]

43

Effect of Shear Stress

Shear stresses exist between the contact surfaces of two beams glued together. If there is no friction between the two beams, the upper beam will slide.

[Mechanics of Structures and Materials]

44

Shearing Stress in a Rectangular Beam Bending Moment relates to Q0 the Normal stress  Q0 mx y   0 Jz

mx0

mx1 y

Q1

t s

Q1

1

x

Shearing Force Q also exists. How is the shearing stress distributed? Select the upper part of the Beam from an arbitrary y to t, and then take the unit

q0

[Mechanics of Structures and Materials]

0

P

1

q1

x 45

Shearing Stress in a Rectangular Beam y

q0

0

y

 P 1

t

q1

y

b

t

y t

y

z

s

x For this unit we have:  Fx = 0 t

t

P    0  b  dy    1  b  dy  0 t

y P    1   0   b  dy    mx1  mx 0   b  dy Jz y y m x  Q  m x1  m x 0  Qx x [Mechanics of Structures and Materials]

t

Q P   x  y  b  dy Jz y 46

Shearing Stress in a Rectangular Beam t

q0

0

t

y

P

1

x

y

b

q1

P   yx  x  b

 yx

xy

y t

Q P   x  y  b  dy Jz y

z

b yx

s

S zy

QS zy  J zb

t

  y dA

first moment

y

J z   y 2 dA

second moment

A

Complementary Property of Shearing Stress

 xy   yx [Mechanics of Structures and Materials]

   xy

QS zy  J zb 47

Shearing Stress in a Rectangular Beam QS zy is a function of y. The maximal value for Sz occurs  J z b always at y =0. y

b

For a rectangular, beam we have 1 J z  bh3 12

y   max

  2 y 2  1       h   2 3 Q   2y    1     2 bh   h   h2

bh 2 y S z   ybdy  8 y

z

h

We can write for general cases

3Q  2 bh

h/2

 max 

QS z J zb

with

Sz 

 ydA 0

J z   y 2 dA A

[Mechanics of Structures and Materials]

48

Shearing Stress in Beams y

b

y z

h



 max

3Q  2 bh

y A 2r

B

B z

 max

SQ Q(2r 3 / 3) 4Q 4Q     4 2 Ib (r / 4)( 2r ) 3r 3A

Assume the stresses act parallel to the y axis and have constant intensity across the width of the beam from A to B

[Mechanics of Structures and Materials]

49

Two Equations for Beams  



QS zy

Q S zy  Jz b

J zb

Q

Q ly Jz

mx y Jz

l

Q

For a rectangular Beam y y

2l mmax  Ql

Q ly   12  2 bh h

1) Maximal normal stress on the edge; Maximal shearing stress on the neutral axis;





2

3Q   2 y   1     2bh   h  

[Mechanics of Structures and Materials]



h

max 2) The ratio  is very small  max 4l

50

Shearing Stress in a W-Beam W-Beam is frequently used for Application with Bending bf tf tw d

y

z

S zy Q   b Jz

y

z

Flange Web 2  1  d  2     t f   y   b tw 2  2  

S zy 2  1  d  2     y  b 2  2  

S zy

[Mechanics of Structures and Materials]

bf

2 2 1  d   d   1       t f    t f d  t f   2  2   2 b   2

S zy

51

Shearing Stress in a W-Beam bf

As Shearing stress on the Web is much larger than that on Flange

y

tf

As Approximation, we can use tw

d



z

Flange Web

[Mechanics of Structures and Materials]

Q Q   Aweb t w d

52

Example f

A W15018-shape beam is to be used for the structure. For the rolledsteel we have []=170MPa and []=100MPa. From the handbook we

C A

can find d=153mm, tw=5.8mm,

1.5m

D 3m

1.5m

B

Jz=9.2  10-6m4, Sz=1.2  10-4m3. Determine the allowable intensity of load in the middle of the beam. Description : The problem is for design

tf

bf

tw

z

d

considering normal and shear stress. First of all we need to find the maximal bending moment [Mechanics of Structures and Materials]

53

Example

f

(1) Supporting Force

R A  RB  3 f

3 R A  RB  f 2

(2) In the part AC

Q

3 f 2

mx 

mx 3 fx 2

A

Q

RA

C

1.5m

D

3m

1.5m

RB

f x A

C

(3) In the part CD RA 1.5m 3  3  x RA Q     x   f  3  x  f 2  2  2 3 1  3 1 9  m x  fx  f  x     3 x  x 2   f mx 2 2  2 2 8  (4) In the part DB Q 6-x 3 3 Q f m x  f 6  x  2 2 [Mechanics of Structures and Materials]

mx Q

B RB 54

Example (5) From the shearing force and the bending moment diagrams we find the maximal shearing force 3f/2N and the maximal bending moment 27f/8N.m. (6) Design considering Normal Stress y d 27  max  max m x  f    Jz 2J z 8

Q 3f/2

x

mx

27f/8

-3f/2 9f/4

8 2J z 16 9.2 10 6    170 10 6 f  27 d 27 0.153

 6057 Nm 1 (7) Design considering Shearing Stress

 max 

x QS z 3 S z  f    bJ z 2 t w J z

2 tw J z 2 5.8 10 3  9.2 10 6 6 1    100  10  29640 Nm f  3 Sz 3 1.2 10  4 [Mechanics of Structures and Materials]

f  6kN / M 55

Parallel Axis Theorem 



Suppose we have calculated the values of Ixx (i.e., about the centroid). The // axis theorem allows us to make use of this result to calculate Ix’x’ about any other given axis without starting from scratch. A (area) The same is also true for Iyy

y' I x ' x '  I xx  A  l y

2

I y ' y '  I yy  A  l x

2

I p'  I p  A  d 2 I x ' y '  I xy  A  l x l y

y lx d

x ly x' 56

Example: 2nd Moments of Area 

Calculate I xx (i.e., about the centroid) yy’

Ans:

3

bh I xx  12

h 2

x’ x

0

h 2 b 2

b 2 57

Example: 2nd Moments of Area 

Ans:

Calculate I x'x'

I x ' x '  I xx  A  l y

2

bh 3 h 2   (hb)( )  I xx  (Area)  (shifted y : distance) 2 12 2 y y’

h

x x’

0 b

58

2nd Moments of Area for Composite Area:



Provided the 2nd moments of area of each part is known, then the moments of inertia for the composite area equals the algebraic sum of the moments of area of all its parts. 59

Example: Composite Area 

Determine Ixx of the composite section.

A1

I xx  A y 2 dA  A y 2 dA  A y 2 dA  A y 2 dA 1

2

A1

3

1 3 2    6  1  6  1  4.5     12 1 1  3  2  8    6  13  6  1 4.52  12  12 

 122  85.33  122  329.33in

A2

4

A3 60

Example 

These two sections have the same areas but different shapes. Calculate and compare their 2nd moment of area about the x axis.

61

1 I xx ,1  A y dA   144.2  144.2 3  36031288 mm 4 12 1 2 3 2 I xx , 2   y dA    200  40  200  40  80   A  12  2

1 1 3 3 2  40  120    200  40  200  40  80  12  12 

 110293333 mm

4

 I xx, 2  I xx,1 62

Neutral axis and second moment 

To find the neutral axis (N-A), the crosssection is divided into three parts, we have

yA  y A

N

250mm

C

20mm A 200mm

2[0.1m](0.2m)(0.015m)  [0.01](0.02m)(0.25m) 15mm 2(0.2m)(0.015m)  0.02m(0.25m)  0.05909m  59.09mm 



The second moment of the area using the parallel-axis theorem applied to the three parts

1  I   (0.25m)(0.02)3  (0.25m)(0.02m)(0.05909m  0.01m) 2  12  1  2  (0.015m)(0.2m)3  (0.015m)(0.2m)(0.1m  0.05909m) 2   42.26(106 )m4 12  [Mechanics of Structures and Materials]

63

Stage Summary  Shearing Force yields Shearing Stress in the Cross Section. Its Maximal Value is at the Neutral Axis and proportional to the First Moment of the Area, but inversely proportional to the Product of the Second Moment of the Area and the Breadth of the Neutral Plane  The ratio of the Shearing Stress to the Normal Stress is small, except in the W-shape Beam. bf y tf

t

QS  max  z J zb

S z   ydA 0

tw d

z



J z   y 2 dA A

[Mechanics of Structures and Materials]

64

Deflection of Beam

max

3

PL  3 EJ

[Mechanics of Structures and Materials]

65

Deflection of Beam (Animation)

[Mechanics of Structures and Materials]

66

Deflection of Beam

Geometric Relationships and resulting equations:

d  ds

d    ds 1

[Mechanics of Structures and Materials]

dv  tan  dx 67

Deflection in the Beam 

Assuming small deflections and rotations

ds  dx

d    dx 1

dv  tan    dx

d d 2 v    2  dx dx 1



So that



Now, from a previous derivation



d 2 v d  2 dx dx

v is vertical displacement

[Mechanics of Structures and Materials]

M d 2v  2    EI dx 1

68

Deflection of Beam (S represents vertical displacement)

mx d 2s  2 dx EJ Z

1 s EJ z

 m dxdx x

For any Part with only Concentrated Load mx   x  

s

1 EJ z

 3  2   x  x  C1 x  C 2  2  6

For any Part with Distributed Load of Constant Intensity f 2

mx   x   x  

Problem is

s

1 EJ z

 4  3  2   x  x  x  C1 x  C 2  6 2   12

how to simply determine the unknown C1 and C2 (integration constants) [Mechanics of Structures and Materials]

69

Deflection of Beam

Method to Determine the unknown C1 and C2 1) Continuity Condition At the Boundary between two Parts two Continuity Conditions have to be satisfied ds Both s and   must change continuously dx

s from Left  s from Right

ds ds  dx from Left dx from Right

n parts  n-1 boundaries  2(n-1) equations [Mechanics of Structures and Materials]

70

Deflection of Beam

Method to Determine the unknown C1 and C2 2) Restraint Condition Pin Support or a Roller Support can restrict Displacement, so that s=0 is satisfied  2 equation for two ends. Fixed Support can restrict Both Displacement and Rotation, so that s=0 and =0 are satisfied  2 equations. From supports we always have 2 equations. [Mechanics of Structures and Materials]

71

Deflection of Beam From Continuity 2(n-1) equations From supports 2 equations. Totally 2n equations for n curves s(x) in n Parts All unknown C1 and C2 can be found. For practical application, the Maximal Deflection and Maximal Change of Angle are of interest.

[Mechanics of Structures and Materials]

72

Example F q

A

l

x

s

Q

for x
C B F

1 1 EJ z s   Fx 3  Fqx 2  c1 x  d1 6 2

for x >q m x  0 EJ z s  c2 x  d 2 At the point A x=0, s=0, ds/dx=0

x

d1  0 and c1  0

At the point C ds 1 2 EJ  Fq from left z

mx x  Fq

mx   x  

EJ z s  

mx   F q  x 

 3  2 x  x C xC 6

2

1

2

dx 2 1 2 ds c  Fq from right EJ z  c2 2 2 dx 1 3 EJ s from left z  Fq 3 1 3 1 3 from right EJ z s  Fq  d 2 d 2   6 Fq 2

[Mechanics of Structures and Materials]

73

Example F q

A

l

C B 

s

smax

 1 2 3q  x  x  q Fx  x  6 EJ z s 1  Fq 2 3 x  q  x  q  6 EJ z

2 Fq 2 3l  q   B  Fs  6 EJ z 2 EJ z

[Mechanics of Structures and Materials]

74

Deflection in Beam 

Governing equation for the elastic curve



EI is known as the flexural rigidity and is constant for a

prismatic beam, integrating both members in the above equation in x, we have



The constants C1 and C2 are determined from the boundary conditions

[Mechanics of Structures and Materials]

75

Deflection of Beam q

q B

A

A

L

x

D

Drawing the free-body diagram of the portion AB of the beam and taking moments about D 1 1 M  qLx  qx 2 2 2 Substituting for M into the governing equation for elastic curve, we write d2y 1 1 EI 2   qx 2  qLx 2 2 dx

Integrating twice in x, we have EI

dy 1 1   qx3  qLx 2  C1 dx 6 4

[Mechanics of Structures and Materials]

EIy  

1 4 1 qx  qLx3  C1 x  C2 24 12 76

Deflection of Beam Observing that y=0 at both ends of the beam, we first let x=0 and y=0 in the previous equation and obtain C2=0. We then make x=L and y=0 in the same equation and write 1 4 1 4 qL  qL  C1 L 24 12 1 3 C1   qL 24 1 4 1 1 3 3 The equation of elastic curve is EIy   qx  qLx  qL x 24 12 24 0

Check the slope of the beam is zero for x=L/2 and that the elastic curve has a maximum at the midpoint, we have So that

q L4 L3 5qL4 3 L yC  (  2 L  L )   24 EI 16 8 2 384 EI 4 5qL y max  384 EI

[Mechanics of Structures and Materials]

77

Symmetric Bending (plane bending) Bending Moment (vector) : perpendicular to the y Axis

y z

x

Bending moment lies in the x-y plane [Mechanics of Structures and Materials]

78

Unsymmetric Bending & Biaxial Bending In the situation where the bending moment does not act in plane of symmetry of the member, either because •They act on the different plane (Fig. a and b) •Or the member does not possess any plane of symmetry (Fig. c)

We cannot expect 1) the member to bend in the plane where the moment acting on, (plane bending) 2) or the neutral axis of the section to coincide with the axis of the moment. [Mechanics of Structures and Materials]

79

Unsymmetric Bending (Anination)

[Mechanics of Structures and Materials]

80

Unsymmetric Bending & Biaxial Bending Consider an arbitrary shape section (The neutral axis coincides with the axis of the couple M, thus the formula used in plane bending can be utilized here) From equilibrium equation, we get

  x dA  0    z x dA  0   ( y x dA)  M

x components moments about y axis moments about z axis

Assume the stresses to remain within proportional limit of the material, we have  x   m y / c

And substitute it into the second equation, we have my ) dA  0 or  yzdA  0  z ( c

[Mechanics of Structures and Materials]

81

Unsymmetric Bending & Biaxial Bending Based on the definition of product of inertia I yz  Thus I yz  0

 yzdA

• Iyz equals to zero only when the y and z axes are the principal centroidal axes of the cross section. • Since in this case, the M coincides with neutral axis z, and the above conclusion indicates that axis z is also one of the principal centroidal axis. Thus, only when the couple vector M is directly along one of the principal centroidal axes of the cross section, the neutral axis and the axis of the couple M acting on are coincided. [Mechanics of Structures and Materials]

82

Unsymmetric Bending & Biaxial Bending •The principal centroidal axis reflects the geometry property of the member, which is easy to be identified; while the neutral axis is related to the force acting on the section, which may require tedious calculation. •Therefore, the above conclusion makes us easy to find whether the axis of couple M acting on coincides with the neutral axis. Because only when we assure this, the formulas for plane bending can be used.

[Mechanics of Structures and Materials]

83

Centroidal Axes of Various Cross Sections For the cross section which are symmetric at least one of the coordinate axes, the y and z axes are the principal centroidal axes of the section

If neither of the coordinate axes is an axis of symmetry, the coordinate axes are not principal axes. The principal axes can be determined analytically.

[Mechanics of Structures and Materials]

84

Unsymmetric Bending & Biaxial Bending •In the above cases (couple vector M is directly along one of the principal centroidal axes of the section, the neutral axis will coincide with the axis of the couple), the equations derived in the above sections for the symmetric members can be used to determine the stresses. •In other cases of unsymmetrical bending which are more usual, the principle of superposition can be used. The moment is first divided into components directly along the principle axes. The results from the separate components are finally superposed. The details will be presented as follows. [Mechanics of Structures and Materials]

85

Unsymmetric Bending & Biaxial Bending Consider a member with a vertical plane of symmetry, which is subjected to bending couples M and M’ acting in a plane forming an angle θ with the vertical plane.

The couple vector M representing the forces acting on a given cross section will form the same angle θ with the horizontal z axis [Mechanics of Structures and Materials]

86

Unsymmetric Bending & Biaxial Bending Resolving the vector M into component vectors Mz and My along the z and y axes, respectively M y  M sin  M z  M cos 

Since the y and z axes are the principal centroidal axes of the cross section, we have stresses resulting from each component of the couple represented by Mz and My Mz y x   Iz

x  

My z

[Mechanics of Structures and Materials]

Iy 87

Unsymmetric Bending & Biaxial Bending The distribution of the stresses caused by the original couple M is obtained by superposing the stresses determined above, Mz y My z x    Iz Iy Note: •The above equation is valid only if the conditions of applicability of the principle of superposition are met. •In other words, it should not be used if the combined stresses exceed the proportional limit of the material, or if the deformations caused by one component of the couple significantly affect the distribution of the stresses due to the other. [Mechanics of Structures and Materials]

88

Unsymmetric Bending & Biaxial Bending As indicated previously, the neutral axis of the cross section will not, in general, coincide with the axis of the bending couple. Since the normal stress is zero at any point of the neutral axis, The equation defining that axis can be obtained by setting  x  0

Thus

Mz y My z   0 Iz Iy

Iz y  ( tan  ) z Iy

Iz tan   tan  Iy

Note: from the equation, we note that    when I z  I y and    when I z  I y Thus, the neutral axis is always located between the couple vector M and the principal axis corresponding to the minimum moment of inertia [Mechanics of Structures and Materials]

89

Exercise A 16 N.m couple is applied to a wooden beam, of rectangular cross section 1.5cm by 3.5cm, in a plane forming an angle of 30°with the vertical. Determine: (a) the maximum stress in the beam, (b) the angle that the neutral surface forms with the horizontal plane

[Mechanics of Structures and Materials]

90

Exercise (a) Maximum stress

M z  1600 cos 30  13.86 N.m M y  1600 sin 30  8.00 N.m

1 bh 3  5.359 cm 4 12 1 Iy  hb 3  0.9844 cm 4 12 Iz 

The largest tensile stress due to Mz occurs along AB and is M y 1386  1.75 1  z   4.526 MPa Iz 5.359 The largest tensile stress due to My occurs along AD and is M y z 800  0.75 2    6.095 MPa 0.9844 Iy

The largest tensile stress due to combined loading at A [Mechanics of Structures and Materials]

 max   1   2  10.62 MPa 91

Exercise (b) Angle of Neutral surface with horizontal plane tan  

5.359 Iz tan   tan 30  3.143 0.9844 Iy

  72.4

The distribution of the stresses across the section [Mechanics of Structures and Materials]

92