23/4/19 22:16 Kylinsoft MOM-1-1

5 Stress and strain analysis

• structure—components—inner force—stress—strength;

• Stresses, distribution of inner force, are used to determine the dangerous location in the dangerous cross section;

• Stress distribution, mechanical properties, and strength theories are introduced in this chapter.

23/4/19 22:16 Kylinsoft MOM-1-2

5 Stress and strain analysis

Divide body as part A and B with plane C

S

p

Slim

0

x y

z

AC

B

n

p

nS

Decompose stress alone nominal and tangent direction

n is the normal

S

pn

Sn

lim0

S

pt

S

lim0

Normal stress

Shear stress

5.1 Normal stress and shear stress

23/4/19 22:16 Kylinsoft MOM-1-3

If normal of cross section C is the direction of coordinate y

Subscripts of normal stress：• direction of the stress component Subscripts of shear stress：• first indicating the direction of the normal to the plane• the second indicating the direction of the component of the stress

x y

zC

yy yz

yx

yz

yx

y

23/4/19 22:16 Kylinsoft MOM-1-4

Notification for stresses• Normal stress: same as axial load• Shear stress: it is positive if its moment to every point of element is

clockwise.

x y

zC

yy yz

yx

yz

yx

y

x

y

z

23/4/19 22:16 Kylinsoft MOM-1-5

tdx

dy

Conclusions for shear stress

mz = 0 (dy t )dx = ’(dx t )dy

’

•Shear stresses on opposite face of an element are equal in magnitude and opposite in direction.•Shear stresses on perpendicular faces of an element are equal in magnitude and have directions such that both stresses point toward, or both point away from, the line of intersection of the faces.

23/4/19 22:16 Kylinsoft MOM-1-6

应力分析Stress state in a point

• There are infinite planes passing through a point• Stresses in different plane passing through a point are

called stress state of a point

Stress components

yyx

yzz zy

zx x

xyxz

z

zxzy

x

xzxy

y yx

yzx xy xz

yx y yz

zx zy z

xy

z

23/4/19 22:16 Kylinsoft MOM-1-7

Shear Stresses

Pin connection

P

P/2

P/2

n

n

P

Pn

n

P

V

n

n

23/4/19 22:16 Kylinsoft MOM-1-8

Discussion about strains

• Normal Strain m

u

x 0

= limx

u du

x dx

• Shear Strain 00

= lim ( B'A'D' )2x

y

23/4/19 22:16 Kylinsoft MOM-1-9

x=

y=

xy=

Discussion about strains

x=

y=

xy=

x=

y=

xy=

x=

y=

xy=

s

s

0

0h

2w

1

0

0

s

0

0

s

23/4/19 22:16 Kylinsoft MOM-1-10

There is no any wrap in cross section of loaded members considered in this course. Cross sections remain plane and normal to longitudinal axis.

Plan assumption

23/4/19 22:16 Kylinsoft MOM-1-11

Prismatic bar loaded by axial force at the centroid of ends (Axially loaded members).

Axial force produce a uniform stretching of the bar.

L

PP

L+

PP

Prismatic bar in tensionIf the forces are reversed in direction, the bar is in

compression

Tension and Compression

23/4/19 22:16 Kylinsoft MOM-1-12

For stress acts in a direction perpendicular to the cuted surface, it is referred to as a normal stress

Sign convention for inner force and normal stress ：tensile as positive, and compressive as negative

Stress σ: the intensity of internal force

A

N A

dAN A

Assuming: that axial force acts through the centroid of the cross sectional area, cross section will keep flat during tension, stress has a uniform distribution.resultant of stress N

23/4/19 22:16 Kylinsoft MOM-1-13

Saint-Venant’s Principle

23/4/19 22:16 Kylinsoft MOM-1-14

For case of that the bar has a straight axis and varied cross section area

)(

)()(x

x

A

Nx

x

xA(x)

N(x)

How to get N(x)?

23/4/19 22:16 Kylinsoft MOM-1-15

B

A

Example The diameter of bar AB is d=20mm, the load Q=15kN. Determine the normal stress in the bar AB.

P

Solution:

12310123)1020(

4

1038.7 6

23

3

MPaPaA

N

1. Internal force: P=Q/sin sin =0.8/(0.82+1.02)1/2=0.388 N=P=15/0.388=38.7 kN

2. Stress

23/4/19 22:16 Kylinsoft MOM-1-16

Normal Strain

L

PP

L+l

PP

l

L1

l

l

L

LL 1

23/4/19 22:16 Kylinsoft MOM-1-17

Normal Strain

normal strainε is a dimensionless quantity.

Tensile strain (positive)

bar is in tension, representing an elongation or stretching of material

Compressive strain (negative)

bar is in compression, representing a shorten of material

L

LL 1

23/4/19 22:16 Kylinsoft MOM-1-18

cos

AA

Stresses On Inclined Sections

P

Normal stresses acting on cross section PP

k pP

Pp

k

k

A

P

A

N

Area of the inclined section k-k

Distributed stresses of the inclined section k-k

coscos

cos

A

P

A

P

A

Pp

k

k

23/4/19 22:16 Kylinsoft MOM-1-19

Stresses On Inclined Sections

cosp

Pp

k

k

2coscos p

sin

cos sin

sin 22

p

Shear stress on inclined section

0 max

4

2max

Due to

Normal stress on inclined section

- 1

- 0. 5

0

0. 5

1

1. 5

- 90 - 45 0 45 90

23/4/19 22:16 Kylinsoft MOM-1-20

Deflections of axially loaded members

PP

l

A

Nl

lStrain

Hooke’s law

When <p = E

EA

lNl

l

b

EA

lPElongation

l+l

b1

Stress

23/4/19 22:16 Kylinsoft MOM-1-21

example Bar with changes in cross section, A1=2cm2, A2=4cm2,

P1=5kN, P2=10kN. E=120×103MPa. Determine its elongation.

Solution:

N1=-5kN; N2=-5kN; N3=5kN

321 llllAB

3

33

2

22

1

11

EA

lN

EA

lN

EA

lN

5kN

5kNN

m4

49

3

49

3

49

3

1005.110410120

5.010510410120

5.0105

10210120

5.0105

P2 P1

A C DB

50 50 50

shortened

23/4/19 22:16 Kylinsoft MOM-1-22

23/4/19 22:16 Kylinsoft MOM-1-23

横向应变 Lateral Strain

Lateral strain and Poisson’s ratio

b

b

If <p

：泊松比 Poisson’s Ratio

l+l

PP

ll

b1 b

, E are material properties

E is about 200 GPa for steel

is about 0.25 to 0.35 for many metal

23/4/19 22:16 Kylinsoft MOM-1-24

Volume change

o

y

z

x

a

c

b

aεcμε

bμε σ

0V abc

(1 ) (1 ) (1 )fV a b c

0

0

(1 2 ) (1 2 )fV VVe

V V E

Final volume

Unit volume change

Initial volume

23/4/19 22:16 Kylinsoft MOM-1-25

materials              

Structural steelLow alloy steel(16Mn)cast ironaluminum alloy(LY12)rubberwoodconcreterock

2.092.00.6-1.620.710.0050.09-0.120.152-0.360.46-0.58

0.280.25-0.300.23-0.270.330.450.09-0.120.16-0.180.21-0.26

E/105(MPa) μ

E and μ of common materials

23/4/19 22:16 Kylinsoft MOM-1-26

23/4/19 22:16 Kylinsoft MOM-1-27

Calculation of truss displacement

mmAA y 461.122x

23/4/19 22:16 Kylinsoft MOM-1-28

Mechanical Properties of Material

The mechanical properties of materials are the behaviors of loaded materials in strength and deformation. It can be showed with - plot.

23/4/19 22:16 Kylinsoft MOM-1-29

23/4/19 22:16 Kylinsoft MOM-1-30

Tensile test of low carbon steelP

l

d

P

Tensile testing specimen

Material testing system

Gage length l

Standard specimen(GB) ： l=5d ；l=10d

Static test: A cylindrical test specimen is placed in the middle of the load assembly. As the specimen is pulled very slowly, the load and the elongation over the gage length are measured and recorded.

23/4/19 22:16 Kylinsoft MOM-1-31

Stress equals P/A

After performing a tension test and determine the stress and strain at various magnitudes of load, we can plot a diagram of stress versus strain, that is Stress-Strain Diagram

23/4/19 22:16 Kylinsoft MOM-1-32

Stress-strain diagram of low carbon steel

CA

B E

P

l

D

A

P

)(

)()(

tA

tPt

Nominal Stress

True Stress

23/4/19 22:16 Kylinsoft MOM-1-33

Stress-strain diagram for soft steel

linear elastic

P()

l

ultimate stressb

A

fracture

yielding hardening necking

C

D

Belastic limit yield stress s

proportional limit P

E

F

elastic

23/4/19 22:16 Kylinsoft MOM-1-34

Stress-strain diagram for soft steel

线性弹性 linear elastic

强度极限 b ultimate stress

断裂 fracture

屈服 yielding

强化 hardening

颈缩 necking

弹性极限 e elastic limit

屈服应力 s yield stress

比例极限 P proportional limit

弹性 elasticity

23/4/19 22:16 Kylinsoft MOM-1-35

Procedural of tensile test of low carbon steel

P

l

d

P

elastic hardening necking fracture

23/4/19 22:16 Kylinsoft MOM-1-36

延伸率 Percent elongation

%1001

l

lll: original gage length

l1: distance between the gage marks at fracture

截面收缩率 Percent reduction in area

%1000

0

A

AA fA0: original cross-sectional area

Af: final area at the fracture section

23/4/19 22:16 Kylinsoft MOM-1-37

• Ductility of a material in tension can be characterized by its percent elongation and by the reduction in area at the cross section where fracture occurs.

• Brittle Materials fail in tension at relatively low values of strain after the proportional limit is exceeded.

O

b

p

• Ductile materials ： >5% such as: carbon steel ， copper ， aluminum alloy

• Brittle materials ： <5% such as: cast iron ， glass ， stone, concrete, ceramic materials

• Steel may have an elongation in the range of 10% to 40%. For structural steel, values of 25% or 30% are common, and its reduction in area is about 50%.

23/4/19 22:16 Kylinsoft MOM-1-38

mechanical properties of other ductile materials in tension

H62CA

B

A3

无屈服阶段

四个阶段

四个阶段

无屈服 无颈缩阶段

无屈服阶段

T10A

20Cr

16Mn

23/4/19 22:16 Kylinsoft MOM-1-39

offset yield stress 0.2 名义屈服极限

When a material does not have an obvious yield point and undergoes large strain after the proportional limit is exceed, a line is drawn on the stress-strain diagram parallel to initial linear part of the curve but is offset by some standard amount of strain,such as 0.2%.

A

O

0.2%

0.002 offset

offset yield stress 0.2

The intersection of the offset line and the stress-strain curve defines the yield stress

23/4/19 22:16 Kylinsoft MOM-1-40

Unloading

Plasticity:For a higher loading level, unloading line is parallel to the initial portion of the loading curve. When the load has been entirely removed, a residual strain remains in the material.

Elasticity: It returns to its original dimension during unloading.

Residual strain

CA

B

D

G

O

loading

Reloading: material has a higher yield stress

Hunloading

23/4/19 22:16 Kylinsoft MOM-1-41

Stress-strain Curve for Steel in Compression

CA

B

D

E

A

P

Nominal Stress

)(

)()(

tA

tPt

True Stress

23/4/19 22:16 Kylinsoft MOM-1-42

Stress-Strain Curve for Cast Iron

There are no yielding in Stress-Strain Curve for Cast Iron.Break face is orient at 45~55° with axis in compression and 90 ° in tensionBrittle materials usually reach much higher ultimate stresses in compression than in tension.

O

b B

b

23/4/19 22:16 Kylinsoft MOM-1-43

Mechanical properties of materialsName Type s(MPa) b(MPa) s(%)

Structure steel A3

A5

216~235

255~275

373~461

490~608

26~27

19~21

Carbon steel 40

45

333

353

569

598

10

16

Low alloy steel 16Mn

15MnV

274~343

333~412

471~510

490 ~ 549

19~21

17~19

Alloy steel 20Cr

40Cr

539

785

834

981

10

9

Cast steel ZG35 275 441 5

Cast iron HT15-33 98.1 ~274(Tension)

637(Compression)

23/4/19 22:16 Kylinsoft MOM-1-44

Which one has best ductility and which one breaks first among the three loaded prismatic bars with different materials.?

23/4/19 22:16 Kylinsoft MOM-1-45

Which one has highest ultimate limit and which one breaks first among the three loaded prismatic bars with different materials.?

23/4/19 22:16 Kylinsoft MOM-1-46

23/4/19 22:16 Kylinsoft MOM-1-47

Hooke’s Law

BC

A

O

In linear elastic state OA (<P), stress is proportional to strain,

P

P

E

E ： Modulus of Elasticity (MPa; GPa) or Young’s Modulus One of the properties of material

23/4/19 22:16 Kylinsoft MOM-1-48

Hook’s Law in Shear

The properties of a material in shear can be determined experimentally from direct-shear tests or torsion tests.Stress-strain diagrams in shear are similar in shape to that in tension.

23/4/19 22:16 Kylinsoft MOM-1-49

G

The initial part of the shear stress-strain diagram is a straight line also.

12

EG

G: Shear Modulus of Elasticity, GPa

Hook’s Law in Shear

23/4/19 22:16 Kylinsoft MOM-1-50

Stress-strain diagram of low carbon steel

CA

B E

P

l

D

A

P

)(

)()(

tA

tPt

Nominal Stress

True Stress

23/4/19 Kylinsoft MOM-6-51

xx E

Ex

xzy

x

y

x

Hooke’s law for uniaxial stress

Hooke’s law for pure shear

G

Hooke’s law for triaxial stress

23/4/19 Kylinsoft MOM-6-52

2

3

1

3211

1 E

1

2

3

1 E1

E2

E3

Hooke’s law for triaxial stress

23/4/19 Kylinsoft MOM-6-53

2

3

1

3211

1 E

1322

1 E

21331 E

Triaxial stress

23/4/19 Kylinsoft MOM-6-54

)]([1

zyxx E

Gxy

xy

)]([1

xzyy E

)]([1

yxzz E

Gyz

yz

Gzx

zx

x y

z

xy yx

yz

zy zx xz

Triaxial stress

23/4/19 Kylinsoft MOM-6-55

Example for Hook’s law

MPaA

Pm

N 60106001.0

1062

62

3

3

332

11 10376.0

EE

0312

2 EE

321

33 10764.0

EE

A aluminum block in confined between plane parallel rigid wall, P = 6kN, E=70GN/m2, =0.33, obtain the three principal stress and deformations of the block. P

10

10

10

01

MPa8.196033.032

02

l1 = 0.01 1 = 3.76 10 -5m l2 = 0 l3 = -7.64 10 -5m

1

3

2

23/4/19 22:16 Kylinsoft MOM-1-56

许用应力和许用载荷Allowable Stress and Allowable load

安全系数 factor of safety

实际强度 (actual strength)所需强度 (required strength)

n= 1

许用应力 allowable stress allow or []

yield stress

factor of safetyFor ductile materials

For brittle materials

n

s

n

b ultimate limit

factor of safety

23/4/19 22:16 Kylinsoft MOM-1-57

Determination of a Factor of Safety• The probability of accidental overloading of the structure;• The type of loads (static, dynamic, repeated) and how

accurately they are known;• The probability of fatigue failure;• Inaccuracies in construction;• Quality of workmanship;• Variations in properties of materials;• Deterioration due to corrosion or other environmental

effects;• Accuracy of the methods of analysis;• Whether failure is gradual or sudden;• Consequences of failure;• Other such considerations.

23/4/19 22:16 Kylinsoft MOM-1-58

许用应力 ,许用载荷和强度条件Allowable Stress and Allowable load

A

Nmaxmax

Strength criterion in tension or in compression

Using above strength criterion, we can do

Check the strength

Design section dimension

Determine allowable load

A

Nmaxmax

ANmax

maxN

A

23/4/19 22:16 Kylinsoft MOM-1-59

30sin45sin 12 NN 0X

PNN 45cos30cos 210Y

PP

N 732.031

21

P

PN 518.0

31

22

A

NmaxDue to strength ：

11 AN P1154.4kN

Allowable load P = min(P1,P2)

P = 97kN

P297.1kN 22 AN N1 N2

P

30°45°

Example Bar AB and AC are made of same material ，[]=160MPa ， cross section area A1=706.9mm2 ， A2=314mm2 ， determine Pmax

Solution: node A

30° 45°

PA

B C

12

23/4/19 22:16 Kylinsoft MOM-1-60

Stress Concentration 应力集中

Stress Concentration Factor

maxk

P

P

P P

P

P

The abrupt geometric changes, such as hole, grooves( 槽 ), notches( 切口 ), shoulders, threads, etc, in structural components will cause high stresses in very small region around those changes.

23/4/19 22:16 Kylinsoft MOM-1-61

23/4/19 22:16 Kylinsoft MOM-1-62

Torsion of circular shaft

Square

Parallelepipedrhomboid

element

23/4/19 22:16 Kylinsoft MOM-1-63

Pure shear

Torsion of circular bar

angle of twist

T T

dx

p q

Pure torsion of circular bar•cross sections rotate as rigid bodies about longitudinal axis •radii remain straight•cross sections remain plane and circular•the length of the shaft keeps unchanged.

d

e

e'

Shear strain A measure of the distortion, or change in shape, of the element

element

23/4/19 22:16 Kylinsoft MOM-1-64

d

d

ca'

b'

a

b

dx

Rdx

dR

ad

aa

d

dx

Shear strain

θ—angle of twist

per unit length

x

d

e

e'

dx

p q

T T

Shear strain in a circular shaft varies linearly with the radial distance from the center, and has the maximum values on an element at the outer surface.

23/4/19 22:16 Kylinsoft MOM-1-65

G G

From the Hook’s law of shear

Static Equilibrium Relationship

dATA 2

AG dA PGI

AP dAI 2Polar moment of inertia

P

T

GI

PI

T

d

dR

d

m

x

T

d

23/4/19 22:16 Kylinsoft MOM-1-66

PI

TRmax

tW

T

R

IW P

t Section Modulus of Torsion

max

maxMaximum shear stress

PI

T

23/4/19 22:16 Kylinsoft MOM-1-67

Polar Moment of Inertia

R

IW P

t 162

33 DR

R

IW P

t 44

16dD

D

d D

dA

For a circle of radius R and diameter D

AP dAI 2

ddR

2

0 0

2

322

44 DR

d

dR

d

For a hollow circular shaft

AP dAI 24 4( )

32

D d

23/4/19 22:16 Kylinsoft MOM-1-68

Deformation of circular bar in Torsion

P

T

GI

l lP P

T Ld dx T

GI GI

GIP/L: Torsional stiffness

PGI

TL

23/4/19 22:16 Kylinsoft MOM-1-69

Nonuniform Torsion

x

)(xTT )(xII PP

dxxGI

xTd

l Pl )(

)(

N1

N2 N3

1 1

n ni i

ii i i Pi

T L

G I

23/4/19 22:16 Kylinsoft MOM-1-70

T

max

min

max

min

max

max

T

max

max

T

Distributions of shear stress on cross section

23/4/19 22:16 Kylinsoft MOM-1-71

ExampleShaft AB, transmitted power N = 7.5kW ， rotation speed n = 360 rpm. Cross sections are showed in following fig. D = 3cm ， d = 2cm 。 Determine the stresses on the outer edge of cross section of segment AC and the stresses on the outer edge and inner edge of cross section of segment CB.

torque mN1999549 n

Nm

T = m = 199N·m

polar moments of inertia of cross section

44

1 cm95.732

D

IP

4442 cm38.6

32 dDIP

D dD

mm

A C

B

23/4/19 22:16 Kylinsoft MOM-1-72

44

1 cm95.732

D

IP

4442 cm38.6

32 dDIP

Shear stresses

6out

1

37.5 10 Pa 37.5MPa2

AC

P

T D

I

6in

2

31.2 10 Pa 31.2MPa2

CB

P

T d

I

6

2

46.8 10 Pa 46.8MPa2

CBout

P

T D

I

D dD

mm

A C

B

T = m = 199N·m

23/4/19 22:16 Kylinsoft MOM-1-73

Strength criterion and Stiffness criterion in torsion

tW

Tmaxmax

Strength criterion

Stiffness criterion

PGI

Tmaxmax

(rad/m)

180max

maxPGI

T(°/m)

[θ]—Allowable angle of twist

per unit length

[τ]—Allowable shear stress

23/4/19 22:16 Kylinsoft MOM-1-74

Check or

chose materials

Design section dimension

Determine allowable load

tW

Tmaxmax

Strength criterion and Stiffness criterion

Using above strength criterion, we can do

PGI

Tmaxmax

tW

Tmaxmax

PGI

Tmaxmax

max

t

TW

maxP

TI

G

max tT W max PT GI

23/4/19 22:16 Kylinsoft MOM-1-75

Example

Gear and clutch shafts C, E, HP1 ＝ 14kW, P2= P3=P1/2n1=n2=120rpm, z1=36, z3=12d1=70mm, d 2=50mm, d3=35mm. Determine max. shear stresses.P1=14kW, P2= P3= P1/2=7 kWn1=n2= 120rpm

13 1

3

36120 rpm 360rpm

12

zn n

z

＝ ＝ ＝

T1=1114 N.mT2=557 N.mT3=185.7 N.m

3

609549 ( )

2

P PT N m

n n

23/4/19 22:16 Kylinsoft MOM-1-76

16.54MPaPa1070π

111416E

9-31P

1max

W

T

.69MPa22Pa1050π

55716H

9-32P

2max

W

T

.98MPa12Pa1053π

7.18516C

9-33P

3max

W

T

3

T1=1114 N.m

T2=557 N.m

T3=185.7 N.m

23/4/19 22:16 Kylinsoft MOM-1-77

Example P ＝ 7.5kW, n=100rpm,[τ]=40MPa, α=d2/D2 = 0.5. Determine weight ratio of two shafts.

Torque

Solid shaft 31 6

16 716 20 045m

π 40 10d

.

.

7 59549 9549 716 2N m

100

PT

n

..

max1 31 1

1640MPa

πP

T T

W d

d2 ＝ 0.5D2=23 mm 32 4 6

16 716 20 046m=46mm

π 1- 40 10

..D

max2 3 42 2

1640MPa

π 1P

T T

W D

Hollow shaft

22 2 3

2 222 3

1 1

1 46 101 0.5 8

45 10

DA

A d

0. 7＝ What’s your choice and why?

23/4/19 22:16 Kylinsoft MOM-1-78

If α =3/4 2

1

0.564A

A

What will happen if we increase α?

Wrinkling or Buckling of the wall

23/4/19 22:16 Kylinsoft MOM-1-79

N

Inner force

A

N

stress

PI

T

??

Bending Stresses

FAyV

M

23/4/19 22:16 Kylinsoft MOM-1-80

Pure bending refers to flexure of beam under a constant bending moments,

nonuniform bending refers to flexure in presence of shear forces.

•Pure Bending

23/4/19 22:16 Kylinsoft MOM-1-81

Normal strains in beam

Under the action of the couple M, the beam deflects in the xy plane and its axis is bent into a circular curve.

y dx

M M

O

y

M

M

d

a a

Cross sections of the beam remain plane and normal to longitudinal line of the beam.

23/4/19 22:16 Kylinsoft MOM-1-82

Normal strains in beamNeutral surface: As a result of the bending deformations, the longitudinal fibers on the convex side of the beam are elongated, whereas those on concave side are shortened. Somewhere between the top and bottom of the beam is a surface in which the longitudinal fibers do not change in length. Neutral axis: the intersection of neutral surface and cross sectional plane.

Neutral axis

Neutral surface

23/4/19 22:16 Kylinsoft MOM-1-83

Normal strains in beam

y

d

ddy

)(

—radius of curvature

O

y

M

M

d

a a

23/4/19 22:16 Kylinsoft MOM-1-84

Stress-strain relation

Uniaxial stress state

Each longitudinal fiber of the beam is subjected only to

tension or compression.

Using Hooke’s law

x xE x

Ey

Normal stresses acting on the cross section vary linearly with distance y from the neutral axis.

M

x

23/4/19 22:16 Kylinsoft MOM-1-85

Resultant of the normal stress

z

y

x

dA

y

z

MA dA

0 A dAyM

AdAy

Ey

0 AAydA

EydA

E

The first moment of the area of the cross section with respect to the z axis is zero.

The neutral axis passes through the centroid of the cross section when material of the beam follows Hook’s law.z

A

zEIdAy

E

2 M

X = 0

AdAy

A

dAyI 2Z The moment of inertia of the cross sectional area

with respect to the z axis

23/4/19 22:16 Kylinsoft MOM-1-86

Criterion of strength for bending normal stress

maxy

IW zSection Modulus

W

M

I

My

z

maxmax

• maximum normal stress on cross section

yE

zI

My

zI

My

zEI

M

1

23/4/19 22:16 Kylinsoft MOM-1-87

A

dAyI 2Z

max

ZZ y

IW

64

4

Z

dI

32

3

Z

dW

)1(64

44

Z

DI

)1(32

43

Z

DW

IZ and WZ

12

3

Z

bhI

6

2

Z

bhW

1212

3300

Z

bhhbI

)2//()1212

( 0

3300

Z hbhhb

W

Shaped Steel

23/4/19 22:16 Kylinsoft MOM-1-88

σIyM

σz

maxmaxmax

For brittle materials

tt max, cc max,

Criterion of strength for bending normal stress

23/4/19 22:16 Kylinsoft MOM-1-89

RA=13.75kN

RB=21.25kN

M

2.5

13.75

V 13.75

16.25

5From diagrams of shear force and bending moments, we find

30kN 5kN/m

1m 1m 1mRA RB

Mmax=13.75kN-m

ExampleChose shape of cross section of beam from rectangular (h/d=2) ， wide-flange, circular and circular tube(D/d=2) . []=160MPa.

23/4/19 22:16 Kylinsoft MOM-1-90

366

3

1094.8510160

1075.13m

W

M maxmax

maxM

W

rectangular 632

1094.856

4

6

bbhW

circular 63

1094.8532

d

W

b =5.05cm h = 10.1cm A=51.cm2

d =9.57cm A=71.93cm2

Circular tube

644

44

1094.8532

2

642

dDDD

dD

D

IW z

D =9.78cm d = 4.89cm A=56.34cm2

14# wide-flange W=102cm3 A=21.5cm2 Appendix C Table 3

23/4/19 22:16 Kylinsoft MOM-1-91

Reasonable shape of cross section of beam

To minimize the weight of a beam, or smallest cross-sectional area, the shape of cross section should be optimized for required section modulus,.

23/4/19 22:16 Kylinsoft MOM-1-92

example Cast iron, [T]=40MPa and [C]=80MPa. Check the safety of the beam

Centroid, Iz=7.64×106mm4

ReactionsBending moment diagram

5kN

1m 1m1m

12kN

AB

CD

3.5

5

M

B

MB

C

MC

z52

88

Dangerous positionsBottom of section B loaded by max. compressive stressBottom of section C loaded by max. tensile stress

Dangerous cross sections, C and B

80 20

120

20

23/4/19 22:16 Kylinsoft MOM-1-93

6max

max 6

5 10 8857.6MPa [ ]

7.64 10C B

Cz

M y

I

6max

max 6

3.5 10 8840.3MPa [ ]

7.64 10T C

Tz

M y

I

max. tensile stress at bottom of section C

Discussion:

Why we often use T shape cross section for brittle materials?

Best way to put cross section, T or ⊥ ?3.5

5

M

B

MB

C

MC

52

88

80

z

20

120

20

It’s dangerous

max. compressive stress at bottom of section B

23/4/19 22:16 Kylinsoft MOM-1-94

Shear Stresses in Rectangular Beam

23/4/19 22:16 Kylinsoft MOM-1-95

dx

V

z x

y

Shear Stresses in Rectangular Beam

dx

y

Shear stresses act parallel to the shear force

Distribution of the shear stress is uniform across the width of the beam

23/4/19 22:16 Kylinsoft MOM-1-96

h/2dx

y

h/2

x

Shear Stresses in Rectangular BeamM V M+dM

V+dVdx dx

MM+dM

N1 N2

N3

1

1 AdAN

y

h/2

h/2

b

z

y1

y

dA

A1

dAI

MyA

z

1

1

z

z

Az I

MSdAy

I

M *

11

*

2z

z

M dM SN

I

’

bdxN '3

23/4/19 22:16 Kylinsoft MOM-1-97

Shear stress

M V M+dM

V+dVdx dx

MM+dM

y

h/2

h/2

b

z

y1

y

dA

A1

z

z

I

MSN

*

1

z

z

I

SdMMN

*

2

dAyS

Az 1

1*

0231 NNN

0

*'

*

z

z

z

z

I

SdMMbdx

I

MS

X = 0

* *' V

=z z

z z

S SdM

dx I b I b

h/2

dx

y

h/2

xN1 N2

N3’

bdxN '3

23/4/19 22:16 Kylinsoft MOM-1-98

V: shear force in cross section

Iz: The moment of inertia of the cross sectional area with respect to

the z axis

Sz*: first moment of the cross-sectional area above the level y1 at

which the shear stress acts

b: width of cross section

Shear Stresses in Rectangular Beam*

' V z

z

S

I b

bI

VS

z

z*

Shear stress in cross section

'

1

*1z A

S y dA

23/4/19 22:16 Kylinsoft MOM-1-99

Shear stress in a rectangular beam varies quadratically with the distance y1 from the neutral axis

zI

Vh

8

2

max bh

V

2

3

Shear Stresses in Rectangular Beamy

h/2

h/2

b

z

y1

y

dA

1

1AdAyS

2

2

42y

hb

12

1 dybyh

y

2

2

42y

h

I

V

z

yc

C

CyA 1

max

A1

23/4/19 22:16 Kylinsoft MOM-1-100

ExampleWooden rectangular beam AB, b=100mm, h=150mm . a=0.5m. []=11MPa , []=1.2MPa, determine Pallow

a P P aA

B

RA RB

h

b

z

y

o

V

P

P

Mx

Pa

Reactions, diagrams of V, MVmax= P Mmax= Pa

][

6

2max

max bhPa

W

M

a

bhP

][

6

2

max

kNN 25.882505.06

101115.01.0 62

bI

SV

z

z*

maxmax

3

][2max

bhP kNN 1212000

3

102.115.01.02 6

][2

3

2

3 maxmax bh

V

A

V

Pmax=8.25kN

23/4/19 22:16 Kylinsoft MOM-1-101

Shear Stress in the Web

yH/2 h/2

B

bH/2 h/2bI

VS

z

z*

2211*

CCz yAyAS

yC2

yC1

A1 A2

y

hyy

hb

hHhhHB

22

1

2222

1

222

2

222

428y

hbhH

B

23/4/19 22:16 Kylinsoft MOM-1-102

Shear stress in the webs of beams with flanges

yH/2 h/2

B

bH/2 h/2

bI

VS

z

z*

yC2

yC1

A1 A2

2

222*

428y

hbhH

BSz

2

222

428y

hbhH

B

bI

V

z

88

22

max

hbBH

B

bI

V

z

88

22

min

BhH

B

bI

V

z

hb

V

The shear stresses in the web account for 90% to 98%

In design work, it’s common to calculate an approximate of the maximum shear stress

23/4/19 22:16 Kylinsoft MOM-1-103

ExampleA beam with an overhang, flange and web are welded together. []=120MPa, []=60MPa, check the strength of beam.

RA= 25kN RB= 105kN

Mmax= 40kN-m Vmax= 65kN

zC = 82mm Iz = 3.97107mm4

1m

90kN

2m1m

20kN/m

ADC B

RA RB

25 40

65

V

M 25

40

20 20

200

140 20

82

C

a b

zI

yM maxmaxmax

][9.1181097.3

11810407

6

MPa

23/4/19 22:16 Kylinsoft MOM-1-104

Example(cont’d)

35max

* 1075.21182

1118202) mmSz （

][4.111097.3202

1075.21065)(7

53max

*

max

MPabI

SV

z

z

1m

90kN

2m1m

20kN/m

ADC B

RA RB

25 40

65

V

M 25

40

20 20

200

140 20z1

82

C

23/4/19 22:16 Kylinsoft MOM-1-105

Example(cont’d) At welded surface

5* 10016.27214020) abzS（

*

3 5

7

( )

65 10 2.016 108.36 [ ]

2 20 3.97 10

z abab

z

V S

bI

MPa

20 20

200

14020

82

C

a b

23/4/19 22:16 Kylinsoft MOM-1-106

Built-up beam

* ?zS *z

z

VS

bI

Box beam glued beam

Plate girder

*z

z

VSf b

I

23/4/19 22:16 Kylinsoft MOM-1-107

How to increase the strength of beams

Decreasing bending moments

6a

P

4a

P

aa

PaaPM 5.162

1

2

1max PaaPM 4

2

1

2

1max

Z

maxmax W

M ][

23/4/19 22:16 Kylinsoft MOM-1-108

F

F

F

23/4/19 22:16 Kylinsoft MOM-1-109

Add number of supports

23/4/19 22:16 Kylinsoft MOM-1-110

L

Pq

PLM4

1max PLqLM

8

1

8

1 2

max

P

F

23/4/19 22:16 Kylinsoft MOM-1-111

2

6Z

bhW 1

2

6Z

hbW 2

Reasonable design of cross section

23/4/19 22:16 Kylinsoft MOM-1-112

23/4/19 22:16 Kylinsoft MOM-1-114

Fully Stressed Beam

P

l

A

B

hB

bx

hx

Design the cross section of a cantilever beam to make the maximum normal stress over all length of the beam unchanged, if its width is a constant.

M

Pl

M = Px 6

2xbh

xW

22

6

6][

xx bh

Px

bh

Px

xW

M

][

6

b

Pxhx

][

6

b

PlhB

l

xhh Bx

23/4/19 22:16 Kylinsoft MOM-1-115

b

xh

23/4/19 22:16 Kylinsoft MOM-1-116