CH.2 Tension & COMPRESSION LIU Jiemin @SJZU 20081.

LIU Jiemin @SJZU 2008 1CH.2 Tension & COMPRESSION


Chapter 2 Stress under axial loading,

Mechanical properties of materials



2.1 Introduction

2.2 Axial force and axial diagrams

2.3 Stress in an axially loaded rod

2.4 Mechanical properties of materials

2.5 Stress concentration

2.6 Allowable stress and strength criteria

2.7 Stress and strength in connections

LIU Jiemin@SJZU 2008 3CH.2 TENSION & COMPRESSION

Characteristic of loads ：Acting line of external forces coincides with the axle

Tension and compression along axle

Characteristic of deformation ：Elongation or compression along axle, accompany contraction or dilation

F F

F F

2.1 Introduction






2.1 Introduction








2.2.1 Axial force — FN, Internal force acting on cross-section along the axle ， determined by cross-section method.

0FF:0F Nx 2.2.2 Symbol of axial force

FFA

A

NFF

FN<0

FNFNFN FN

FN>0



N xF F F Axial force is equal to sum of all axial forces acting on one side of the section

FFA

A

NFF

2.2.3 Determine FN by simple cross-section method



① 反映出轴力与截面位置变化关系，较直观；② 确定出最大轴力的数值及其所在横截面的位置，即确定危险截面位置，为强度计算提供依据。

2.2.4 Axial force diagram

意义意义

FFA

AFN

xF

+



Ex.1 The rod is subjected to forces of 5kN, 8kN, 4kN and 1kN at points A, B, C and D as shown. Drawing the axial-force diagram.

Solution ： FN1 in portion OA ：

0 xF 0 1485F 1N

2kNF 1N

A B C D

5kN 8kN 4kN 1kN

O

A B C D

5kN 8kN 4kN 1kN

FN1

Simple section method,please!



By the same token ， axial forces in portion AB 、 BC 、CD:

xFN/kN2

3

5

1++–

FN2= –3kNA B C D

5kN 8kN 4kN 1kN

O 2

2

FN3= 5kNA B C D

5kN 8kN 4kN 1kN

O 3

3

FN4 =1kNA B C D

5kN 8kN 4kN 1kN

O 4

4Axial force diagram:







2.1 Introduction








2.3.1 Stress on cross-section

FFN(x)

A

xFN )(

AN dAxF )(

Assumption: Stress distributs uniformly on crose-section

F F

2.3 Stress in an axially loaded bar,

Saint-venant’ principle


Plane assumotion: Cross-sections transit along axle and hold as plane

2.3.2. Plane assumotion

d ´a´c´

b´

a bc d

F F

Axial strains are the same ， and normal stresses distribute uniformly




2.3.3 Stress on oblique-section

F=F

A

Fp

cos cos

AA

A

A

coscos 0

A

F

A

Fp

A rod is subjected to F, find stresses on the k-k section

k

k

FF

FF pSolution:




cos0p

20coscos p

0 22

p sin sin

min90 0 0,, ,

00: , =0 ,max

Maximum normal stress on cross-section ， maximum normal stress on 45° inclined-section

0 045 2, = 2,max, / /

p

FF p

Stresses on typiocal sections




MPamax / . / . 0 2 127 4 2 63 7

0 127 4( ) ( ) MPa

2 2

.

oσ 1+ cos2a 1+ cos60 95.5

MPa.

sin . 0 127 4

sin2α 60 55 22 2

MPaF

.A .

0 2

4 10000127 4

3 14 10

Ex.4 A rod of diameter d =1 cm ， F =10 kN ， determine maximum sheari

ng stress ， and normal stress and shearing stress on 30°inclined section.

Solution:

k

k

FF








2.1 Introduction








Conditions ： ☆ Commen temprature(20℃) ☆ Static load

• Strength• Stiffness• Stability

• External forces• Size of member• Configration

Mechanical properties

Standard Specimen

l

d

2.4 Mechanical Properties of Mild Steel


2.4.1 Tension testing of Mild Steel （ Q235 ）



F

OP- L curve△

F

F

△L

A

F,

L

ΔL

2.4.2. Curve of F-ΔL for Mild Steel



O -curve

2.4.3 Stress-Strain curve (Elastic portion)

Hooke LAWE ，

ea

p

b


ob: elasticity portion

oa: proportional portion

ab: non-proportional portion


-curve

O

pe

b c

Ultimate stress b

e

d

2.4.3 Stress-Strain curve (Yielding and Strenthening portions)

f

s

s Yield stressb


bc : Yield portion

ce : Strengthening portion

ef : Necking portion


O

c

f

-curve

d′ g

f′ h

Strain hardening

d

Unloading law

ep

2.4.4 Unloading law and hardening



100%1l - lδ =

l

2. Percent reduction in area ：

3. Brittle and ductile materials

1. Percent elongation

2.4.5 Plastic indexes and strngth indexes

Yielding Stress: s

Ultimate Strength: b

Proportional Limit: p

235

235MPa

380MPa

200MPa

s

b

p

Q :

δ=5% as bound



O

Hard Alu.

20.

0.2% Mild Steel C.

O

Com.

F

F

s

Tension

2.4.6 σ0.2 and compression of mild steel







2.1 Introduction








在集中力和截面尺寸突变处，变形和应力急剧变大的现象。

外力的分布方式只影响作用区域附近的应力分布，离开载荷作用区域一定距离，应力分布与大小几乎不受外载荷作用方式的影响。

b c

Fc

aStrain Distribution

Stress Distribution

F bamax

n

max

k

FF

2.5 Stress concentration;







2.1 Introduction








2.6 Allowable Stress and Strength Design 　　　　　　　　　

uσ > σ (Ul timate stress)

[]—Allowable stress, n-Safety factor

②Design:][

maxmin

,NFA

][ AFN

2.6.1 Failure and Allowable stress

max①Check ：

③Allowable load ：

n

u

2.6.2 Stength Condition )( maxmax A

FN

Failure


Ex. 4 A rod of diameter d =14mm is subjecte to a axial load F =

25 k N, []=170MPa ， check the strength of the rod.

Solution ：① Axial force ： FN = F =25kN

MPa1620140143

1025442

3

2

..πd

P

A

FNmax②Stress ：

③Check ： 170MPa162MPamax

④Conclusion ： Safety.

l

dF F



Ex. 5 A simple crane is shown in Fig.. Knowing that length of the rigid beam AC, L=2m, h=1m = 45◦ and allowable stress []=160 MPa. Find allowable load [F].

L

h

F

A

BC

D



M , ( ) ( )A L N0 F sinθ hctgθ F 0N

FLF =

hcosθ

N

LF A , F A

hcos

Find [F]

Find FN

FAy

FAx

FN

L

F

A B

C

Solution:

kN

d hcosF .

L

2 220 1 160 235 5

4 4 2 2



Ex. 5 A simple crane is shown in Fig.. Find angle at such state, rod BD is lightest. Knowing that total weight of the body F and allowable stress [].

xL

h

F

A B

C

D



M , ( ) ( )A N0 F sinθ hctgθ Fx 0

cosh

LFF max,N

hcos

LFFA maxN ， Area of BD

：

For DC ， Axial force FN

FAy

FAx

FN

xL

F

A B

C


Solution ：


③Determine minimum value of VBD ：

;sin [ ] sin

BDBD

L BD

A h 2FV = A L

θ σ 2θ

][ L2F

V , mino 时45

sin/BDL h θBD BDV = A L



[ 例 6] 已知三铰屋架如图，承受竖向均布载荷，载荷的分布集度为： q =4.2kN/m ，屋架中的钢拉杆直径 d =16 mm ，许用应力 []=170M Pa 。试校核刚拉杆的强度。

钢拉杆

4.2m

q

8.5m



① 整体平衡求支反力解：

钢拉杆

8.5m

q

4.2m

FRA RB

FA

kN. Fm

FX

ARB

A

5190

00



③ 应力：

④ 强度校核与结论：　　 MPa 170 MPa 131 max

此杆满足强度要求，是安全的。

MPa1310160143

103264

4

2

3

..

.

d

P

A

F

2N

max

② 局部平衡求轴力： q

RC

HC

FN

kN. Fm NC 3260

FRA

FA



例 7 图示拉杆沿 mn 由两部分胶合而成 , 受力 P ，设胶合面的许用拉应力为 []=100MPa ；许用剪应力为 []=50MPa ，并设杆的强度由胶合面控制 , 杆的横截面积为 A= 4cm² ，试问 : 为使杆承受最大拉力 , 角值应为多大 ?( 规定 : 在 0~60 度之间 ) 。

kN50,6.26 BB P

联立 (1)、 (2)得：

P Pm

n

解：)1( ][cos2

A

P

)2( ][cossin A

PP

6030

B



( ) kNP A / cos sin / . 260 60 60 4 50 10 4 3 46 2

kN50maxP

(1)、 (2)式的曲线如图 (2)，显然， B 点左侧由正应力控制杆

的强度， B点右侧由切应力控制杆的强度，当 =60°时，由 (2)

式得

kN44.553/41060460sin/60/cos 260,1

APB

kN44.55maxP

解 (1)、 (2)曲线交点处：

kN4.54;3111 BB P

?;MPa60][ max P讨论：若

P

6030

B







2.1 Introduction








2.8.1 Charactristics of external forces ang deformations

1. Connections

Connections connecting other members

FF

Bold connection


FF

Rivet


m

Shaft

Key

Gear wheel

Charactristic ： Trasnsmission torque 。



①External forces ： Two parallel forces are equal in magnitude , oposite in sence, and the distance between the forces is very small.

②Deformation ： The member has trend of relative movement about interface of the two parallel forces 。

n nF

F



③Shearing surface ： n– n

④Internal force in the shearing surface:Shear force Fs ， parallet to the shear surface

F

F

n nn n

Fs

Shear surface

F


n n


①Shear failure

②Bearing failure

③Tension failure

FF

n nF

F

F3. Three types of failure ：



1. Shear surface-As ： Shear force--Fs ：

2 Nomonal shear stress- ：

3 Strength condition of shear

4. Assumption calculation of shear

n nFs

Fs

s

s

A

F

][ s

s

A

F

n nFs

Shear surfae

Fs



5. Bearing stressσbs

Bearing force―Fbs (bearing surface)

FbsFbs

Effective area of the bearing surface =dt

][ bsbs

bsbs td

F

A

F


Effective area

Effective area


Ex.2.7 Determine diameter d of the bold. Knowing:F=200kN ，t=20mm ， [] =80MPa ， [bs]=200MPa.

Solution ：

Fs

Shear surface

Fs

F/2F

t/2

t

F/2 t/2

Fs

][

2

s

s

d

4

2

F

A

F

40mm1080

1020022Fd

6

3

][

][ bsbs

bsbs td

F

A

F

50mm20020

200000

t

F'd

bs

][

50mmd


CH.2 Tension & COMPRESSION LIU Jiemin @SJZU 20081.

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