交变载荷

No reasonable reasonable
2 . Impact toughness ( material constant )
specimen

W A

Gh h
1
2
Α
unit: J/mm2
① Ductile material
k
>> brittle material
k
② Standard specimen : U--notch ; V--notch ;

Suddenly stop rotating :
1057MPa d max

Stop within 10 seconds :
' 2.67MPa d max
' 396
d max
d max
P brake
A P
B
l

10.4 Practical Examples Improving Impact Resistance
K 1 1 2H 1 1 T
d

U
( Perpendicular impact )
st
s
K v2 T
d
g
U
st
s
( Horizontal impact )
, U
st
s
K d
Example
(2)
(1)
(3)
st
st
st
K K K (2)
(1)
d
不依赖于
A.
② n=360 r/min , D=4 m , =75 kN/m3
n=720 r/min , D=4 m , =75 kN/m3
43.5MPa d
174MPa d
例 1. 旋转轴 AB : n =100r/min , d =100 mm ,
飞轮: Ix=0.5kN·m·s2, 忽略旋转轴 AB的质量 , 10秒
d
d
d
d st
d
d st
② 自由落体 T 1 W v2 , v2 2gH 2g
T WH
K 1 1 2H
d

st
W l
W
Wl
H Wl 3
H
st EA
st 3EI
W
l
a
A l/2
H
B Wl 3
l/2
st 48EI
W H
a
4Wa3 st 3EI
③ If force W is applied suddenly
d
d st
动载荷系数
b
q R
③ 另一个例子
T
b l
R
静载荷下的许可应力
q A st
q A A a
d
g
A ( l b)l
st 2W 4 z
(1 a )
d
st
g
2 . 匀速旋转的圆环

t D
D
qd
d dS
O
d

Nd
Nd
q A. D 2 AD 2
Critical position : The lowest position A1B1
q A A r 2 A(1 r 2 )
g
g
M

ql 2

Al 2
(1
r 2
)
max
8
8
g
A an r
A1
q
bl h qd an
B an r
B1

M max
Al 2 r 2
(1 )

d

error 10 0 0
st
st
⑥ Strength condition :
[ ] d
⑦ Approximate method : plastic deformation,
heat energy, local deformation , sound energy
[ ] s

A
A
l
B l1
C
C
d
Using T + V = Ud , we get :
K d
l 2 2
g st
Using Mohr’s theorem , We have


Ql (l

l )2 1
st
3EI
Section B is critical section ,


M

Q(l

l) 1
st
hbl 2
[1
r (nπ)2 ] 107MPa
d max
W
8W
g 8bh3 6 g 30
z
z
10.3 冲击载荷下的应力
1 . 定义 : 构件在非常短的时间内，速度发生很大的变化， 这种现象称为冲击或撞击 .
A large acceleration 2. 例子 : Pile (打桩), Forge （锻造）, Bullet.
d
neglecting the mass of the bar AC .
Solution : kinetic energy :
T 1 Q (l )2
2g
stain energy :
U 1 P 1 K 2 Q
d
2d d
2 d st
potential energy : V 0
n d
more safety
⑧ The general steps of determining dynamic load
Example 3 . The bar AC rotates on the horizontal plane.
, Q , l , suddenly seizing up at the point B . Determine
③
k
is
related to
temperature
v1
F1 v2
t

B
md 0
Example 2 . 车轮 : n=300r/min , 连杆 AB : h =5.6cm ,
b =2.8cm, l =2m , r =25cm , =76.5kN/m3 ,确定连杆 max
和失效位置
Solution : Rod AB : Translational motion
W
W
z
z
K 3EIlQ
d
d st
W
g
z
Note: σd does not depend on l1.
K l 2 2
d g st

A A
A
C
l
B l1
C
d
Q BC
st
K
d
d st
Example 4 . The shaft AB suddenly stops rotating at end A .
(3)
d
d
d
( K )
d
d st
When
d 2
d 0.8, 1
l 2
l 0.1 1
(3) 0.82 (1) 1.25 (2)
d
d
d
(1) 1.52 (2)
d
d
(1)
(3)
(2)
d
d
d
d1 d2
(1) l2
(2)
(3) l1
W
v
W
v
W
v
1 . Practical example No reasonable reasonable reasonable
Determine the maximum dynamic stress .
Solution : potential energy : V = 0
stain energy :
U
1T

T 2l d
d 2 d d 2GI
p
kinetic energy : T 1 I 2
2x
P brake
Using T + V = Ud , We get :
Pd
U

2 d
W
d 2
st
T V T W d
2
2

2T st
0
d
st d
W
(1 1 2T )
d
st
W
st
K
d 1
1
2T
d
W
st
st
冲击动荷系数
6 . 讨论: ① Kd -----冲击动荷系数
K

d
1
1
2T
d
W
st
st
Βιβλιοθήκη Baidu
P K W; K ; K
W Pd
d
3 . 方法 : 能量守恒定律
4 . 模型
m
l
l
Δl
P P
l Pl P EA EA/ l
ml m
GI p
GI / l p
P
A
C
B
l
y Pl 3 P C 48EI 48EI / l3
胡克定律 P
P K
刚度系数
5 . 计算
Assuming that T V U d
H 0
K 2 d
K 2 d
K 1 1 2H
d

st
④
W
v
l
1W 2g
v2

1 2
Pd d

1 2
K
2 d

W st
W
K d

P d
H
d
d W
v st
2
st
K
l
d g
st
⑤
2H 10,
K 1 d
2H ,
error 5 0 0
st
st
2H 110, K 2H ,
之内转速 n=0，试确定最大剪应力 max
Solution :


1
0

0 nπ
30


π
rad/s2
t
t
3
m I π kN m
P
d
x
6
π
M m m km
t
d
f6
mf
A
P


M t

M t
2.67MPa
W max πd 3 16
动能
势能
应变能
W H
d
P
H
Pd
P st=W Pd
φ
st

d
When max
p
and Pd changes from 0 to Pd , We have :
U 1P
2 d
dd
K
d

P d
d
d W
st
st
U

2 dW
d 2
st
W H
d
T V U
H
d
g
N (x) W Ax (1 a )
d
A
A
g
(1 a )
st
g
静力平衡系统
x x
P
q A
a
N(x)
A a A
g
W
Wa W
g
(1 a )
d
st
g
Discussion : ① Let
K d d
, We get
st
K 1 a
d
g
②
K [ ]
第10章 动载荷
10.1 概述
载荷
静载荷 动载荷
惯性力 ( 内力 ) 冲击载荷
交变载荷 ( 下一章 )
注意 :
p
胡克定律
10.2 惯性力作用下应力的确定
达朗贝尔原理 :
质点系原力系系统
惯性力系统
1 . 物体 W匀加速上升
N (x) W W a Ax A ax
g
g
(1 a )(W Ax )
d g2
2g
Y 0,
qπ
0d
D d
2
sin
2N d

0
N

qD d

AD2
2

A
v2
d
2
4g
g


N d
D2 2

v2
d A 4g
g


N d
D2 2


v2
d A 4g
g

t D
D
qd
d dS
O
d

Nd
Nd
Discussion : ①
GI I
T
px
d
l
A P
B
l


T d

GI I
p x
2GI x
W d max
LW 2
Al
t
t
I (p
2)
W2 A
t


T d

2GI x
W d max
Al
t
Discussion : ① Volume Al↑
d max
② d =100 mm , l = 1m , Ix = 0.5 kN·m·s2 , n = 100 r/min