Mechanics of Materials
CHAPTER 2 SHEAR
§ 2－ 1 STRENGTH CALCULATION ABOUT SHEAR
AND BEARING OF CONNECTING MEMBERS
EXCERCISE LESSONS OF SHEAR AND BEARING

§ 2－ 1 连接件的剪切与挤压强度计算
Production of the shearing stress

§ 2－ 1 STRENGTH CALCULATION ABOUT SHEAR AND
BEARING OF CONNECTING MEMBERS
1,Characteristics of loads and deformation of connecting members,
1)Connecting member
The structure member that connects one member to another is called the
connecting member,Such as,bolts,rivets,keys etc,The connecting
member is small,but it plays the role of passing loads,
Characteristic,It can
pass general loads and can be
dismounted,
P P
bolt
§ 2－ 1 连接件的剪切与挤压强度计算

1、连接件

P

P P
rivet Characteristic,
It can pass general loads,but can not be dismounted.for example,
the truss in a bridge is connected by it,
no gap
Characteristic,It can pass torques,
m
Shaft
Key
Gear
m
P P

m

m
n n
（ Resultant）
（ Resultant）
P
P
Use a rivet as an example,
The rivet is subjected to two equal and
opposite forces,The acting lines of these
two forces are very close,
② Characteristic of deformation,
Two parts subjected to two equal and
opposite forces tend to shift over one another
along the junction plane of two forces,
2、受力特点和变形特点,
n n
（合力）
（合力）
P
P

② 变形特点,

n n
（ Resultant ）
（ Resultant
P
P
③ Shearing plane,
The plane along which two parts of the
member tend to shift over one another,Such
as n– n,
④ Internal force on shearing plane,
Internal force — Shearing force Q, its
acting line is parallel to the shearing plane,
P
n n
Q
Shearing
plane
n n
（合力）
（合力）
P
P
③ 剪切面,

n– n 。
④ 剪切面上的内力,

P
n n
Q 剪切面
n n
(Resultant )
(Resultant ）
P
P
3)Three kinds of breakage at joint,
① Failure due to shear
Snip along the shearing plane of the rivet,
such as along section n– n,
② Breakage due to bearing
Fail due to mutual bearing between the
rivet and the steel plate in their connecting plane,
③ Breakage due to tension
P
n n
Q
Shearing
plane
The steel plate is weakened in the section in
which the rivet holes exist and stress in the
weakened section increases so that the steel plate is easily broken due to
tension at the connecting position,
n n
（合力）
（合力）
P
P
3、连接处破坏的三种形式,
① 剪切破坏

② 挤压破坏

③ 拉伸 破坏
P
n n
Q 剪切面

2,Practical calculation of shear
Method of the practical calculation,According to possibility of breakage of
the member some assumptions by which basic characteristic subjected to force
actions can be reflected and calculations can be simplified are used,Then
calculate its nominal stress,determine the corresponding permissible stress in
accordance with the result of direct test,At last do the strength calculation,
Applying range,volume of the member is not large and real stress is quite
complex,Such as the connecting pieces etc,
Assumption of practical calculation,Assume that shearing stress is distributed
uniformly in the shearing plane and equal to the average shearing stress,

1)Shearing plane--AQ, Shifting plane,
Shearing force--Q,Internal force
on the shearing plane,
QA
Q??
2)Nominal shearing force--?,
3)Strength condition of shear,
? ??? ??
QA
Q ? ?
n
jx?? ?
n n
（ Resultant ）
(Resultant ）
P
P
P
n n
Q
Shearing
plane
Working stress should not exceed the permissible stress,
,where
1、剪切面 --AQ, 错动面。

QA
Q??
2、名义剪应力 --?,
3、剪切强度条件（准则）,
? ??? ??
A
Q
? ?
n
jx?? ?:其中
n n
（合力）
（合力）
P
P
P
n n
Q 剪切面

3,Practical calculations of bearing
1)Bearing force― Pjy, The resultant force acting on the touching plane,
Bearing,The phenomenon that there is pressure on the partial area of a member,
Bearing force,The resultant force acting on the bearing plane,designated by
Pjy,
Assumption,Bearing stresses are distributed uniformly over the effective
bearing plane,

1、挤压力 ― Pjy,接触面上的合力 。

2)Bearing area,Area of the projection plane of the touching plane in the
direction perpendicular to Pjy
? ?jy
jy
jy
jy A
P
?? ??
3)Strength condition of bearing,
Working bearing stress should not exceed the permissible bearing stress
Bearing area
dtA jy ?
2、挤压面积：接触面在垂直 Pjy方向上的投影面的面积。
? ?jy
jy
jy
jy A
P
?? ??
3、挤压强度条件（准则）,

jy ?
4、
Applications
][ ][ jyjy ???? ?? ；
1)Check the strength
][
][ jy
jy
jyQ
P
A
Q
A
??
?? ；
2)Select the cross section area
][
][
jyjyjy
Q
AP
AQ
?
?
?
][ ][1 jyjy ???? ?? ；、校核强度：
][
][
2
jy
jy
jyQ
P
A
Q
A
??
?? ；、设计尺寸：
][ ][3 jyjyjyQ APAQ ?? ?? ；、设计外载：

M P a952.010
3512
40 7 ??
?
???
bh
P
A
Q
Q
?
M P a 4.710
125.4
40 7
??
?
???
cb
P
A
P
jy
jy
jy?
Example 1 A wooden tenon joint is shown in the figure,Knowing that the
quantities are a = b =12cm,h=35cm,c=4.5cm and P=40KN,Try to determine
the shearing stress and bearing stress for the joint,
Solution,?,Free body diagram is
shown in the figure
?,Shearing stress and bearing
stress
PQbhA Q ?? ;
Shearing force is
Bearing force is
PPcbA jyjy ?? ;
P
P P
P
P P b
a c
h
QA
jyA
h
M P a952.010
3512
40 7 ??
?
???
bh
P
A
Q
Q
?
M P a 4.710
125.4
40 7
??
?
???
cb
P
A
P
jy
jy
jy?
[例 1] 木榫接头如图所示,a = b =12cm,h=35cm,c=4.5cm,
P=40KN，试求接头的剪应力和挤压应力。

? 剪应力和挤压应力

P
P P
P
P P b
a c
h
QA
jyA
h
PQbhA Q ?? ;
PPcbA jyjy ?? ;
Solution,?Free body diagram of the key is shown
in the figure
Example 2 A gear and a shaft are connected by a key（ b× h× L=20 × 12 × 100).The
torque that the key can transmit is m=2KNm.Knowing the diameter of the shaft is d=70mm
,the permissible shearing stress and the permissible bearing stress of the key are respectively
[?]= 60 MPa and [?jy]= 100 MPa,Try to check the strength of the key,
kN57
07.0
222 ????
d
mP
2
h
m
b
h
L m
d
P

[例 2] 齿轮与轴由平键（ b× h× L=20 × 12 × 100）连接，它传递

Pa,许用挤压应力为 [?jy]= 100M Pa，试校核键的强度。
kN57
07.0
222 ????
d
mP
2
h
m
b
h
L m
d
P
According to the above calculation,strength conditions of the key are satisfied,
? ??? ??
?
???? M P a6.28
1 0 020
1057 3
bL
P
A
Q
Q
?Check the strength of shear and bearing PPQ
jy ??
? ?jy
jy
jy
jy hL
P
A
P
?? ??
?
???? M P a3.95
6100
1057
2
3
b
h
L
d
m
Q

? ??? ??
?
???? M P a6.28
1 0 020
1057 3
bL
P
A
Q
Q
?剪应力和挤压应力的强度 校核 PPQ
jy ??
? ?jy
jy
jy
jy hL
P
A
P
?? ??
?
???? M P a3.95
6100
1057
2
3
b
h
L
d
m
Q
Solution,?Free body diagram of the key is
shown in the figure
[Example 3] A gear and a shaft are connected by a key（ b =16mm,h =10mm).The
torque that the key can transmit is m = 1600Nm,Knowing the diameter of the shaft is d
=50mm,the permissible shearing stress and the permissible bearing stress of the key are
respectively [?] = 80MPa and [?jy] = 240MPa.Try to design the length of the key,
kN6405.0 1 6 0 022 ?????? d mPQP jy
b
h
L
2
h
m
m
m
d
P

[例 3] 齿轮与轴由平键（ b=16mm,h=10mm，）连接，它传递的

Pa,许用挤压应力为 [? jy]= 240M Pa，试设计键的长度。
kN6405.0 1 6 0 022 ?????? d mPQP jy
b
h
L m
d
P
2
h
m
m
b
h
L
?Strength conditions of the shearing stress and the bearing stress
? ? mm50)m(108016
64][ ][ 3
1 ???????
?
?
?
b
QL
Lb
Q
mm3.53)m(10
24010
642
][
2
][ ][
2 3
2 ???
????? ?
jy
jy
jy
jy
h
P
L
Lh
P
?
?
?According to the
above calculation ? ? ? ? ? ?? ? mm3.53,m a x 21 ?? LLL
d
m
Q
b
h
L
?剪应力和挤压应力的强度条件
? ? mm50)m(108016
64][ ][ 3
1 ???????
?
?
?
b
QL
Lb
Q
mm3.53)m(10
24010
642
][
2
][ ][
2 3
2 ???
????? ?
jy
jy
jy
jy
h
P
L
Lh
P
?
?
?综上
? ? ? ? ? ?? ? mm3.53,m a x 21 ?? LLL
d
m
Q
Solution,?Free body diagram
is shown in the figure
[Example 4] A riveted tie-in acted by force P=110kN is shown in the figure,Knowing the
thickness is t =1cm,width of it is b =8.5cm,The permissible stress is [? ]= 160MPa,
Diameter of the rivet is d =1.6cm and the permissible shearing stress is [?]= 140MPa, the
permissible bearing stress is [?jy]= 320MPa,Try to check the strength of the riveting,
(Assume the force acted on each rivet is equal.）
4
PPQ
jy ??
b
P P
t
t
d
P
P
P
1
1 2
2
3
3
P/4

[例 4] 一铆接头如图所示，受力 P=110kN，已知钢板厚度为
t=1cm，宽度 b=8.5cm,许用应力为 [? ]= 160M Pa ；铆钉的直径
d=1.6cm，许用剪应力为 [?]= 140M Pa,许用挤压应力为 [?jy]=
320M Pa，试校核铆接头的强度。（假定每个铆钉受力相等。）
4
PPQ
jy ??
b
P P
t
t
d
P
P
P
1
1 2
2
3
3
P/4
?The sections 2—2 and 3—3 of the steel plate are the critical sections,
?The strength conditions of shear and bearing
? ??
?
? ???
?
??? M P a8.13610
6.114.3
110 7
22d
P
A
Q
Q
? ??? ???
???
??
?
? M P a7.15510
)6.125.8(4
1103
)2(4
3 7
2 dbt
P
? ?jy
jy
jy
jy td
P
A
P
?? ???
??
??? M P a9.17110
6.114
110
4
7
? ??? ???
??
?
?
? M P a4.15910
)6.15.8(1
110
)(
7
3 dbt
P
Therefore,the tie-in
is safe,
t
t
d
P
P
P
1
1 2
2
3
3
P/4
?钢板的 2--2和 3--3面为危险面
?剪应力和挤压应力的强度条件
? ??
?
? ???
?
??? M P a8.13610
6.114.3
110 7
22d
P
A
Q
Q
? ??? ???
???
??
?
? M P a7.15510
)6.125.8(4
1103
)2(4
3 7
2 dbt
P
? ?jy
jy
jy
jy td
P
A
P
?? ???
??
??? M P a9.17110
6.114
110
4
7
? ??? ???
??
?
?
? M P a4.15910
)6.15.8(1
110
)(
7
3 dbt
P

t
t
d
P
P
P
1
1 2
2
3
3
P/4
1,Internal-force and
axial-force diagrams of
the rod in tension and
compression
1)Expression of
Axial force?
2)Method to determine axial force?
3)Positive and negative of axial
force?
Why do we plot the axial-force diagram? What should we pay attention when
we do it?
4)Axial-force diagram,
Expressed by the diagram of N = N(x)?
P
A
N
A
Simple sketch
B C
P P
N
x
P
+

1、轴力的表示?
2、轴力的求法?
3、轴力的正负规定?

4、轴力图,N=N(x)的图象表示?
P
A
N
B C

A P P
N
x
P
+
Simple method to determine axial forces,
① Take the left part of the section x as the object,the axial force on the section x
can be calculated by following formula,
Where“?P(?)”and“?P(?)”express the sum of left direction forces and the sum
of right direction forces of the left part of the section x,
② Take the right part of the section x as the object,the axial force N(x) of point x
can be calculated by the following formulate,
Where,, and,,denote the sum of right direction forces and the sum
of left direction forces of the right part of section x,
?? ???? )()()( PPxN
?? ???? )()()( PPxN
? ?)(P ? ?)(P

① 以 x点左侧部分为研究对象,x点的轴力 N(x)由下式计算,

② 以 x点右侧部分为研究对象,x点的轴力 N(x) 由下式计算,

?? ???? )()()( PPxN
?? ???? )()()( PPxN
[Example 1] Forces 5P,8P,4P and P are acted at points A,B,C and D on
the rod respectively,their directions are shown in the figure,Try to plot the
axial-force diagram of the rod,
A B C D O
5P 4P P 8P
N
x
–3P
5P
P
2P

[例 1] 图示杆的 A,B,C,D点分别作用着 5P,8P,4P,P

A B C D O
5P 4P P 8P
N
x
–3P
5P
P
2P

Positive and negative of stress?
1)Stress on the cross section,
A
xN )( ??
2,Stress of the rod in tension or compression
Critical section and maximum working stress?
?
?
?
??
?
?
?
??
?
?
?
?
?
?
?
?
2s i n
2
)2c o s(1
2
0
0
2)Stress on the inclined section
Saint-Venant principle? Stress concentrations?
? N(x) P
? ?
x

1、横截面上的应力,
A
xN )( ??

?
?
?
??
?
?
?
??
?
?
?
?
?
?
?
?
2s i n
2
)2c o s(1
2
0
0
2、拉压杆斜截面上的应力
Saint-Venant原理?

? N(x) P
? ?
x
3,Strength design criterion,
1) Strength design criterion
? ? )
)(
)(m a x (
m a x ?? ?? xA
xN
? ? m a x ?? ?
① Check strength,
② Design the cross section area,
? ?
m a x
m in ?
NA ?
③ Design the load,? ?; m a x ?AN ? ? ? )( m a xNfP ?

1、强度设计准则?
? ? )
)(
)(m a x (
m a x ?? ?? xA
xN
? ? m a x ?? ?
① 校核强度,
② 设计截面尺寸,
? ?
m a x
m in ?
NA ?
③ 设计载荷,? ? ;
m a x ?AN ?
? ? )( m a xNfP ?
EA
NL
EA
PLL ???
1)Elastic law of the rod with equal axial forces
2)Elastic law with variable internal forces
3)Elastic law in uniaxial stressed state 1 ??
E
?
?? ???? LL xEA xxNxL )( d)( )d(
?
?
??
n
i ii
ii
AE
LN
L
1
4,Deformation and strain of the rod in tension or compression N （ x ）
x
d x
N(x)
dx
x
P P
EA
NL
EA
PLL ???
1,等轴力拉压杆的弹性定律
2、变内力拉压杆的弹性定律
3、单向应力状态下的弹性定律 1 ??
E
?
?? ???? LL xEA xxNxL )( d)( )d(
?
?
??
n
i ii
ii
AE
LN
L
1

x
d x
N(x)
dx
x
P P
4)Possion’s ratio（ or lateral deformation factor）
?
?? ??
5)The enlarged-deformation diagram and the method to determine the displacement
C'
A B
C
L1 L
2
P
C"
1L?2L?
4、泊松比（或横向变形系数）
?
?? ??
5、小变形放大图与位移的求法
C'
A B
C
L1 L
2
P
C"
1L?2L?
Assemble stress——initial stress
Temperature stress
① Equilibrium equations；
② Geometric equations—compatibility equations of deformation；
③ Physical equations—elastic laws；
④ Complementary equations,getting from geometric equations and
physical equations；
⑤ Solving the combined equations including of equilibrium equations
and complement equations,
6)The steps to solve the statically indeterminate problem

① 平衡方程；
② 几何方程 ——变形协调方程；
③ 物理方程 ——弹性定律；
④ 补充方程：由几何方程和物理方程得；
⑤ 解由平衡方程和补充方程组成的方程组 。
6、超静定问题的处理方法步骤,
5,Mechanical properties of the materials in tension and
compression
3)Unloaded law； Cold hardening； Cold-drawn time effect,
1)Elastic law
??? tg ; ?? E
E
? ?bsjx ????,,2.0?
4)Residual relative elongation
0
0
0
01 1 0 0???
L
LL?
5)Permanent relative reduction of area
0
0
0
10 1 0 0???
A
AA?
? ?
n
jx?? ?
2)Limit stress
6)Permissible stress

3、卸载定律；冷作硬化；冷拉时效。
、许用应力6
、极限应力2
1,弹性定律
??? tg ; ?? E
E
? ?bsjx ????,,2.0?
4、延伸率
5、面缩率
? ?
n
jx?? ?
0
0
0
01 1 0 0???
L
LL?
0
0
0
10 1 0 0???
A
AA?
1)Practical calculation of shear
? ??? ?? AQ
6,Practical calculation of shear and bearing of the connecting member of
the rods in tension and compression
n n
（ Resultant）
（ Resultant ）
P
P P
n n
Q
Shearing
plane
2)Practical calculation of bearing
? ?jy
jy
jy
jy A
P
?? ??
1、剪切的实用计算
? ??? ?? AQ

n n
（合力）
（合力）
P
P P
n n
Q 剪切面
2、挤压的实用计算
? ?jy
jy
jy
jy A
P
?? ??
Bearing area
dtA jy ?
][ ][ jyjy ???? ?? ；
][
][ jy
jy
jyQ
P
A
Q
A
??
?? ；
][ ][ jyjyjyQ APAQ ?? ?? ；
Check the strength
Design the dimension

jy ?
][ ][ jyjy ???? ?? ；校核强度：
][
][ jy
jy
jyQ
P
A
Q
A
??
?? ；设计尺寸：
][ ][ jyjyjyQ APAQ ?? ?? ；设计外载：
[Example 2] The rods AB,CD,EF and GH in the structure are shown in
the figure Each of them consists two of rolled unequal-legs steels.Knowing
[?]=170MPa,E =210GPa, Rods AC and EG may be seen as rigid rod,Try to
select the section dimension of each rod and to determine the displacements of
points A,D and C,
P=300kN
0.8m 3.2m 1.8m 1.2m
2m
3.4m
1.2m
A
B
C
D
F H q
0=100kN/m
Solution,① Determine the internal force.The free body diagram is shown
in the figure,
E G
[例 2] 结构如图,AB,CD,EF,GH都由两根不等边角钢组

P=300kN
0.8m 3.2m 1.8m 1.2m
2m
3.4m
1.2m
A
B
C
D
F H q
0=100kN/m

E G
kN1 8 6?EN
kN2403004 2.3 ???AN
kN603004 8.0 ???DN
kN1 7 4?GN
D
q0=100kN/m
E G
A C
NG
NC NA
NE
ND
=ND P=300kN
② Determine the area by strength condition
][?
i
i
NA ?
23 cm12.1410
1 7 0
2 4 0 ??? ?
ABA
2cm5.3?CDA
2cm9.10?EFA
2cm2.10?GHA
kN1 8 6?EN
kN2403004 2.3 ???AN
kN603004 8.0 ???DN
kN1 7 4?GN
D
q0=100kN/m
E G
A C
NG
NC NA
NE
ND
=ND P=300kN
② 由强度条件求面积
][?
i
i
NA ?
23 cm12.1410
1 7 0
2 4 0 ??? ?
ABA
2cm5.3?CDA
2cm9.10?EFA
2cm2.10?GHA
2
1 cm212.72),55690(2,????? ABAAB
2
1 cm89.12),32540(2,????? CDACD
2
1 cm609.52),54570(2:)( ????? EFAGHEF
③ Determine the types of the rods reference to the table
④ Determine the deformation
i
ii
i EA
LNL ??
mm67.210
54.141.2
4.32 4 0 4
1
??
?
???? ?
AB
ABAB
AB EA
LNL
mm91.0?? CDL mm74.1?? EFL mm63.1?? GHL
2
1 cm212.72),55690(2,????? ABAAB
2
1 cm89.12),32540(2,????? CDACD
2
1 cm609.52),54570(2:)( ????? EFAGHEF
③ 试依面积值查表确定型钢号
④ 求变形
i
ii
i EA
LNL ??
mm67.210
424.141.2
4.3240 4
1
??
?
???? ?
AB
ABAB
AB EA
LNL
mm91.0?? CDL mm74.1?? EFL mm63.1?? GHL
⑤ Determine the displacement.The deformation is shown in the figure,
mm61.2?????? CDDC L mm61.2???? ABA L
A
B
D
F H
E G
mm70.1????????? GHGHEFD LDGEG LL
C
C1 A1
E1 D
1 G1
⑤ 求位移,变形图如图
mm61.2?????? CDDC L mm61.2???? ABA L
A
B
D
F H
E G
mm70.1????????? GHGHEFD LDGEG LL
C
C1 A1
E1 D
1 G1
[Example 3] Diameters of rod AC and BD in the structure shown in the figure are respectively d1
=25mm,d2 =18mm,Knowing [?]=170MPa,E =210GPa, Rod AB may be seen as a rigid rod,Try to (1)
check the strength of each rod and to determine the displacements △ A and △ B of points A an B,(2)
determine the displacement △ F′of point F when the force P is acted on point A,△ F′=
△ A is a general law,which is called theorem of conjugate displacement,
B
NB P=100kN NA
A
A B
C D P=100kN
1.5m 3m 2.5m
F
A?
B?
F?
Solution,① Determine the internal force
The free body diagram is shown in the fingure,
kN7.661 0 05.4 3 ???AN kN3.33?BN
[例 3] 结构如图,AC,BD的直径分别为,d1 =25mm,d2 =18mm,已

B
NB P=100kN NA
A
A B
C D P=100kN
1.5m 3m 2.5m
F
A?
B?
F?

kN7.661 0 05.4 3 ???AN kN3.33?BN
② Check the strength
? ??
?
? ??? 24
i
ii
i d
N
A
N
? ??? ???
?
?? M P a8.13510
2514.3
7.664 9
2A
? ??? ?? M P a1 3 1B
③ Determine the deformation and the displacement
i
ii
i EA
LNL ??
mm62.110
251.214.3
5.27.664 2
2 ????
????? ?
AC
A
AC EA
LNL
mm56.1?? BDL
m m,56.1 m m,62.1 h av e w eSo ???? BA
② 校核强度
? ??
?
? ??? 24
i
ii
i d
N
A
N
? ??? ???
?
?? M P a8.13510
2514.3
7.664 9
2A
? ??? ?? M P a1 3 1B
③ 求变形及位移
i
ii
i EA
LNL ??
mm62.110
251.214.3
5.27.664 2
2 ????
????? ?
AC
A
AC EA
LNL
mm56.1?? BDL
m m,56.1 m m,62.1 ????? BA
④ Determine displacement △ F′ of point F when force P is acted on
point A
mm62.1??????? ACF L
AB
BFL
0 ;kN1 0 0 ???? BA NN
mm43.210
251.214.3
5.21004 2
2 ????
????? ?
ACL
FAC LL ?????
P=100kN
1.5m 3m 2.5m
A??
F??
A F B
C D
④ 求当 P作用于 A点时,F点的位移△ F′
mm62.1??????? ACF L
AB
BFL
0 ;kN1 0 0 ???? BA NN
mm43.210
251.214.3
5.21004 2
2 ????
????? ?
ACL
FAC LL ?????
P=100kN
1.5m 3m 2.5m
A??
F??
A F B
C D
2
221
1
11 )()(
EA
LGGP
EA
LGPL ??????
[Example 4] The structure is shown in the figure,Knowing [?]=2MPa,
E=20GPa,unit weight of concrete ? = 22kN/m3,Try to design the areas of the two
parts and to determine the displacement △ A of point A
? ? ? ???
11m a x
1
GPNA ???
Solution,Determine the area in accordance
with the strength condition,
? ? ? ???
212m a x
2
GGPNA ????
?? ???? LL xEA xxNxL )( d)( )d(
P=100kN
12m
12m
A
2
221
1
11 )()(
EA
LGGP
EA
LGPL ??????
[例 4] 结构如图，已知材料的 [?]=2MPa, E=20GPa,混凝土容

? ? ? ???
11m a x
1
GPNA ???

? ? ? ???
212m a x
2
GGPNA ????
?? ???? LL xEA xxNxL )( d)( )d(
P=100kN
12m
12m
A
Chapter 2 Exercises
1,What is the difference between bearing and compression？
2,Knowing the relation between the allowable shearing stress[τ]and the
allowable tensile stress[σ]of the bolt material is [τ]=0.6[σ],Try to
determine the reasonable ratio of the diameter d of the bolt to the height h
of the bolt cap。
Solution,
? ???? ??? 24 dPAN
? ???? ??? dhPAQ
Q
? ?
? ? 6.0
4/
2 ?? d
P
dh
P
???
?
4.2/ ?? hd

? ???? ??? 24 dPAN
? ???? ??? dhPAQ
Q
? ?
? ? 6.0
4/
2 ?? d
P
dh
P
???
?
4.2/ ?? hd

the safe pin will be snipped,Knowing for the safe pin,the average
diameter is d=5mm and the limit shearing stress is, Try to
determine the maximum force couple M that can be transmitted by the
Solution,
M P ajx 3 7 0??
jxQ
d
D
MAQ ??? ???
4//
2
mNDdM jx ??? 3.1454 2 ??

M P ajx 370??
jxQ
d
D
MAQ ??? ???
4//
2
mNDdM jx ??? 3.1454 2 ??
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