材料力学(双语)Torsion

∑m
x
= 0, T =m
T −m =0
m
m
m 3). Sign conventions for internal torque (扭矩的符号规定)： 扭矩的转向与截面外法线方向满足右手螺旋规则为正，反之为负。 8
T
x
4). Internal torque diagram(扭矩图)： 表示沿杆件轴线各横截面上扭矩变化规律的图线 ① 扭矩变化规律 ②|T|max值及其截面位置，可用于确定危险截面及强度计算
A0：Area of the circle with an average radius
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3. 剪应力互等定理：
a dy
γ
τ´
b
∑ mz = 0
τ ⋅ t ⋅ dxdy = τ ′ ⋅ t ⋅ dxdy τ = τ ′ (a) Thus,
τ τ´
c z dx d t
τ
Formula (a) is called theorem of conjugate shearing stresses (剪应力互等定理). 该定理表明：在单元体相互垂直的两个平面上，剪应力必 然成对出现，且数值相等，两者都垂直于两平面的交线， 其方向则共同指向或共同背离该交线。 单元体的四个侧面上只有剪应力而无正应力作用，这种应 力状态称为纯剪切应力状态。
=
π
32 4 πD 4 = (1 − α ) 32
(D − d )
4 4
d (α = ) D
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④ Distribution of the shearing stresses (应力分布)
（solid section）
（hollow section）
In engineering the members with hollow section are widely used to increase the strength, save materials and decrease the weight of structures.
D
10
②求扭矩（扭矩按正方向设） m 2
1
m3
2
m1
3
m4
x
∑m
x
= 0 , T1 + m2 = 0
A 1 B 2 C n 3 D
T1 = −m2 = −4.78kN ⋅ m
T2 + m2 + m3 = 0 , T2 = −m2 − m3 = −(4.78 + 4.78 ) = −9.56kN ⋅ m − T3 + m4 = 0 , T2 = m4 = 6.37 kN ⋅ m
剪应变（γ）：直角的改变量
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工 程 实 例
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3.2 External Torque of Shaft · Internal Torques and its Diagrams
1. External torque of a transmission shaft The relation between the transmission power, revolution and external torque of the transmission shaft： 传动轴的外力偶矩: 传递轴的传递功率、转数与外力偶矩的 关系
1、Experiment： 1). Preparation：
①绘纵向线，圆周线 ②施加一对外力偶 m。
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2. After deformation ：
①圆周线不变 ②纵向线变成斜直线
3.Conclusions：
①圆筒表面的各圆周线的形状、大小和间距均未改变， 只是绕轴线作了相对转动。 ②各纵向线均倾斜了同一微小角度 γ 。 ③所有矩形网格均歪斜成同样大小的平行四边形
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目录
§3.1 概述 §3.2 传动轴的外力偶矩 · 扭矩及扭矩图 §3.3 薄壁圆筒的扭转 §3.4 等直圆杆在扭转时的应力 · 强度条件 §3.5 等直圆杆在扭转时的变形 · 刚度条件 §3.6 等直圆杆的扭转超静定问题 §3.7 等直非圆杆自由扭转时的应力和变形
3
3.1 Introduction
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4. Hooke’s law of shear (剪切虎克定律)：
l Relation between ϕ and γ ：
γ ⋅L =ϕ ⋅R
So γ = ϕ ⋅ R L
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T=m
τ
ϕ
γ
T (τ ⋅ 2 A 0 t )
∝
ϕ
(γ ⋅ L ) R
τ ∝γ
剪切虎克定律：当剪应力不超过材料的剪切比例极 限时（τ ≤τp)，剪应力与剪应变成正比关系。
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③绘制扭矩图
m2
m3
m1 n
m4
A T
B
C
D
6.37
⊕ – 4.78 – 9.56
x
T max = 9.56 kN⋅ m
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3.3 Torsion of Thin-walled Tube
Thin-walled Tube (薄壁圆筒)：
Thickness of the wall
1 t ≤ r0 （r0: average radius） 10
G1G′ ρ ⋅ dϕ γ ρ ≈ tgγ ρ = = dx dx
dϕ γρ = ρ dx
距圆心为 ρ 任一点处的γρ与该点 到圆心的距离ρ 成正比
dϕ ——扭转角沿长度方向变化率 dx
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2). Physical relation (物理关系)： Hooktituting it into the preceding formula we get (代入上式得)：
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τ = G⋅ γ
式中：G是材料的一个弹性常数，称为剪切弹性模 量，因γ 无量纲，故G的量纲与τ 相同，不同材料 的G值可通过实验确定，钢材的G值约为80GPa。 剪切弹性模量、弹性模量和泊松比是表明材料弹 性性质的三个常数。对各向同性材料，这三个弹 性常数之间存在下列关系：
E G= 2(1 + μ )
A
B
O
A m
γ
O ϕ B m
4
A m
γ
O ϕ B m
The angle of twist (ϕ)：The angle of rotation of one section with respect to another .
扭转角（ϕ）：任意两截面绕轴线转动而发生的 角位移
Shearing strain (γ )：The change of a right angle between two straight lines.
dϕ dϕ τρ = G ⋅γ ρ = G ⋅ ρ = ρ ⋅ G dx dx
dϕ τρ = ρ G dx
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3). Static relation (静力学关系):
dA τp
T = ∫ A dA ⋅ τ ρ ⋅ ρ dϕ dϕ = ∫ A Gρ dA = G ∫ A ρ 2 dA dx dx
2
ρ
Shaft：In engineering the members of which deformations are mainly torsion. Such as transmission shafts in machines, drill rods in oil-drilling rigs etc. 轴：工程中以扭转为主要变形的构件。 Torsion：Resultant of the external forces is a force couple and its acting plane is perpendicular to the axis of the shaft. Under this case the deformation of the rod is torsion. 扭转：外力的合力为一力偶，且力偶的作用面与直杆的 轴线垂直，杆发生的变形为扭转变形。
P = m ⋅α / t
P m = 9.55 (kN ⋅ m) n
where：P – power(功率), unit: kilowatt (kW) (千瓦) n – rotational speed(转速), unit: r/min or (rpm) (转/分)
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2. Internal torque and its diagram 1). Internal torque (扭矩): 构件受扭时，横截面上的内力偶 矩，记作“T” 2). Determine the internal torque by the method of section (截面法求扭矩):
1.观察圆直杆扭转试验
a.横截面变形后仍为平面 b.轴向无伸缩 c.纵向线变形后仍为平行
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2. Stresses on the cross section of the circular bar in torsion 等直圆杆扭转时横截面上的应力 1). Geometric deformation relation (变形几何关系)：
4. 公式讨论： ① 仅适用于各向同性、线弹性材料，在小变形时的等圆截面 直杆。 ② 式中：T—横截面上的扭矩，由截面法通过外力偶矩求得。
ρ —该点到圆心的距离。
Ip—截面极惯性矩，纯几何量，无物理意义。
I p = ∫ A ρ2dA
Unit：mm4，m4。
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③ 尽管由实心圆截面杆推出，但同样适用于空心圆截面杆， 只是Ip值不同。 a. For solid circular section (实心圆截面)：
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3.4 Stresses in Circular Bar under Torsion · Strength Conditions
Stress on the cross section of the circular bar
①变形几何方面 Geometric deformations ②物理关系方面 Physical relations ③静力学方面 Statics
Chapter 3 Torsion
MECHANICS OF MATERIALS 材料力学（双语）
Content
§3.1 Introduction §3.2 External Torque of Shaft · Internal Torques and its Diagrams §3.3 Torsion of Thin-walled Tube §3.4 Stresses in Circular Bar under Torsion · Strength Conditions §3.5 Deformations of Circular Bar under Torsion · Rigidity Conditions §3.6 Statically Indeterminate Problems of Circular Bar in Torsion §3.7 Stresses and Deformations of Noncircular Bar under Free Torsion
I p = ∫ A ρ 2 dA = ∫ ρ ⋅ 2 ρ ⋅ π ⋅ dρ
2 D 2 0
dρ
ρ
O
D
=
πD
32
4
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b. For a hollow circular section (空心圆截面):
dρ
I p = ∫ A ρ dA
2
ρ
d O D
= ∫ ρ ⋅ 2 ρ ⋅ π ⋅ dρ
2
D 2 d 2
O
Let
I p = ∫ A ρ2dA
dϕ T = GIp dx
dϕ T = dx GI p
After substituting into the physical relation (代入物理关系式)
dϕ τρ = ρ G dx
we get
T ⋅ρ τρ = Ip
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T ⋅ρ τρ = Ip
—横截面上距圆心为ρ处任一点剪应力计算公式。
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a 微小矩形单元体如图所示： ①无正应力 ②横截面上各点处，只产生垂直于 半径的均匀分布的剪应力τ ，沿周 向大小不变，方向与该截面的扭矩 方向一致。 c dy
γ
τ´
b
τ τ´
d dx
τ
τ
τ
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2. 薄壁圆筒剪应力τ 大小
∫
A
τ ⋅ dA ⋅ r0 = T
τ
τ
So τ ⋅ r0 ⋅ ∫ AdA = τ ⋅ r0 ⋅ 2π r0 ⋅ t = T T T = τ= 2 2π r0 t 2 A 0 t
T
⊕
x
9
Example 1 已知：一传动轴， n =300r/min，主动轮输入 P1=500kW，从动轮输出 P2=150kW，P3=150kW，P4=200kW， 试绘制扭矩图。 Solution：①计算外力偶矩
m2
m3
m1 n
m4
P m1 = 9.55 1 n A B C 500 = 9.55 ⋅ = 15.9(kN ⋅ m) 300 P2 150 m2 = m3 = 9.55 = 9.55 ⋅ = 4.78 (kN ⋅ m) n 300 P4 200 m4 = 9.55 = 9.55 ⋅ = 6.37 (kN ⋅ m) n 300