胜录国的具些质 Appendix Properties of Plane Areas
Appendix Ⅰ Properties of Plane Areas
截面的几何性质(Properties of Plane Areas) 附录I截面的几何性质 (Appendix I Properties of plane areas) 1-1截面的静矩和形心(The first moments of the area& centroid of an area) 1-2极惯性矩惯性矩惯性积(Polar moment of inertia Moment of inertia Product of inertia) 1-3平行移轴公式(Parallel-Axis- theorem) 1-4转轴公式(Rotation of axes)
(Properties of Plane Areas) 附录Ⅰ 截面的几何性质 (Appendix Ⅰ Properties of plane areas) §1-1 截面的静矩和形心(The first moments of the area & centroid of an area) §1-4 转轴公式 (Rotation of axes) §1-2 极惯性矩 惯性矩 惯性积 (Polar moment of inertia Moment of inertia Product of inertia) §1-3平行移轴公式 (Parallel-Axis theorem)
國的肌何质( Properties of Plane Areas)國□ §1-1截面的静矩和形心 CThe first moment of the area centroid of an area 静矩( The first moment of the area) 截面对y,z轴的静矩为 zdA i dA ydA 静矩可正,可负,也可能等于零
(Properties of Plane Areas) §1-1 截面的静矩和形心 (The first moment of the area & centroid of an area) 一、静矩(The first moment of the area ） O y z dA y z 截面对 y , z 轴的静矩为 静矩可正，可负，也可能等于零. = A S y zdA = A Sz ydA
國的肌何质( Properties of Plane Areas)國□ 二、截面的形心( Centroid of an area) ZdA s J da da S J A A S,=Az S= Ay (1)若截面对某一轴的静矩等于零,则该轴必过形心. (2)截面对形心轴的静矩等于零
(Properties of Plane Areas) y z O dA y z 二、截面的形心(Centroid of an area) C z y A S A z A z A y = = d A S A y A y A z = = d S Az y = Sz = Ay （2）截面对形心轴的静矩等于零. （1）若截面对某一轴的静矩等于零，则该轴必过形心
國的肌何质( Properties of Plane Areas)國□ 三、组合截面的静矩和形心 CThe first moments ¢roid of a composite area) 由几个简单图形组成的截面称为组合截面 截面各组成部分对于某一轴的静矩之代数和,等于该截 面对于同一轴的静矩 用
(Properties of Plane Areas) 三、组合截面的静矩和形心 (The first moments ¢roid of a composite area) 由几个简单图形组成的截面称为组合截面. 截面各组成部分对于某一轴的静矩之代数和，等于该截 面对于同一轴的静矩
國的肌何质( Properties of Plane Areas)國□ 1.组合截面静矩( The first moments of a composite area) S=∑Az;S:=∑4 其中A1-第个简单截面面积 (z;,y—第诠个简单截面的形心坐标 2.组合截面形心( Centroid of a composite area) ∑ ∑Ay ∑A ∑A
(Properties of Plane Areas) 其中 Ai —第 i个简单截面面积 1.组合截面静矩(The first moments of a composite area) 2.组合截面形心(Centroid of a composite area) S A zi n i y i = = 1 = = n i z i i S A y 1 (zi , yi )—第 i个简单截面的形心坐标 = = = n i i n i i i A A z z 1 1 = = = n i i n i i i A A y y 1 1
國的肌何质( Properties of Plane Areas)國□ 例题1试确定图示截面形心C的位置 解:组合图形,用正负面积法解之 方法1用正面积法求解将截面分为,20 两个矩形 取z轴和y轴分别与截面的底边和左边缘 重合 ∑A a Any,+ A2 y2 ∑A A1+A2 O 90 i=1 A11+A22 图(a) A1+A2
(Properties of Plane Areas) 解：组合图形，用正负面积法解之. 方法1 用正面积法求解. 将截面分为1，2 两个矩形. 例题1 试确定图示截面形心C的位置. 取 z 轴和 y 轴分别与截面的底边和左边缘 重合 A A A y A y A A y y n i i n i i i 1 2 1 1 2 2 1 1 + + = = = = A A A z A z z 1 2 1 1 2 2 + + = 10 10 1 2 O z y 90 1 y 1z 2 z 2 y 图(a)
國的肌何质( Properties of Plane Areas)國□ 矩形1A1=10×120=1200mm2 D=5mm 1=60mm 矩形2A2=10×80=800mm2 80 J2=10+ =50mm 2 2 =5mm O 所以p=41+A2 23mm 90 A1+A2 z1+ 2=38mm A1+A2
(Properties of Plane Areas) 矩形 1 矩形 2 2 A1 = 10120 = 1200mm y1 = 5mm z1 = 60mm2 A2 = 1080 = 800mm 50mm 2 80 10 2 y = + = z 2 = 5mm 所以 38mm 23mm 1 2 1 1 2 2 1 2 1 1 2 2 = + + = = + + = A A A z A z z A A A y A y y 10 10 1 2 O z y 90 1 y 1z 2 z 2 y
國的肌何质( Properties of Plane Areas)國□ 方法2用负面积法求解,图形分割及坐标如图(b) 负面积c1(00J ∑A_141+卫242 C2(5,5) A1+A2 J 5×(-80×110) 22 120×90-80×110 图(b)
(Properties of Plane Areas) 方法2 用负面积法求解，图形分割及坐标如图(b) 图(b) C1（0,0） C2（5,5） C2 负面积 C1 y z = + + = = 1 2 1 1 2 2 A A y A y A A y A y i i 22 120 90 80 110 5 ( 80 110) = − − −
國的肌何质( Properties of Plane Areas)國□ §1-2极惯性矩、惯性矩、惯性积 (Polar moment of inertia, Moment of inertia, Product of inertia) 、惯性矩( Moment of inertia) dh Z dA VdA 二、极惯性矩( Polar moment of inertia) =Jpa4p2=x2+y2=「 dA 所以1P=12+1y
(Properties of Plane Areas) §1-2 极惯性矩、惯性矩、惯性积 (Polar moment of inertia、Moment of inertia、Product of inertia) y z O dA y z 二、极惯性矩 (Polar moment of inertia） 一、惯性矩(Moment of inertia) = = A z A y I y A I z A d d 2 2 = A IP dA 2 z y 2 2 2 = + = A IP dA 2 所以 z y I = I + I P