意画应 4.几何方程: A,B,-ABA,B,-00, de b AB OO1 B (p+ y)d0-pd8 y B od e E.= y 中性轴 o(M)
A1 B1 O O1 4. 几何方程： ...... (1) y x = a b c d A B dq x y 1 1 A1 B1 OO1 AB A B AB x − = − = ) ) ) OO1 ) q y q q y = + − = d ( )d d
STRESSESINBENDING (2) Physical relation Assume there is no extrusion between the longitudinal fibers, therefore an arbitrary point in the beam is at an uniaxial stressed state Ey o=Ea (2) (3) Static relations: E3 E ∑N=Jou-∫ad4=∫ ydA ES2=0 Z Neutral axis S.=0 so z(neutral)axis is through d(M2) th e center of the section Y
（2）Physical relation： Assume there is no extrusion between the longitudinal fibers, therefore， an arbitrary point in the beam is at an uniaxial stressed state. ......(2) s Ey x = E x = sx sx （3）Static relations： = d = d = d = = 0 s z A A A x ES y A E A Ey N A th e center of the section. S 0 so z (neutral) axis is through z = Neutral axis Y x z
意画应 (二)物理关系: 假设:纵向纤维互不挤压。于是,任意一点均处于单项应 力状态。 a,=Ea Ey (2) (三)静力学关系: ∑N.=」od4=J E dA= el.ydA ES NA =0 中性轴 S.=0∴z(中性)轴过形心 o(M,)
（二）物理关系： 假设：纵向纤维互不挤压。于是，任意一点均处于单项应 力状态。 ......(2) s Ey x = E x = sx sx （三）静力学关系： = d = d = d = = 0 s z A A A x ES y A E A Ey N A Sz = 0 z(中性)轴过形心
STRESSESINBENDING ∑M-J(a=4= El yd4=x≡0 Symmetric plane ∑ E E M2=(od4)y= da dg EL =M A 1m (3) El- flexural rigidity P El of the beam M e。。 (4) Z Neutral axis 12 f(M) Y
= ( d ) = d = d = 0 s yz A A A y EI yz A E A Eyz M A z （Symmetric plane） M EI y A E A Ey M A y z A A A z = = = = = (sd ) d d 2 2 z z EI M = 1 … …(3) EIz flexural rigidity of the beam ...... (4) z x I s = M Neutral axis Y x z
意画应 ∑M,=(oa4)=JaEd= J√d4= ≡0 (对称面) 2M:=L(odA)y=lsy dA=LydA=2=M 1m ∴(3) P El EL2一杆的抗弯刚度。 ONE My 中性轴 d(M2)
= ( d ) = d = d = 0 s yz A A A y EI yz A E A Eyz M A z （对称面） M EI y A E A Ey M A y z A A A z = = = = = (sd ) d d 2 2 z z EI M = 1 … …(3) EIz 杆的抗弯刚度。 ...... (4) z x I M y s =