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B.
=ax2 ,
X=0, Y=0
• 满足相容方程 4 =0 • 由下式求出应力分量 x=2/y2=0 y= 2/x2=2a xy=-2/xy=0 • 对和坐标轴平行的矩形板求出面力分量知 =ax2 能解决矩形板y向受均匀拉力(a>0)或 均 匀压力(a<0)的问题。P37 Fig.3.1.1(a)
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E.
=ay3 ,
X=0, Y=0
• It satisfies the compatibility equation 满足相容方程 4 =0 • Find the stress components by 由下式求出应力分量 x=2/y2=6ay y= 2/x2=0 xy=-2/xy=0 • For a rectangular plate shown in fig.1, find the surface force components shown in fig.1对矩形 板求出面力分量，如图1所示。 • Solve the problem of pure bending of a rectangular beam. 矩形板纯弯曲问题
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C.
=2 ,
X=0, Y=0
• It satisfies the compatibility equation 4 =0 • find the stress components by x=2/y2=2c y= 2/x2=0 xy=-2/xy=0 for a rectangular plate with its edges parallel to the coordinate axes, find the surface force components by (lx+m yx)s=X (my+lxy)s =Y • the stress function =cy2 can solve the problem of uniform tension (c>0) or uniform compression (c<0) of a rectangular plate in x direction. P37 Fig.3.1.1(c)
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Integration of geometrical equations x=u/x y=v/y rxy=u/y+v/x 几何方程的积分 • substitution of strains into the geometrical equations (2.4.6) yields 将应变代入几何方程 u/x =My/(EI) u=Mxy/(EI)+f(y) v/y = -My/(EI) v= -My2/(2EI)+g(x) u/y+v/x=0 -df(y)/dy=dg(x)/dx+Mx/(EI)
8
C.
=cy2 ,
X=0, Y=0
• 满足相容方程 4 =0 • 由下式求出应力分量 x=2/y2= 2c y= 2/x2=0 xy=-2/xy=0 • 对和坐标轴平行的矩形板求出面力分量知 =cy2 能解决矩形板x向受均匀拉力(c>0)或 均 匀压力(c<0)的问题。P37 Fig.3.1.1(c)
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E.
• Fig 1
=ay3 ,
X=0, Y=0
• Fig 2 • statically equivalent systems 静力等效 a=2M/h3 • x=6ay=12My/h3=My/I I=h3/12 y= 0 xy=0
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3.2 Determination of displacements when x=My/I y= 0 xy=0 位移的确定 • In the case of plane stress, substitution of stresses into the physical equations(2.6.4) yields 平面应力问题，将应力代入物理方程得 应变 x=[x- y]/E =My/(EI) y=[y- x]/E = - My/(EI) rxy=xy/G =0
4
B.
=ax2 ,
X=0, Y=0
• It satisfies the compatibility equation 4 =0 • find the stress components by x=2/y2=0 y= 2/x2=2a xy=-2/xy=0 • for a rectangular plate with its edges parallel to the coordinate axes, find the surface force components by (lx+m yx)s=X (my+lxy)s =Y • the stress function =ax2 can solve the problem of uniform tension (a>0) or uniform compression (a<0) of a rectangular plate in y direction. P37 Fig.3.1.1(a)
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Separation of variables 分离变量 -df(y)/dy=dg(x)/dx+Mx/(EI)= • -df(y)/dy= f(y)=- y+u0 • dg(x)/dx+Mx/(EI)= g(x)= - Mx2/(2EI)+x+v0 • u=Mxy/(EI)- y+u0 (3.2.5) v= -My2/(2EI)-Mx2/(2EI)+x+v0 (3.2.6) =u/y= Mx/(EI)- ----the cross section remains plane after bending 横截面弯曲变形后仍 为平面。 1/ = 2v/x2 =- M/(EI) ----all the longitudinal lines will have the same curvature曲率相同
2
A.
=a+bx+cy ,
X=0, Y=0
• It satisfies the compatibility equation 4 =0 • find the stress components by x=2/y2=0 y= 2/x2=0 xy=-2/xy=0 • find the surface force components by (lx+m yx)s=X=0 (my+lxy)s =Y=0 • a linear stress function corresponds to the case of no surface forces and no stress . • The superposition of a linear function to the stress function for any problem does not affect the stresses.
3
A.
=a+bx+cy ,
X=0 Y=0
• 满足相容方程 4 =0 • 由下式求出应力分量 x=2/y2=0 y= 2/x2=0 xy=-2/xy=0 • 由下式对给定坐标的物体求出面力分量 (lx+m yx)s=X=0 (my+lxy)s =Y=0 • 确定所设定的 能解决的问题为：任意物体无 体力，无面力，无应力。 • 应力函数加或减一个线性项不影响应力。
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Review: Rigid-body displacements-displacements corresponding to zero strains 刚体位移--应变为零时的位移 • u= - y +u0 v= x+v0
• u0--the rigid-body translation in the x direction x向刚体平动 • v0--the rigid-body translation in the y direction y向刚体平动
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D.
=bxy ,
X=0, Y=0
• It satisfies the compatibility equation 4 =0 • find the stress components by x=2/y2=0 y= 2/x2=0 xy=-2/xy=-b • for a rectangular plate with its edges parallel to the coordinate axes, find the surface force components by (lx+m yx)s=X (my+lxy)s =Y • the stress function =bxy can solve the problem of a rectangular plate in pure shear. P37 Fig.3.1.1(b)
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3.3 Bending of a simple beam under uniform load简支梁在均布荷载作用下的弯曲
• A simple beam ,length 2L, depth h, uniform load q. P41 fig.3.3.1
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Semi-inverse method: 半逆解法 assume 假定y= 2/x2=f(y)
• Successive integration with respect to x yields 对x连续积分二次得 /x=xf(y) +f1(y) =x2 f(y)/2 +xf1(y)+f2(y) • substitution of into compatibility equation (4/x4+24/x2y2 +4/y4)=0 (2.12.11) yields 将代入相容方程得： x2/2 f(4)(y)+xf1(4)(y)+f2(4)(y)+2f”(y)=0