已知条件(见上图)： Known conditions (see above) : a. Material = PC，许用应变 4% A. Material = PC b. Length (L) = 19mm B. Length (L) = 19mm c. Width (b) = 9.5mm C. Width (b) = 9.5 mm d. Undercut (y) = 2.4mm (y) = 2.4 mm e. Angle of inclination (a) = 30° E. Angle of inclination (a) = 30 ° 求解： Solution: a. 在充分变形y下，厚度h允许的应变为材料许用应变的1/2 A. under the full deformation y, the strain of h allowed to be applied to the material with a strain of 1/2 b. 变形力P B. the deforming force P c. 装配力W C. assembly force W 解答： Answer: a. 确定厚度h A. determining thickness h 材料许用应变为：ε pm=4%， The material can be applied to: epsilon PM = 4%, 该处允许的应变为ε=1/2 ε pm=2% The allowable strain is epsilon = 1/2 epsilon PM = 2% 变形公式参照上篇文章的表一，type 2，shapeA： The deformation formula is based on table 1 of the previous article, type 2, shapeA: Y=1.09 (ε l2)/h àh=1.09 ε l2 /y = 1.09 x 0.02 x 192 / 2.4 = 3,28 mm Y = 1.09 (epsilon l2)/h ah = 1.09 epsilon l2 / Y = 1.09 x 0.02 x 192/2.4 = 3 28 mm b. 确定变形力P B.determine the force P 同样参照上篇文章的表一，crosssection A, And also in the form of the previous one, crosssection A, P= (bh2/2) (Es ε / L ) 从材料属性图，在ε=2%时，Es = 1815 Mpa P = (bh2/2) (Es epsilon/L) from the material properties diagram, at epsilon = 2%, Es = 1815 Mpa P=(9.5 x 3.282 /6) (1815 x0.02 /19) = 32.5 N P = (9.5 x 3.282/6) (1815 x0.02/19) = 32.5 N c. 确定装配力 C. determine the assembly force W= P (u + tanα)/ (1- tanα) W = P (u + tan alpha)/(1 - tan alpha) 摩擦系数:上篇表2，u=0.5 x 1.2 = 0.6 Friction factor: in table 2, u = 0.5 x 1.2 = 0.6 从上篇图十五，u=0.6，α= 30°， From the last figure 15, u = 0.6, alpha = 30 °, W = 32.5 x 1.8 =58.5 N W = 32.5 x 1.8 = 58.5 N