材料力学第二章剪切详解.ppt

* * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * [例2] 结构如图，AB、CD、EF、GH都由两根不等边角钢组成，已知材料的[?]=170 M P a ，E=210 G P a。 AC、EG可视为刚杆，试选择各杆的截面型号和A、D、C点的位移。 P=300kN 0.8m 3.2m 1.8m 1.2m 2m 3.4m 1.2m A B C D F H q0=100kN/m 解：①求内力，受力分析如图 E G D q0=100kN/m E G A C NG NC NA NE ND =ND P=300kN ②Determine the area by strength condition D q0=100kN/m E G A C NG NC NA NE ND =ND P=300kN ②由强度条件求面积 ③Determine the types of the rods reference to the table ④Determine the deformation ③试依面积值查表确定型钢号 ④求变形 ⑤Determine the displacement.The deformation is shown in the figure. A B D F H E G C C1 A1 E1 D1 G1 ⑤求位移,变形图如图 A B D F H E G C C1 A1 E1 D1 G1 [Example 3] Diameters of rod AC and BD in the structure shown in the figure are respectively d1 =25mm, d2 =18mm. Knowing [?]=170MPa，E =210GPa . Rod AB may be seen as a rigid rod. Try to (1) check the strength of each rod and to determine the displacements △A and △B of points A an B. (2) determine the displacement △F′of point F when the force P is acted on point A. △F′= △A is a general law, which is called theorem of conjugate displacement. B NB P=100kN NA A A B C D P=100kN 1.5m 3m 2.5m F Solution：①Determine the internal force The free body diagram is shown in the figure. [例3] 结构如图，AC、BD的直径分别为:d1 =25mm, d2 =18mm,已知材料的[?]=170 M Pa ，E=210 G Pa,AB可视为刚杆，试校核各杆的强度;求A、B点的位移△A和△B。(2)求当P作用于A点时,F点的位移△F′，△F′= △A是普遍规律：称为位移互等定理。 B NB P=100kN NA A A B C D P=100kN 1.5m 3m 2.5m F 解：?求内力，受力分析如图 ②Check the strength ③Determine the deformation and the displacement ②校核强度 ③求变形及位移 ④Determine displacement △F′ of point F when force P is acted on point A P=100kN 1.5m 3m 2.5m A F B C D Compared with Q (3) ④求当P作用于A点时,F点的位移△F′ P=100kN 1.5m 3m 2.5m A F B C D [Example 4] The structure is shown in the figure. Knowing [?]=2MPa ，E=20GPa, unit weight of concrete ? = 22kN/m3. Try to design the areas of the two parts and to determine the displacement △A of point A Solution：Determine the area in acco