④求多余约束反力 将上述结果代入变形协调方程得 11P P X135P =0X1 (f)( 3 B BEI 48EI C 16 5网 ⑤求其它约束反力 16 16 由平衡方程可求得A端反 5Pl 力,其大小和方向见图(。 Filem (g) ⑥作弯矩图,见图(g)。 3Pl 16 ⑦求梁中点的挠度 16
16 ④求多余约束反力 将上述结果代入变形协调方程得 0 48 5 3 3 3 1 − = EI Pl EI X l X P 16 5 1 = ⑤求其它约束反力 由平衡方程可求得A端反 力，其大小和方向见图(f)。 C P A ( f ) B 16 5P 16 11P 16 3Pl ⑥作弯矩图，见图(g)。 (g) + – 16 3Pl 32 5Pl ⑦求梁中点的挠度
STATICALLYINDETERMINATE STRUCTURES Select the primary structure(see Fig. b)) as our study object. The structure subjected to the unit load is shown in Fig. ( h) P From Mohr's theorem we get (b) B 15P(2+x)-Px1(-x).dxA EⅠJ0-16 X 7P3 () 768E Attention: For the same statically h)AA kx-C B indeterminate structure. if we select different P redundant constraints, the primary structures (i) are different too. In the above example if we B select the rotational constraint at the fixed as X the redundant one, the primary structure is the simplely supported beam as shown in Fig. (i) 7
17 Select the primary structure (see Fig.( b)) as our study object. The structure subjected to the unit load is shown in Fig.(h). P A B C X1 (b) x 1 A B C (h) From Mohr’s theorem we get ( ) 768 7 ) ] ( ) d 2 ( 16 5 [ 1 3 2 0 = = + − − EI Pl x Px x x l P EI y l C Attention：For the same statically indeterminate structure,if we select different redundant constraints,the primary structures are different too. In the above example if we select the rotational constraint at the fixed as the redundant one, the primary structure is the simplely supported beam as shown in Fig.(i) C P A (i) B X1
选取基本静定系(见图(b)作为计算对象。单位载荷如图(h)。 用莫尔定理可得 P (b) 。P(+x)PxH(-x)dxn B yC EIJ0 16 2 X 7PI 768EI dA kx-c B 注意:对于同一静不定结构,若选 P 取不同的多余约束,则基本静定系 也不同。本题中若选固定端处的转 B X 动约束为多余约束,基本静定系是 1 如图()所示的简支梁。 18
18 选取基本静定系( 见图( b)) 作为计算对象。单位载荷如图(h) 。 P A B C X1 (b) x 1 A B C (h) 用莫尔定理可得 ( ) 768 7 ) ]( )d 2 ( 16 5 [ 1 3 2 0 = = + − − EI Pl x Px x x l P EI y l C 注意：对于同一静不定结构，若选 取不同的多余约束，则基本静定系 也不同。本题中若选固定端处的转 动约束为多余约束，基本静定系是 如图(i)所示的简支梁。 C P A (i) B X1
STATICALLYINDETERMINATE STRUCTURES 2 Canonical equation of the force method Rewrite the compatibility equation of deformation with the unknown force as the unknown quantity in above example into the following expression 1X1+△1p=0 It is the standard form of the compatibility equation of deformation. That is the so-called canonical equation of the force method X1-Redundant unknown quantity; the displacement at the acting point of x, that is induced by the unit force Xi and is along the direction of X on the primary structure AlP-the displacement at the acting point of X, that is induced by the original load and is along the direction of X, on the primary struture 19
19 2、Canonical equation of the force method Rewrite the compatibility equation of deformation with the unknown force as the unknown quantity in above example into the following expression. 11X1 +1P =0 It is the standard form of the compatibility equation of deformation. That is the so-called canonical equation of the force method. X1——Redundant unknown quantity； 11——the displacement at the acting point of X1 that is induced by the unit force X1 and is along the direction of X1 on the primary structure; 1P——the displacement at the acting point of X1 that is induced by the original load and is along the direction of X1 on the primary struture.
二、力法正则方程 上例中以未知力为未知量的变形协调方程可改写成下式 61X1+△1=0 变形协调方程的标准形式,即所谓的力法正则方程。 X1多余未知量 61在基本静定系上,X1取单位值时引起的在X1作用点沿 H1方向的位移; △1在基本静定系上,由原载荷引起的在X作用点沿 X1方向的位移; 20
20 二、力法正则方程 上例中以未知力为未知量的变形协调方程可改写成下式 11X1 +1P =0 变形协调方程的标准形式，即所谓的力法正则方程。 X1——多余未知量； 11——在基本静定系上， X1取单位值时引起的在X1作用点沿 X1方向的位移； 1P——在基本静定系上， 由原载荷引起的在X1作用点沿 X1方向的位移；