��1 Exercise 4.2 For a given planar force system find: I An equivalent coupe-force system at point A. II. The simplest equivalent force system. Draw the result of reduction on the figure. q1=4 kN/m g�=3 kN/m2 10 kN y 1m x A 3��2 kN 2m 17 kNm 15 kN q2=2 kN/m 14 kN 2m 4m The origin of the coordinate system is positioned at point A I. An equivalent couple-force system at point A �� Replacing the distributed loads by their resultants. 1 q1=4 kN/m W1 g�=3 kN/m2 F1=10 kN y 1m 2 x *l2 3 W3 F2 A F3 l2 3��2 kN W2 2m 15 kN 1 *l2 F6 3 a� F5 q2=2 kN/m M1=17 kNm F4=14 kN 2m 4m Project  The development of the didactic potential of Cracow University of Technology in the range of modern construction is co-financed by the European Union within the confines of the European Social Fund and realized under surveillance of Ministry of Science and Higher Education. 2 W1 =� q1 �� l1 (q1 =� 4[kN / m], l1 =� 4[m]) W1 =� 4[kN / m]�� 4[m] =� 16[kN] 1 W2 =� q2 �� l2 (q2 =� 2[kN / m], l2 =� [3m]) 2 1 W2 =� �� 2[kN / m]�� 3[m] =� 3[kN] 2 W3 =� g� �� area (g� =� 3[kN / m2 ], area =� 1[m]�� 2[m] =� 2[m2 ]) W3 =� 3[kN / m2 ]�� 2[m2 ] =� 6[kN] �� Resolving the oblique forces into components F2 =� sin(45��) �� 3 2[kN] , F3 =� cos(45��) �� 3 2[kN] F2 =� F3 =� 3[kN] 4 3 Where: cos(a�) =� , sin(a�) =� (see the figure) 5 5 F5 =� cos(a� ) ��15[kN] =� 12[kN] F6 =� sin(a� ) ��15[kN] =� 9[kN] �� Determining the sum vector S =� (Sx , S ,O) y Sx =� F1 +� F2 -� F5 -�W2 Sx =� -�2[kN] S =� -�F3 +� F4 +� F6 -�W1 -�W3 S =� -�2[kN] y y S =� (-�2[kN],-�2[kN],O) �� Detrmining the total moment about point A + - W1 rW1 F1 W3 rW3 rF1 y 1m x 2m F2 A F3 rF3 rW2 rF5 2m W2 1m F6 rF6 M1 F5 F4 2m 2m 2m Project  The development of the didactic potential of Cracow University of Technology in the range of modern construction is co-financed by the European Union within the confines of the European Social Fund and realized under surveillance of Ministry of Science and Higher Education. 3 M =� (0,0, M ) (for a planar force system) A Az M =� -�F1�� rF1 +� F3 �� rF 3 -� F5 �� rF 5 +� F6 �� rF 6 -�W1 �� rW 1 -�W2 �� rW 2 +�W3 �� rW 3 +� M1 Az Remark: the moment of F2 and F4 about A equal zero, because their lines of action pass through that point. M =� -�10[kN]��1[m] +� 3[kN]�� 2[m] -�12[kN]�� 2[m] +� 9[kN]�� 4[m] -�16[kN]�� 2[m] -� 3[kN]��1[m] +� Az +� 6[kN]��1[m] +�17[kNm] =� -�4[kNm] M =� (0,0,-�4[kNm]) A �� Answer The planar system of forces can be reduced at point A to a coupe-force system comprising one force S =� (-�2[kN],-�2[kN],O) applied at point A, and one couple with a moment M =� (0,0,-�4[kNm]) , A II. The simplest equivalent system �� S =� (-�2[kN],-�2[kN],O) , M =� (0,0,-�4[kNm]) hence the parameter of the system k =� S o� M =� 0 A A S �� 0 �� k =� 0 the system can be reduced to a resultant force. �� The equation of a central axis $�P(x, y, z);M =� 0 P M =� M +� S �� AP p A point A =� (0,0,0) point P =� (x, y,0) 0 =� M +� S �� AP A vector AP =� (x, y,0) S =� (-�2[kN],-�2[kN],0) (0,0,0) =� (0,0,-�4[kNm]) +� (0,0,-�2[kN]�� y +� 2[kN]�� x) �� AP =� ( x , y ,0) 0 =� -�4[kNm] -� 2[kN]�� y +� 2[kN]�� x �� y =� x -� 2[m] S ��AP =�(0,0,-�2[kN]�� y+�2[kN]�� x) y =� x -� 2[m] the central axis (or the line of action of the resultant force.) �� Answer The given planar system of forces can be reduced to a resultant force equal to the sum vector, acting along the central y =� x -� 2[m] Project  The development of the didactic potential of Cracow University of Technology in the range of modern construction is co-financed by the European Union within the confines of the European Social Fund and realized under surveillance of Ministry of Science and Higher Education. 4 q1=4 kN/m g�=3 kN/m2 10 kN y 1m x 3��2 kN A 2m W=S(-2[kN],-2[kN]) 17 kNm 15 kN q2=2 kN/m 14 kN 2m 4m Project  The development of the didactic potential of Cracow University of Technology in the range of modern construction is co-financed by the European Union within the confines of the European Social Fund and realized under surveillance of Ministry of Science and Higher Education. ] m [ 2 - x = y