Equilibrium of the General Coplanar Force Group

Abstract

In this chapter, we will learn how to write down the equilibrium equations for a general coplanar force group. We mainly introduce two cases, i.e., the first one is on the one rigid body system and the other is on the rigid multi-body system, which is more complicated.

Exercises

• 5.1 As shown in Fig. 5.12, two bars are hinged at point A. A horizontal force is exerted at point D. The length AB = AC = BC = 2AD. Please give the reactive forces at point B and C.

  Fig. 5.12

  

  Two bars hinged together

• 5.2 Please give the reactive forces at point A and C, as shown in Fig. 5.13.

  Fig. 5.13

  

  Two bars hinged together

• 5.3 Please give the reactive forces at point A and C, as shown in Fig. 5.14.

  Fig. 5.14

  

  Two curved bars hinged together

• 5.4 As shown in Fig. 5.15, there is a simply supported beam with the total length 8 m. A concentrated load P = 20 kN and a distributed load q = 10 kN/m are acting on the beam. Please calculate the reaction forces of point A and point B.

  Fig. 5.15

  

  One beam under external forces

• 5.5 As shown in Fig. 5.16, two bars AB and BC are hinged together at point B, where a force F is applied at point D. The magnitude of the force F = 11 kN. The length AD = 2L, AB = BC = 4L. Please give the reactive forces at point A and C.

  Fig. 5.16

  

  Two bars hinged together

• 5.6 As shown in Fig. 5.17, a structure includes three bars. The length for each segment is marked in the figure. The weight for each bar is ignored, and the external forces are the uniformly distributed force q, concentrated force P, and couple M. Please give the reactive loads at the fixed end.

  Fig. 5.17

  

  A structure including three bars

• 5.7 As shown in Fig. 5.18, the bar AB and BE are linked at point B, and point A is the fixed end. The bar CD is linked with AB at point C, and linked with BE at point D. There is a uniformly distributed force acting on the DE segment, with the magnitude q = 3 kN/m. The lengths are AB = BE = 2AC = 2BD = 4 m. Please solve the reactive forces at point A and the internal force of the bar CD.

  Fig. 5.18

  

  A structure including three bars

Answers

• 5.1 $X_{B} = - \frac{1}{4}F,Y_{B} = - \frac{\sqrt 3 }{4}F,X_{C} = - \frac{3}{4}F,Y_{C} = \frac{\sqrt 3 }{4}F$

• 5.2 $X_{A} = - \frac{1}{4}P,Y_{A} = - \frac{3}{4}P,X_{C} = \frac{1}{4}P,Y_{C} = - \frac{1}{4}P$

• 5.3 $X_{A} = \frac{1}{4}ql,Y_{A} = \frac{3}{4}ql,X_{C} = - \frac{1}{4}ql,Y_{C} = \frac{1}{4}ql$

• 5.4 $X_{A} = 0,Y_{A} = 0\,\text{kN},Y_{B} = 35 \, \text{kN}$

• 5.5 $X_{A} = \frac{11\sqrt 3 }{6}\text{kN},Y_{A} = \frac{11}{2}\text{kN},X_{C} = - \frac{11\sqrt 3 }{6}\text{kN},Y_{\text{C}} = \frac{11}{2}\text{kN}$

• 5.6 $X = \frac{\sqrt 2 M}{4a} - P,Y = \frac{\sqrt 2 M}{4a} + 2qa,M_{end} = Pa + 2qa^{2}$

• 5.7 $X_{A} = - 6\,{\text{kN}},Y_{A} = 0,M_{A} = 6\,{\text{kN}}\,{\text{m}},F_{CD} = 9\sqrt 2 \, {\text{kN}}$