1-14 Section 1 Example 4 A load of 7 kN may be placed anywhere within A and B in the trailer of negligible weight. Determine the reactions at the wheels at D, E, and F, and the force on the hitch H that is mounted on the car, for the extreme positions A and B of the load. The mass of the car is 1500 kg, and its weight is acting at C (see Figure 1.2.16). Solution. The scalar method is best here. Forces and Moments in Beams Beams are common structural members whose main function is to resist bending. The geometric changes and safety aspects of beams are analyzed by first assuming that they are rigid. The preceding sections enable one to determine (1) the external (supporting) reactions acting on a statically determinate beam, and (2) the internal forces and moments at any cross section in a beam. FIGURE 1.2.16Analysis of a car with trailer. Put the load at position A first For the trailer alone, with y as the vertical axis ∑M F = 7(1) – H y (3) = 0, H y = 2.33 kN On the car H y = 2.33 kN ↓Ans. ∑F y = 2.33 – 7 + F y = 0, F y = 4.67 kN ↑Ans. For the car alone ∑M E = –2.33(1.2) – D y (4) + 14.72(1.8) = 0 D y = 5.93 kN ↑Ans. ∑F y = 5.93 + E y – 14.72 – 2.33 = 0 E y = 11.12 kN ↑Ans. Put the load at position B next For the trailer alone ∑M F = 0.8(7) – H y (3) = 0, H y = –1.87 kN On the car H y = 1.87 kN ↓Ans. ∑F y = –1.87 – 7 + E y = 0 E y = 8.87 kN ↑Ans. For the car alone ∑M E = –(1.87)(1.2) – D y (4) + 14.72(1.8) = 0 D y = 7.19 kN ↑Ans. ∑F y = 7.19 + E y – 14.72 – (–1.87) = 0 E y = 5.66 kN ↑Ans. Fj O =100 N M O ∑ =0 MMrWrW OCOAAOBB ++× () +× () =0 Mijk ijk jij j O +−+ () ⋅+ −+ () ×− () +− () ×− () =100 20 50 06 03 04 60 02 40 0 Nm m N m N . Mijkkik O +⋅−⋅+⋅−⋅+⋅−⋅=100 20 50 36 24 40 0 Nm Nm Nm Nm Nm Nm Mijk O =− + + () ⋅124 20 26 Nm . j O +−+ () ⋅+ −+ () ×− () +− () ×− () =100 20 50 06 03 04 60 02 40 0 Nm m N m N . Mijkkik O +⋅−⋅+⋅−⋅+⋅−⋅=100 20 50 36 24 40 0 Nm Nm Nm Nm Nm Nm Mijk O =− + + () ⋅ 124 20 26 Nm . 7(1) – H y (3) = 0, H y = 2. 33 kN On the car H y = 2. 33 kN ↓Ans. ∑F y = 2. 33 – 7 + F y = 0, F y = 4.67 kN ↑Ans. For the car alone ∑M E = 2. 33(1 .2) – D y (4) + 14. 72( 1.8) = 0 D y = 5.93 kN. mounted on the car, for the extreme positions A and B of the load. The mass of the car is 1500 kg, and its weight is acting at C (see Figure 1 .2. 16). Solution. The scalar method is best here. Forces