POSSI ad P TER 2 sent sent the Steel, Pa). a of Principles of Mechanics P 0 = 75 (Q1) 1000 lb P60 P45 2P 1.5P P 30 0 = 45 Z t 2P 1 (Q5) 12 ft. w= 2 kips/ft W www (Q2) P 2000 lb 15A+ (Q3) 4P P (Q4) 2P ft 15 ft (Q6) (Q7) (Q8) 5 kips w= 2 kips W 4 ft 4 ft 12 ft L (Q12) 5 105 (Q9) L/3 + 2L/3 (Q10) P (Q11) L L L FL2LI2 LI2LI2. vely y of sed. (Q13a) (Q13b) + L/2 + L12+ + L12+ L12- (Q13c) FIGURE 2.59 Questions. (Q13d) EXAMPLE Find the elongation caused by a tensile force of 7500 lb (33,364 N) in a steel member 144 in. (3658 mm) long having a cross-sectional area of 0.785 in. (506.5 mm). Note that E=29.6 x 106 lb/in. (204.000 MPa). = 45 Answer: R = 600 lb and RB RA = 400 lb. 2.6. Determine the reactions for the structure shown in Figure 2.59(Q6). = Answer: RA, P/2 and RB, = 5P/2. 2.7. Determine the reactions for the structure shown in Figure 2.59(Q7).