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Truss Problem

1a. Truss Problem. Isolate joint C: Draw the force of AC Draw the force of BC Draw the force of F 1. Figure 2. C. F AC. 30 °. F BC. F 1 = 100 lbs. Figure 1. Figure 2. A. 4 ft. C. 30 °. B. 1b. Truss Problem.

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Truss Problem

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  1. 1a Truss Problem • Isolate joint C: • Draw the force of AC • Draw the force of BC • Draw the force of F1 Figure 2 C FAC 30° FBC F1 = 100 lbs.

  2. Figure 1 Figure 2 A 4 ft C 30° B 1b Truss Problem B. Calculate the length of truss member BC. (answer precision = 0.000) G: AC = 4 ft, θ= 30° U: BC E: Cos θ= adjacent/hypotenuse S: 0.866 = BC/4ft S: BC = 3.464 ft***(length only here…not force!)

  3. C FAC 30° FBC F1 = 100 lbs. 1c Truss Problem C. Using joint C, determine the magnitude and type of force (tension or compression) that is being carried by truss member BC. (answer precision = 0.0) G: F1= 100 lbs. U: FBC E: ∑FCY = 0 = F1 + FBCY S: ∑FCY = 0 = -100 lbs. + -(FBCsin30°) ∑FCY = 0 = -100 lbs. + -(FBC • 0.5) S: FBC = -200 lbs.

  4. Take-off angle = 35° Take-off speed = 36.99 ft/sec 36.99 ft/sec 35° VIX 2a Trajectory Motion Vertical Velocity VIY = VI sinθ Might be good to know, hint, hint a. G: θ= 35°, Take-off speed 36.99 ft/sec U: VIX (horizontal velocity) What equation is used to calculate horizontal velocity? E: VIX = VI cosine θ S: VIX = 36.99 ft/sec cosine 35° VIX = 36.99 ft/sec • 0.82 S: VIX = 30.33ft/sec

  5. 36.99 ft/sec 35° VIX = 30.33ft/sec 2b Trajectory Motion b. G: θ= 35°, Take-off speed 36.99 ft/sec, VIX = 30.33ft/sec U: X E: X = VI2sin(2*θ)/g S: X = (36.99 ft/sec)2 sin(2*35°) / 32.15 ft/sec2 X = 1368.26ft2/sec2sin 70º / 32.15 ft/sec2 X = 1368.26 ft * 0.94 / 32.15 X = 1286.16 ft / 32.15 S: X = 40.00 ft

  6. 3a Calculate the force that the sample experienced at the proportional limit. (answer precision = 0.0) G: AO = 0.2 in2, s=40,000 psi U: F P.I. (Force at the proportional limit) E: F p. l. = A O x sp. l. (Area x Stress) S: F p. l. =0.2 in2 x 40,000 psi S: F p. l. = 8,000 lbs.

  7. 3b 3b. Starting at the origin and ending at the proportional limit, calculate the modulus of elasticity for this material. To solve this problem we are going to use the equation E = s/∈ Using the GUESS Method here is what we have. G: ∈ = 0.005 in/in, s =40,000 psi U: E (modulus of elasticity) E: E = s/∈ (Stress/Strain) S: E = 40,000/ 0.005 S: E = 8,000,000 psi

  8. To measure current with a multimeter, the “leads” must be placed in series. To measure voltage with a multimeter, the “leads” must be placed in Parallel (across the load).

  9. 5a Figure 1 Figure 2 The images in Figures 1 and 2 show the voltmeter configurations that two different POE students used to take a voltage reading within a simple circuit. Only one of the two students was able to measure the voltage value in the simple circuit. Use the information given in the figures to answer the following questions. 5a. Which of the two setups (Figure 1 or Figure2) shows the correct way to measure voltage? ANSWER : Figure 2 (see page before)

  10. Pulley Problem Figure 5 represents a belt driven system. Pulley B, which has a diameter of 16 inches, is being driven by Pulley A, which has a diameter of 4 inches. If Pulley A is spinning at 60 RPMs, then Pulley B is spinning at ______ RPMs. G: DIA of B= 16”, DIA of A=4, A RPM=60 U: B RPM E: DIA-B/DIA- A=RPM-B/RPM-A S: 16/4=60/B RPM S: 15=B RPM

  11. Gear Problem The gear train in Figure 3 consists of a 40-tooth (input), 20-tooth, and 30-tooth (output) gear. If the input gear rotates 10 times, how many times will the output gear rotate? G:40 g, 20 g, 30 g, Nin 10 U: Nout E: Nin*R=Nout*R S: 40*10=Nout*30 S: 13.3= Nout 40/20=2/1 *10= 20 20/30=2/3 *5=20/3

  12. Mechanisms The wheels on a bicycle have a 10” radius. If the bike must travel exactly 2000”, how many revolutions are required? Assume that no sliding or slipping occurs between the wheel and the road. G: Radius 10”, Dist.=2000” U: Revolutions E: Distance/ Circumference = Revolutions (Circumference = π*Dia) S: 2000/ (3.14*20) = Rev. S: 31.8 Revolutions

  13. Properties of Materials Figure 10 shows a 100 lb. normal force being applied to a 12” long x 10” diameter cylinder. What is the resulting compressive stress in the cylinder? G: F=100 lb., Dia.=10” U: Compressive Stress E: 1) A = π * r2 2) S = F/A S: 1) A = 3.14 * 52 2) S = 100 lbs/78.5 in2 S: S = 1.27 psi

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