1.60 m -2.21 m- -2.21 m- B D F 0.60 m 0.60 m E 1.20 m 1.2 kN 1.2 kN H L 0.60 m 0.60 m K 2.97 m 1.20 m 1.2 kN 1.2 kN N 0.60 mi M P 0.60 m R 1.2 kN 1.2 kN S T
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- Solve the internal force JD (in kN to 1 d.p). J K 1m OKN 45.000° 45.000° 45.000° 45.000° 45.000° 45.000° C D 6KN 12.0 1m If your answer is a compressive force state a negative answer. 6KN1- Select the optimum choice for: high speed air craft skins material. Table 1: The properties of metallic materials E, GPa Max. use lemp, 'C | Density, Mg/cm' ay, MPa | Relative cost per unit sheet area Material 2.8 300 2 Aluminum alloy 200 69 200 7,8 350 350 700 Carbon steel Stainless steel 193 7.8 650 6 8.9 600 6 Nickel alloy Titanium alloy 207 1000 116 600 4.5 900 10 Table 2: Weighting factor Specific yield strength: Max, use temperature: Youngs modulus: 4 2F1: 0318, F2: 1,894, M: 8,942
- Hi can someone tell me how he gets 1.75FB=1500 from 3/4FB+FB-1500=0%.00 @ 4G JO 100 I. Hw 1 Name: How many minutes you took ID. to solve this problem? Section: HW-1 Determine the magnitude and coordinate direction angles of the resultant force acting on the bracket. F = 450 N 45 60 F, = 600 NCan you please do P7.41 Thx
- WA 3 46.| Asiacel B/s docs.google.com/forms D 1m 35 N/m 2m 30 N 1m B C 85 N 100 N.m 60 2m 2m 0.5m 1.5 m bläi 3 the concentrated load replaced by distributed load is equal to إجابتكShoulder Load Ratings, KN Fillet Diameter, mm Deep Groove Angular Contact Bore, OD, Width, Radius, mm mm mm mm ds dμ C10 Co C10 Со 10 30 9 0.6 12.5 27 5.07 2.24 4.94 2.12 12 32 10 0.6 14.5 28 6.89 3.10 7.02 3.05 15 35 11 0.6 17.5 31 7.80 3.55 8.06 3.65 17 40 12 0.6 19.5 34 9.56 4.50 9.95 4.75 20 47 14 1.0 25 41 12.7 6.20 13.3 6.55 25 52 15 1.0 30 47 14.0 6.95 14.8 7.65 30 62 16 1.0 35 55 19.5 10.0 20.3 11.0 35 72 17 1.0 41 65 25.5 13.7 27.0 15.0 40 80 18 1.0 46 72 30.7 16.6 31.9 18.6 45 85 19 1.0 52 77 33.2 18.6 35.8 21.2 50 90 20 1.0 56 82 35.1 19.6 37.7 22.8 55 100 21 1.5 63 90 43.6 25.0 46.2 28.5 60 110 22 1.5 70 99 47.5 28.0 55.9 35.5 65 120 23 1.5 74 109 55.9 34.0 63.7 41.5 70 125 24 1.5 79 114 61.8 37.5 68.9 45.5 75 130 25 1.5 86 119 66.3 40.5 71.5 49.0 80 140 26 2.0 93 127 70.2 45.0 80.6 55.0 85 150 28 2.0 99 136 83.2 53.0 90.4 63.0 90 160 30 2.0 104 146 95.6 62.0 106 73.5 95 170 32 2.0 110 156 108 69.5 121 85.0Q.1 A rotary shaft carries distributed load of 250 w/m with a long 3m. The shaft transmits 35 KW at 290 rpm. A shaft is simply supported shaft. The mux. shear stresses 56 MPa and the allowable Compression stress 45 MPa, the safety factor is 1.5. Find the size of the shaft if it subjected to suddenly applied load.
- Identify the equations that go with these concepts: Conservation of Energy (for conservative forces) b. Work-Energy Principle C. Force of gravity for an object near the surface of the Earth Force of gravity for an object far from the surface of the Earth d. T1+ V1 = T2+ V2 T1 + U1->2= T2 F = mg F= (GMm)/r^2In-Class Practice Examples - For P = 2184 N, determine FcD, FCB, and FDB. 0.8 m N B -2 m. A D C 4.8 m 2.1 m 2.1 m Xfrom the equation F/A= explain and discuss for 1- IF x=0. 2- IF 29 P₁₂ - P₂- 3- iF x Pp g TTT (P1-P₂) air 4- iF Z Pp g> P₁ - Pz. Cole* AR√ CdA A2* Pa P1-PL-P * P₁ - Pz