2.XZ kN/m 2.55 kN/m 15.55m 1X.ZX m 10.45m 1V.WY m B
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Kindly refer to the given figure below. Calculate the support reactions at points A and B. kindly show your complete solutions.
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- For the ff. beams:a. Determine the location of the neutral axis from the bottom of the beam.b. Calculate IY.c. Calculate IX using the three approach. (The three approach) • combine the area using bh3/12 • minus the area using bh3/12 • using bh3/3let : L1=2.5m,L2=3.8m,L3=5.0m,P=250NQ5. Crank OA Rotates With Uniform Angular Velocity 0, = 4 Rad/s Along Counterclockwise. Take OA= = 0.5 M, AB=2r, And BC = V2r. For The Instant, P=45, OA Is Horizontal And AB Is Vertical. Determine The Angular Velocity
- The 450N plate is attached to collars which may slide on the vertical rod. If the coefficient of static friction is 04 between both collars and the rod determine the maximum supporting force available from friction when F = 0 and when F = 200N; a = 2.75m , b = 4.5m , c = 2m Show solution, the answer is 618.75A rectangular reinforced concrete beam with width of 300 mm & effective depth of 600 mm is subjected to a service bending moment of 200 kN.m. The beam is reinforced with 4-32 mm diameter bars. n = 8. 1. Compute the distance kd? a. 246.33 mm b. 353.67 mm c. 170.33 mm d. 302.33 mm 2. Which of the following gives the moment of inertia about N. A.? a. 4173.81 x 106 mm⁴ b. 3471.81 x 106 mm⁴ c. 4371.81 x 106 mm⁴ d. 4713.81 x 106 mm⁴ 3. Which of the following most nearly gives the maximum stress in steel in Mpa? a. 100 b. 110 c. 120 d. 130Q: The ball was thrown with a speed (v, = 35 m/sec) and directed at an angle (8,4 = 30°)with the horizontal. Determine the distance R where the ball hits the slope at B.
- please solve 16 with the information from 14....... Asap plzzzFrom the beam shown. Let P = 30 kN, Q = 2 kN/m, L =6m and EI = 200MNm^2Determine the following;C. Slope at L/3. D. Deflection at L/3.sorry，what are these： Step 3 Slope deflection equation In section AB MAB= MFAB+22EIL2θA+θB-3δLMAB= -12.5+4EI5θB MBA= 12.5+4EI52θB In section BC MBC= 0+2EI52θB+θC MCB= 2EI5θB+2θC In section CD MCD= -16.67+4EI52θC MDC= 16.67+4EI5θC arrow_forward Step 4 Equilibrium equations Equilibrium equations MBA+MBC= 0 -(i)MCB+MCD= 0 -(ii) Now, From i12.5+8EIθB5+4EIθB5+2EIθC5= 0EIθB125+EIθC25 = -12.5 -(iii) From ii2EIθB5+4EIθC5+-16.67+8EIθC5=02EIθB5+EIθC125= 16.67 iv Solving equations (iii) and (iv), we get EIθB= -6.548EIθC = 8.037
- For the ff. beams:a. Determine the location of the neutral axis from the bottom of the beam.b. Calculate IY.c. Calculate IX using the three approach. (The three approach) • combine the area using bh3/12 • deducting the area using bh3/12 • using bh3/3will rate immediately pls answer asap wc of the ff gives the deflection at point CDetermine the slope of the x²+y²-6x-4y-21=0 at (0,7)