2. Compute for the moment of inertia about xo and yo of the following figures. a) b) 100 mm 100 mm 75 mm 25 mm 120 mm 75 mm 50 mm 180 mm 1= 11 - 12 |= 11 + 12 Ans. Ixo = 27,963,997 mm", lyo = 12,346,602 mm
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- WDetermine the polar of moment of inertia of the quarter of the circle having a radius of óm shown in the figure relative to the polar axis through one of its corners. * Y x² + y² = r² X 636.173 m* O 381.704 m* O None of the above O 763.407 m* O 508.938 mCalculate the moment of inertia of the shaded area about the x-axis. ... 95 mm 28 mm 49 mm t49 mm - A Ix= 6.12(10^6) mm^4 B) Ix= 6.76(10^6) mm^4 c) Ix= 6.34(10^6) mm^4 D) Ix= 6.29(10^6) mm^4 O O O ODetermine the centroid x-coordinate for the area shown below with respect to the given axis. Given t= 131 mm; a = 393 mm; b = 524 mm and h = 786 mm. tt O A. 559.27 mm O B. 216.65 mm O.C. 166.27 mm O D. 229.25 mm OE. 428.27 mm
- 人工知能を使用せず、 すべてを段階的にデジタル形式で解決してください。 ありがとう SOLVE STEP BY STEP IN DIGITAL FORMAT DON'T USE CHATGPT Determine the position of the center of gravity of the section of the following body Ø10 Ø30 Ø40 40 80Determine the centroid x-coordinate for the area shown below with respect to the given axis. Given a = 39 mm; b = 234 mm and d = 273 mm. b a a 2a За 2a d a a O A. 175.50 mm B. 148.20 mm C. 191.10 mm D. 157.75 mm E. 165.61 mmSHOW ALL YOUR WORK P 150 P 100 -300- Find MI around axis PP 150
- Table 4: Measurement of Horizontal Distance on Level Ground DISTANCE TRIAL MEAN DISCREPANY PRECISION FORWARD BACKWARD A-1 10 10 2-3 10 10 3-4 10 10 4-5 10 10 74.335m 0.005m 1/1486 5-6 10 10 6-7 10 10 7-B 4.34 4.33 2. The discrepancy for the measurement is the difference between the first measurement and the second measurement.- 100 80 9 A3 200 1. What is the value of ybar from the 1-1 line 52.69 2. What is the value of centroidal MOI Ixx? 31543447 mm^4 3. What is the value of centroidal MOI lyy? 16794122 mm^4 >I cannot understand these For point of contraflexural, 40090300=200×x90300 x=2 m θA=0 (fixed support) θDA=θD-θA=θD=Area of MEI Dia between A and D = 19030012×600×3+600×4+12×300×3-12×200×2 =355090300 θD=0.03931 radianθD= 2.252° Deflection at point D ∆D=31-CosθD ∆D=2.3176×10-3 mm Deflection at point D is 2.3176×10-3 mm
- From the figure shown below: |-100 mm--100 mm-–150 mm - 150 mm 150 mm 75 mm 1. Determine the distance of the centroid from the x axis. 450 x10^6 2. Determine the moment of inertia of the shaded area about the x axis. a a. 1.72 x109 mm4 b. 3.23 х109 mm4 c. 2.46 x109 mm4 d. 4.11 x109 mm4 e. none of the above 3. Determine the moment of inertia of the shaded area about the y axis. c a. 1.09 x109 mm4 b. 3.71 х109 mm4 2.03 x109 mm4 d. 4.83 x109 mm4 С. e. none of the above▶ Calculate the moment of inertia of the shaded area about the axis. its centroid. y D 90 mm -50 mm 0:01/2:45 ▪0 30 mm 50 mm 1 I Ix-t₁ Ixx 4+1 +0 I = I₁ - IFA = [1₁+ Adi] - [Ix +Adi] continue 1x E cc (3Part C Compute the unadjusted latitudes. Express your answers in meters to six significant figures and separated by commas. ΕΠΙΑΣΦΗ AB, BC, CD, DE, EA= Submit Part D Submit Compute the linear misclosure. Express your answer in meters to five significant figures. [V]] ΑΣΦΠ vec Part E Request Answer 1: Request Answer Compute the relative precision. Express your answer to five significant figures. ΑΣΦΗ Π vec vec ? ? m ? m, mn, m, mn, m