Activity 1. For the H- section shown, Find the value of h so that I, = Iy. 20mn 100mm h = ? 20mm 140mm
Q: 180 kN 15 kN/m A D В -5 m 5 m -4 m El = constant E = 70 GPa I = 2,340 (106) mm4
A: Find deflection at D
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Q: В A C 120 kN E= 200 GPa A = 1200 mm? D 10m 10m 5m
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Q: 30 kN/m A B 5 m –2 m El = constant E = 200 GPa I = 146(106) mm4
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Q: 35kN/m 180 kN 15 kN/m B -5 m- -5 m El = constant E = 70 GPa I = 2,340 (10) mm
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Q: B 5 m -5 m- El = constant E = 70 GPa I = 1,030 (106) mm4 Determine the horizontal deflection of C in…
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Area Moment of Inertia (Transfer Formula)
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- A line was determined to be 2,651.56 meters when measured with a 30 m. steel tape supported throughout its length under a pull of 4 kg. at a mean temperature of 39 °C. Tape used is of standard length of 15 °C under a pull of 5 kg. Cross-sectional area of the tape is 3.95 mm^2. Determine the TOTAL ERROR of the LINE due to TEMPERATURE in meters. Round off to three decimal places. Input the magnitude only.Given the cross sectional area below. The horizontal neutral axis is 129.0mm from reference corner 1. The moment of inertia around this neutral axis is 1073352100.6mm4.Calculate the following: A.Adjusted value of angle A B.Adjusted value of angle C C.Adjusted value of angle F D.Adjusted value of angle H
- A 100 m tape weighing 3 kg was standardized and found to be 0.02 m short at temperature of 10degC and a 30 N pull when supported throughout. The 1324.55 – meter distance measured is an inclined distance whose difference in elevation between the points is 2 m. What is the correct distance? a. 1 324.28 m b. 1 234.18m c. 1 345.83 m d. 1 432.11 mDetermine the centroidal moment of inertia.**correction in picture L200x200x20 L 200 x 200 x 20theoretical mass: 59.7 kg/marea: 7600 mm2 I: 28.8 x 106 mm4 S (I/C) :202 x 103 mm3 r: 61.6 mmx or y: 57.4 mmAxis z r: 39.3 mmC380 x 74 theoretical mass: 74.4 kg/marea: 9480 mm2 depth: 381 mm flange width: 94 mmflange thickness: 16.5 mm web thickness: 18.2 mm axis x I: 168 x 106 mm4 axis x S (I/C) :881 x103 mm3 axis x r: 133 mm axis y I: 4.60 x 106 mm4 axis y S (I/C) :62.4 x103 mm3 axis y r: 22.0 mmx: 20.3 mm W310 x 500 theoretical mass: 500.4 kg/marea: 63700 mm2 depth: 427 mm flange width: 340 mmflange thickness: 75.1 mm web thickness: 45.1 mm axis x I: 1690 x 106 mm4 axis x S (I/C) :7910 x 103 mm3 axis x r: 163 mm axis y I: 494 x 106 mm4 axis y S (I/C) :2910 x103 mm3 axis y r: 88.1 mmPlease calculate for D and please explain how to input the angle 2°54'13'' in the calculator. D = 1/tan(2°54'13''/2)
- Calculate the value of the integration constant C4. Enter your answer in kNm2 to three decimal places.The distances shown in Fig. 3-11(the attached image) are measured. All measurements are uncorrelated and have the same precision. The measured values are l1= 100.010 m, l2= 200.050 m, l3=200.070 m, and l4= 300.090 m. Use the principle of least squares to find the adjusted distance between A and C.Problem no. 1: The difference in elevation between observed points A and B is 47 m and the horizontal distance between them is 73. The difference in elevation between observed points C and D is 27 m and the horizontal distance between them is 38. With transit at Sta. O, additional information was obtained as shown in the table. Assume K=100 and f=0.25 and c=0.25. Determine the stadia interval at A Determine the stadia interval at B Determine the stadia interval at C Determine the stadia interval at D
- Diameter 48mm Length0.95m1. What is the distance between the stadia hairs on the instrument's scope? (answer in mm)2. What is the instrument's stadia constant? (answer in m)3. What is the horizontal distance between point A and B? (answer in m)Load (N) Deflection (mm) 0 0 4.903034 0.000346 11.416539 0.000802 18.337648 0.001251 25.050375 0.001727 29.707962 0.002173 31.75 0.002634 29.888114 0.003094