3. In the figure shown, lines NQ and PR are established perpendicular to line MNP, and points Q and R are lined up with the distant point M. If NQ = 318.55 m, PR = 475.62 m, and NP = 210.38 m, determine the length of MN which represents the width of the river. 90 90

6. Determine the magnitude of the resultant R, if P2 = 978, P3 = 780.

Please use Double Integration method to determine slope (Ɵ) and deflection (Δ) at point B and D based on the following structure! (E = 2.105 kg/cm2, I = 208.33 cm4)

* 10. Calculate the vertical displacement at C and E with the graphic multiplication method. Know E=2.0x105MPa, 11=6560cm4,12=12430cm4 30 kN 130 kN C 2 m to 2 m 2m

A line is regarded as 485 m long. It is measured with a 500 grams tape which is 30.05 m long at 20°C under a 4.5 kg pull supported throughout its length. During measurement the temperature is 4.5°C and the tape is suspended under a 6.5 Kg pull. The line is measured at 3% grade. What is the true horizontal distance? ◦E=2x106 kg/cm2 ◦v =126.54 cm3/kg Given:   Nominal Tape Length L1 Total weight of tape W Standard Temperature of Tape T1 Temperature during measurement T2 Standard Pull of Tape P1 Pull applied during measurement P2 Measured Distance Di correction due incorrect tape length Ci Ground slope S Cross-sectional area of the tape A   Solution correction per tape length (correction per 30-m length) incorrect tape length CI Question Blank 11 of 20 meter temperature correction CT Question Blank 12 of 20 meter pull correction CP Question Blank 13 of 20 meter sag correction CS Question Blank 14 of 20 meter slope correction (approximate method) CSL…

In the figure, compute the vertical and horizontal components of displacement of joint E produced by the loads. The area of bars AB, BD, and CD = 5 in^2 , the area of all other bars = 3 in^2. E = 30,000 kips/in^2 .   Use Virtual Work Method to find the - Vertical Displacement at Joint E - Horizontal Displacement at Joint E