Use Slope-Deflection Method or Moment Distribution Method to solve for the unknowns on the 120mm x 200 mm beam shown in the figure. E = 80 GPa. For MDM, use at least 4 iterations (up to bal 4). a. Fixed-end moment (FEM) at joint A on member AB. Indicate negative if counter-clockwise. b. Mument at A in kN-m. Indicate negative if counter-clockwise. c. Moment at Cin kN-m. Indicate negative if counter-clockwise. c kN d kN/m "b" m 3m

Use Slope-Deflection Method or Moment Distribution Method to solve for the unknowns on the 120mm x 200 mm beam shown in the figure. E = 80 GPa. For MDM, use at least 4 iterations (up to bal 4). a. Fixed-end moment (FEM) at joint A on member AB. Indicate negative if counter-clockwise. b. Mument at A in kN-m. Indicate negative if counter-clockwise. c. Moment at Cin kN-m. Indicate negative if counter-clockwise. c kN d kN/m "b" m 3m

Structural Analysis

6th Edition

ISBN:9781337630931

Author:KASSIMALI, Aslam.

Publisher:KASSIMALI, Aslam.

Chapter2: Loads On Structures

Section: Chapter Questions

Problem 1P

See similar textbooks

Similar questions

ANSWER THE QUESTION AND DISCUS THE RESULTS. in an oil – water system, the following fluid and sand data are available: h= 75 ft, hp=22.5 ft, po=48.3 lb/ft³, pw=65.76 lb/ft³ , Ho=0.63cp, Bo=1.136, bbl/STB, rw=3 in, re=1000 ft, ko=93.5 md, Distance from the top of the perforations to top of the sand = 30 ft, CALCULATE: THE CRITICAL OIL FLOW RATE USING CHANEY METHOD.
2. A CUBICAL FLOAT 1.2m ON A SIDE, WEIGHS 182kg AND IS ANCHORDED BY ME NS OF A CONCRETE IS TAUT, 22.5cm OF FLOAT ARE SUBMERGED. WHAT RISE IN THE WATER LEVEL WILL LIFT THE CONCRETE OFF THE BOTTOM ? CONCRETE WEIGHS = 2400 kg/m³. 1.2 H 86824 _WS
Q/shows a hycreaulic structur thut was built on opm channel . The residunl head at point f calculated by using Lon's theory is equal to 1.25m. Thi suk o + and G2.5. bydraulic gradient is equal to : 50 Lon's VIs WIL h, = 125 - +4, G.2.5 12465 15m 12m 15m
Juo Situation 1: alqma Isoubilya s no 300 hom63 28W 120 illdsom bsod terco A Sito be A layer of saturated clay 4 m thick is overlain by sand 5 m deep, the water table being 3 m below the surface. The saturated unit weights of the clay and sand are 19 and 20 kN/m², respectively. Above the water table, the dry unit weight of the sand is 17 kN/m³. Determine the following: 169 11 1. Effective stress. 167 kPa b. 71.38 kPa 2. Water pressure. a. 167 kPa b. 71.38 kPa ASYMENT nim ni PTSS.O d. nimal+vsso d almal 2017.O .b 3. Total stress or pressure. 167 kPa 108.14 kPa 58.86 kPa im ni ni vioolov.sgisdbal 108.14 kPa 58.86 kPa nimal XP240 biov odt li visolev ogeqso? TA24.0 niral 201.I d C. 108.14 kPa dribsnodeb. sin 71.38 kPani benimsieb 26 dalam 58.86 kPa vivibubno osbyd SAT bozu sish edT 191emseneg baad gailiste to seu geblso
With the soil profile as shown below, if the water table gradually drops to elevation (E) = -6 m, then the settlement is GWT G.L. El =0 Ya- 19.43 kN/m Yary- 17.53 kN/m Sand El = -10 m my- 0.02 cm/kg cy-0.20 cm/min Y 19.62 kN/m Clay El = -16 m Sand El=-20 m Rock (a) 5.80 cm (c) 2.40 cm (b) 4.20 cm (d) 6.10 cm
1. Determine the following: If w₁=5 kN/m and w₂ = 7 kN/m, Est= 210 GPa, b = 50 mm and h= 150 mm. (BY AREA-MOMEMNT METHOD) O 1.0A = 2. to 3. SB = Wi A 8 m B - 4 m C in in W2
Soil k (mm/s) 0.10 300 тm B 0.03 0.0049 a.Compute the equivalent k in cm/s. A C b.Compute the rate of flow in cm³ /hr. 15 cm15 cm15 cm» Given: Cross-sectional area of the soils A, B and C is 100 mm x 100 mm
Last Two Digit of Roll = 15
8.1
please show written complete and clear solution
The length of the levee perpendicular to the paper is 600 m; the sand on top and bottom have k1=750 ft/day and k2= 40 in/s respectively. The thickness of sands is equal, m = 670 m, n=575 ft, B = 8ft, C = 525 ft. Determine the amount of water flowing to the ditch in in3/min. (See picture below)
Determine H ?