Using Moment Distribution Method, determine all reactions at supports also, determine shear and bending moments with complete labels. Use I = 1/12 bh^3 and E = 2.15x10^7 KPa. Use eight(8) iterations. "Wx(1m)" KN "w" KN/m "Wx1/2" KN/m D 4.0 m 3.0 m 7.0 m b = 0.25m h = 0.50m 4.0 m 4.0 m b= 0.35m h = 0.70m b = 0.45m h = 0.90m

Using Moment Distribution Method, determine all reactions at supports also, determine shear and bending moments with complete labels. Use I = 1/12 bh^3 and E = 2.15x10^7 KPa. Use eight(8) iterations. "Wx(1m)" KN "w" KN/m "Wx1/2" KN/m D 4.0 m 3.0 m 7.0 m b = 0.25m h = 0.50m 4.0 m 4.0 m b= 0.35m h = 0.70m b = 0.45m h = 0.90m

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

A car starts from rest and moves with a constant acceleration of 1.5m/s2 until it achieves a velocity of 25m/s. It then travelleds with constant velocity for 60 seconds. 11. What is the total distance traveled? O 1,708.33m O 208.33m O 1,500m O all of these
At time t = 0 a small ball is projected from point A with a velocity of 119 ft/sec at the angle 52°. Neglect atmospheric resistance and determine the two times t₁ and t2 when the velocity of the ball makes an angle of 27° with the horizontal x-axis. A u 119 ft/sec t₁ = 52° Answers: i t₂ = i 5.76 13.4 sec sec
None
The two cars A and B travel along the circular path at constant speeds V A = 84 ft/s and v B = 116 ft/s, respectively. If they are at the positions shown when t = 0, determine the time when they are 90° apart. Round your answer to 2 decimal place TA = 400 ft TB = 390 ft B B
oil 3 сm 9 cm 6.4 cm water www Temperature Density Dynamic Kinematic Surface Vapor Viscosity Viscosity Tension Pressure °C Kg/m3 Ns /m? m?/s N/m N/m? abs. 999.9 1.792x10-3 1.792x10-6 0.0762 610 1000 1.519x10-3 1.51x10-6 0.0754 872 10 999.7 1.308x10-3 1.308x10-6 0.0748 1130 15 999.1 1.14x10-3 1.141x10-5 0.0741 1600 20 998.2 1.005x10-3 1.007x10-6 0.0736 2340 30 995.7 8.01x10-4 8.04x10-7 0.0718 4240 40 992.2 6.56x10-4 6.61x10-7 0.0701 7380 50 988.1 5.49x10-4 5.56x10-7 0.0682 12,300 60 983.2 4.69x10-4 4.77x10-7 0.0668 19,900 70 977.8 4.09x10-4 4.15x10-7 0.0650 31,200 80 971.8 3.57x10-4 3.67x10-7 0.0630 47,300 90 965.3 3.17x10-4 3.28x10-7 0.0612 70,100 100 958.4 2.84x10-4 2.96x10-7 0.0594 101,300 Given: The temperature of the distilled water is taken to be 26° C. Use interpolation in getting the density of water. Solve: Compute the specific gravity of oil based on the experiment.
Asap please I need it now thanks
A train moving with constant acceleration travels 7.50m during the 10th second of its motion and 5.50m during the 12th second of its motion. Find the distance traveled by the train until it stops. 256.00 m B 144.50 m 240.67 m 108.00 m
where V is velocity in m/s, t time in s ,x distancein m. for x=1.5m and t= 6 s , 1= 2 m find total acceleration Q1 A) V = 2t 21 Q1 B) Find Q in 1/s and find V3 in ft/min V½=2.5 m/s D2= 300 mm Q=? V = 3 m/s D1 = 500 mm V = ? = 200 mm -D3
Answer the ff.
e y a b. a Calculate the y coordinate of the center of gravity of the shaded area given in the figure. a105 b 45 c: 25 d: 210 e 35
I am not sure that I am doing this right up to the point that I scanned...is it supposed to be W_B=-5pi rad/sec or...? Thank you for the help!
Two decimal places only