the continuous beamshown in Fig., if the support B issettled by 5mm and support Csettled by 2mm, by using themoment distribution method,find the following.Take (E=2*105N/mm²) and (I=16*106 mm)1. Fixed end moment of AB (MFAB) is:a)-13.33 kM.mc) -7.33 kM.m2. Fixed end moment of CB (MFCB) is:a) KCD=c)KCD:=a) 13.6 kM.mb) 6.4 km.mc) 3.6 kM.md) 10 kM.m3. The modified stiffness of member CD (KcD) is equal to4EIEI2110 kN/mb) KCDEl-4.0m-d) KCDb)-19.33 kM.md) -6.33 kM.m==B3EIī-2m-3EI2120 kN↓EI+4. For the 2nd cycle, the carry-over moment of CD is equal toa)-0.16 kM.mb) +0.16 kM.mc) +0.65 kM.md) +0.06 kM.m5. The final moment of member BC after 3rd cycle is:a) +8.22 kM.mb)-8.22 kM.mc) +8.62 kM.md) -8.62 kM.m-2m-C10 kNTFFFTimpΕΙ-3m-D

Answered: Image /qna-images/answer/a4add18b-4d09-44a6-bc97-f3e2386aabdc.jpg

pport B is tled by 5mm and support C tled by 2mm, by using the ment distribution method, d the following. ke (E=2*105N/mm²) and (I= 106 mm) 1. Fixed end moment of AB (MFAB) is: a)-13.33 kM.m c) -7.33 kM.m 2. Fixed end moment of CB (MFCB) is: a) KCD = 4EI L = ΕΙ 21 10 kN/m b) Ксо -4.0m- a) 13.6 kM.m b) 6.4 km.m c) 3.6 kM.m d) 10 kM.m . The modified stiffness of member CD (KcD) is equal to 3EI El = b)-19.33 kM.m d) -6.33 kM.m F d) KCD = B 3EI 21 -2m- c)KCD 4. For the 2nd cycle, the carry-over moment of CD is equal to a)-0.16 kM.m b) +0.16 kM.m c) +0.65 kM.m d) +0.06 kM.m 5. The final moment of member BC after 3rd cycle is: a) +8.22 kM.m b)-8.22 kM.m c) +8.62 kM.m d)-8.62 kM.m 20 kN El -2m- C 10 kN + 痧 fimp ΕΙ -3m-

pport B is tled by 5mm and support C tled by 2mm, by using the ment distribution method, d the following. ke (E=2*105N/mm²) and (I= 106 mm) 1. Fixed end moment of AB (MFAB) is: a)-13.33 kM.m c) -7.33 kM.m 2. Fixed end moment of CB (MFCB) is: a) KCD = 4EI L = ΕΙ 21 10 kN/m b) Ксо -4.0m- a) 13.6 kM.m b) 6.4 km.m c) 3.6 kM.m d) 10 kM.m . The modified stiffness of member CD (KcD) is equal to 3EI El = b)-19.33 kM.m d) -6.33 kM.m F d) KCD = B 3EI 21 -2m- c)KCD 4. For the 2nd cycle, the carry-over moment of CD is equal to a)-0.16 kM.m b) +0.16 kM.m c) +0.65 kM.m d) +0.06 kM.m 5. The final moment of member BC after 3rd cycle is: a) +8.22 kM.m b)-8.22 kM.m c) +8.62 kM.m d)-8.62 kM.m 20 kN El -2m- C 10 kN + 痧 fimp ΕΙ -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

For the forces acting on the beam shown in Figure 3.1f, determine the location of the equivalent resultant force with reference to point A in m. Given / = 2.4, P1 = 8, P2 = 8 and w = 2 %3D (Note: Prefix the answer with negative sign "." if the location is to the left of the reference point. E.g. if the location of the equivalent resultant is located between points A and B, or to the left of point B by 1.2m, the answer should be "-1.2" with reference to point B) P, kN W kN/m P, kN C 21 m 21 m Figure 3.1f
Q1/ Determine the resulting displacement of pin B. Assume: A1=A2= 900 mm2, E1=E2= 100 GPa, a1=a2= 12x10/°C, AT= 35 °C, h= 1200 mm, e= 60° (1) (2)
solve for MA,RA, and RB
Find f(4) using forward, backward and central difference of: (3 – 2x)² f(x) : - x2 Step size =02
Determine by integration the centroid (?̅) of theshape in the figure (to the right).
Solve the problem by the moment-area method. The beam has constant flexural rigidity EI. A simple beam AB supports two concentrated loads P at the positions shown in the figure. B C 4. 4 A support C at the midpoint of the beam is positioned at distance d below the beam before the loads are applied. Assuming that d = 12 mm, L = 5.4 m, E = 200 GPa, and I = 193 x 10° mm, calculate the magnitude of the loads P (in kN) so that the beam just touches the support at C. 163.87 x kN
Problem 1: A nonuniform beam is simply supported at both ends. The flexural rigidity of the left part equals EI, whereas the flexural rigidity of the second part equals (s+1)EI where's'is your serial number. (a) Use K functions to derive the characteristic equation; (b) Solve for the characteristic equation to derive natural frequencies. Report the results. (s+4)EI EI L/2 SERIAL NUMBER= 2
Read the question carefully and give me the right solution with clear calculations.
Determine the moment of inertia with respect to y axes through the centroid of the shown area in in. Glven: a = 2 in, b-20 in. d=12 in and R=8 in b Select one: O A. ly = 902.40 OB. ly = 1542.4 %3D OC. ly = 1382.40
SOLVE THE FOLLOWING PROBLEM AND SHOW YOUR COMPLETE SOLUTIONS. EXPLAIN YOUR ANSWER FOR BETTER UNDERSTANDING.
The rigid bar BC is supported by the steel rod AC of cross-sectional area 0.36 in^2. Find the elongation (in inches) of rod AC caused by the 3530-lb load if x = 9ft and theta = 51.8 degrees. Use E = 29167 ksi for steel. Note: Round off the final answer to four decimal places.