In our last update we talked about merit and demerit of statically determinate and indeterminate structures. In this, a continuous beam with simply supported ends will be analysed using Clapeyron’s theorem of three moments.
A continuous beam ABCD, simply supported at A, B, C and D, is loaded as shown below
Find the salient moment and draw bending moment and shear force diagrams.
Considering Joint A, B and C
Considering Joint B’, C’ and D
Solving simultaneously using elimination method
22MB + 5MC = -172.8…………..(1)
5MB + 18MC = -115.2…………..(2)
Looking at the equations critically, you will observe that the equations are not balanced, therefore we will try to balance it.
I will balance MB at the both equations therefore, it will be eliminated. multiply the coefficient of MB (5) of equation (2) in equation (1) and vice versa.
22MB + 5MC = -172.8………..(1)*5. 5MB + 18MC = -115.2…………(2)*22. 110MB + 25MC =-864…………..(3). 110MB + 396MC =-2534.4……..(4). Since the arithmetic signs are the same in both equations, you subtract. 0. -371MC = 1670.4. Dividing both sides by the coefficient of Mc(-371) MC= -4.50KNM
Substituted MC in equation (1). 22MB + (-4.5*5) = -172.8. 22MB -22.5= -172.8. Collect like terms 22MB = -172.8+22.5. 22MB = -150.3. Divide both sides by coefficient of MB(22). MB =-6.83
Since we have gotten our support moments, the next thing to do is, calculation of Support reactions. Now decompose the beam into Members.let us consider beam member AB first
Again let us consider member B’C‘
Lastly, considering member C’D
Total reactions at support B =4.13+5.27=9.4KN. Total reactions at support C = 2.73 + 7.25 =9.98KN. Having gotten all our reactions at various supports we can draw our diagram freely now.
Computation of shear force
Shear force at A=4.86KN. Shear force at E=4.86-9=-4.14KN. Shear force at B=-4.14+9.4=5.26KN. Shear force at F=5.26-8=-2.74KN. Shear force at C=-2.74+9.98=7.24KN. Shear force at D=7.24-3*4=-4.76KN
Computation of salient moment
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