Solving Mathematical Equations Involving Theory of Structures (Coursework Sample)

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Structural Design-Solution
Question1: Moment Distribution Method
Distribution factors at A and D are 1 and 0.
Stiffness coefficient for columns = EI/2.5
Stiffness coefficient for the beam = EI/8
Joint stiffness B and C = 2EI/5 + EI/8 = 21EI/40
Stiffness factors
Joint B
BA: (2EI/5)/ (21EI/40) = 0.762
BC: (EI/8)/ (21EI/40) = 0.238
But BA= CD and BC = CB
Calculation of the fixed end moments for the beam BC
Moment = wL2/12 = 5×82/12 = 26.667 KNM
We then stretch the beam to obtain the figure shown below
Bending moment diagram sketch
Question 2: Portal Method
For the given frame below
First, consider the upper storey
There is an inflection point at the mid-height of every column and the shear force at the interior column is as twice as the shear force at the exterior columns.
Free body diagram for the upper storey
From the above free body diagram
V + 2V + V = 4V = 7.5
V = 1.875 KN
At G
Clockwise Moments = 1.875×1.25 = 2.344KNM
F1 = 2.344/4 = 0.586KN that causes counter clockwise moments
Lower Storey
The total shear force = 7.5 + 15 = 22.5KN
Therefore, shear force for the exterior columns = 22.5/4 = 5.625KN
Shear force at the interior column = 11.25KN
At D
Moments = 1.875*1.25 + 5.625*1.25 = 9.375KNM
F4 = 9.375/4 = 2.34375KN
Therefore F5 = 2.34375 + 0.586 = 2.93KN
At joint E
Moments = 11.25*1.25 + 1.25× 3.75 – 4 × 2.344 = 9.374KNM
Therefore F6 = 9.374/4 = 2.344KN
At joint F
Moments about joint I = 5.625 × 1.25 + 1.875× 1.25 - 4×2.344 = 0
Hence, joint I is in equilibrium
The bending moment diagram for the frame:
Question 3: Combined stress
Stress