VERTICAL DISTRIBUTION OF BASE SHEAR: Example 2

PROBLEM: From Problem 1-2, distribute the base shear to the structure's respective floors.

SOLUTION:

Note that T = 0.5291 seconds which is less than 0.70s, thus Ft=0.

By using a table format, it would be easier to solve the forces applied to each floor.

• Tabulate values for weight at each floor and the relative height of the floors from the base of ground. Disregard foundation and basement.
• Multiply the weight by the height. Compute for the summation of (weight x height).
• Solve for the percentage distribution, Cvx, by taking (w x h) divided by the summation of (w x h).

                                 
                        
                                  
                                  
                                  

• Solve for the force per floor using the base shear previously computed as 2,732.28kN.

                         
                                   
                                   
                                   
                                   

In summary:

FLOORS	WEIGHT	HEIGHT	WxH	Cvx	FORCE
RD	1000	14.0	14000	0.27	738.45
4Flr	1800	10.5	18900	0.36	996.91
3Flr	1800	7.0	12600	0.24	664.61
2Flr	1800	3.5	6300	0.12	332.30
GFlr	1500	0.0	0	0.00	0.00
Σ			51800	1	2732.28

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