PLAIN AND CIVIL: HORIZONTAL DISTRIBUTION: Example 1: REGULAR STRUCTURES

PROBLEM: Continuing Problem 1-1, assuming the columns are oriented as shown, determine the total shear force carried by each frame in the Roof Deck. Rectangular columns [400x600mm] and circular columns [500mm diamter].

SOLUTION:
Roof deck shear to be distributed: 235.21 kN

Compute for the moments of inertia of the columns.
Circular column: $I_{x}=I_{y}=\frac{\pi D^{4}}{64}$
$I_{x}=I_{y}=\frac{\pi \left ( 0.50^{4} \right )}{64}=3.07x10^{-3}$
   Vertical column [400x600mm(LxW)]:
$I_{x}=\frac{bh^{3}}{12}=\frac{0.40\left ( 0.60^{3} \right )}{12}=7.20x10^{-3}$
$I_{y}=\frac{hb^{3}}{12}=\frac{0.60\left ( 0.40^{3} \right )}{12}=3.20x10^{-3}$
Horizontal column [600x400mm(LxW)]:
   $I_{x}=\frac{bh^{3}}{12}=\frac{0.60\left ( 0.40^{3} \right )}{12}=3.20x10^{-3}$
   $I_{y}=\frac{hb^{3}}{12}=\frac{0.40\left ( 0.60^{3} \right )}{12}=7.20x10^{-3}$

A. For Y-Frames [A-B-C-D-E]

1. Computing for ksy. $k_{sy}=\sum I_{x}$

For Frame A: $7.20+3.07+7.20=17.47x10^{-3}$
Frame B: $3\left ( 3.20 \right )=9.60x10^{-3}$
   Frame C:   $3.20+2\left ( 3.07 \right )=9.34x10^{-3}$
Frame D: $3\left ( 7.20 \right )=21.60x10^{-3}$
Frame E:   $7.20+3.07+7.20=17.47x10^{-3}$

2. For direct shear, take the summation of ksy to proceed to the percentage of ksy column.

   $\sum k_{sy} = 75.48x10^{-3}$
   $Percentagek_{sy}@FrameA=\frac{17.47x10^{-3}}{75.48x10^{-3}}=0.23$
$Percentagek_{sy}@FrameB=\frac{9.60x10^{-3}}{75.48x10^{-3}}=0.13$
$Percentagek_{sy}@FrameC=\frac{9.34x10^{-3}}{75.48x10^{-3}}=0.12$
   $Percentagek_{sy}@FrameD=\frac{21.60x10^{-3}}{75.48x10^{-3}}=0.29$
$Percentagek_{sy}@FrameE=\frac{17.47x10^{-3}}{75.48x10^{-3}}=0.23$

3. Direct shear, $V_{d}=Percentagek_{sy}\left ( V_{_{fRD}} \right )$

$V_{d}@Frame A=0.23\left ( 235.21 \right )=54.44kN$
$V_{d}@Frame B=0.13\left ( 235.21 \right )=29.92kN$
$V_{d}@Frame C=0.12\left ( 235.21 \right )=29.11kN$
$V_{d}@Frame D=0.29\left ( 235.21 \right )=67.31kN$
$V_{d}@Frame E=0.23\left ( 235.21 \right )=54.44kN$

Y-frames	ksy x 10-3	%ksy	Vd
A	17.47	0.23	54.44
B	9.6	0.13	29.92
C	9.34	0.12	29.11
D	21.6	0.29	67.31
E	17.47	0.23	54.44
	75.48	1	235.21

4. For torsional shear, a longer procedure is required starting with the location of the center of rigidity in the x-axis. Using the stiffness of columns, one can compute for the center of rigidity location in the x direction.

$K_{sy}C_{rx}=\sum k_{sy}\left ( c_{rx} \right )$

$C_{rx}=\frac{\left (17.47*0 \right )\left ( 9.60*4 \right )\left ( 9.34*9 \right )\left ( 21.6*15 \right )\left ( 17.47*21 \right )}{75.48}=10.78m$ from Fr. A

5. Eccentricity is computed as the difference between the center of rigidity located at 10.78m with the center of gravity (L/2).

   $eccentricity=\left |C_{rx} -\left ( \frac{L}{2} \right )\right |$
   $eccentricity,e=\left | 10.78-\left ( \frac{21}{2} \right ) \right |=0.28m$

6. As per code, the standard eccentricity adopted is equal to the eccentricity and 5% of the length in the direction analyzed.

$e_{x}=e+0.05L$
$e_{x}=0.28+0.05\left ( 21 \right )=1.33m$

7. With eccentricity known, the torsion (Moment) is determined.

$M=V\left ( e_{x} \right )$
$M=235.21\left ( 1.33 \right )=312.83kN-m$

8. From the table, C is the distance of the frame to Crx.
$C_{A}=10.78-0=10.78$
$C_{B}=10.78-4=6.78$
$C_{C}=10.78-9=1.78$
$C_{D}=15-10.78=4.22$
$C_{E}=21-10.78=10.22$
9. Consider using the leftmost frame as the reference to solve for   $\frac{c}{c_{1}}$ . Thus,
   $\frac{c}{c_{1}}@FrameA=\frac{10.78}{10.78}=1$
   $\frac{c}{c_{1}}@FrameB=\frac{6.78}{10.78}=0.63$
   $\frac{c}{c_{1}}@FrameC=\frac{1.18}{10.78}=0.165$
$\frac{c}{c_{1}}@FrameD=\frac{4.22}{10.78}=0.391$
$\frac{c}{c_{1}}@FrameE=\frac{10.22}{10.78}=0.948$
10. $k_{sy}'=k_{sy}\left ( \frac{c}{c_{1}} \right )$

   $k_{sy}'@FrameA=17.47x10^{-3}\left ( 1 \right )=17.47x10^{-3}$
   $k_{sy}'@FrameB=9.60x10^{-3}\left ( 0.629 \right )=6.038x10^{-3}$
   $k_{sy}'@FrameC=9.34x10^{-3}\left ( 0.165 \right )=1.542x10^{-3}$
$k_{sy}'@FrameD=21.60x10^{-3}\left ( 0.391 \right )=8.456x10^{-3}$
$k_{sy}'@FrameE=17.47x10^{-3}\left ( 0.948 \right )=16.562x10^{-3}$

11. $M'=k_{sy}'\left ( c \right )$

   $M'@FrameA=17.47x10^{-3}\left ( 10.78 \right )=0.188$
   $M'@FrameB=6.038x10^{-3}\left ( 6.78 \right )=0.041$
   $M'@FrameC=1.542x10^{-3}\left ( 1.78 \right )=0.003$
   $M'@FrameD=8.456x10^{-3}\left ( 4.22 \right )=0.036$
   $M'@FrameE=16.562x10^{-3}\left ( 10.22 \right )=0.169$

12. Take summation of M' and solve for %M.

$\sum M'=0.437$
$PercentageM@FrameA=\frac{0.188}{0.437}=0.431$
$PercentageM@FrameB=\frac{0.041}{0.437}=0.094$
$PercentageM@FrameC=\frac{0.003}{0.437}=0.006$
$PercentageM@FrameD=\frac{0.036}{0.437}=0.082$
$PercentageM@FrameE=\frac{0.169}{0.437}=0.387$

13. $AppliedM=PercentageM\left ( Moment \right )$

$AppliedM@FrameA=0.431\left ( 312.83 \right )=134.828kN-m$
$AppliedM@FrameB=0.094\left ( 312.83 \right )=29.308kN-m$
$AppliedM@FrameC=0.006\left ( 312.83 \right )=1.965kN-m$
$AppliedM@FrameD=0.082\left ( 312.83 \right )=25.546kN-m$
$AppliedM@FrameE=0.387\left ( 312.83 \right )=121.183kN-m$

14. $Vt_{y}=\frac{AppliedM}{c}$

$Vt_{y}@FrameA=\frac{134.828}{10.78}=12.51kN$
$Vt_{y}@FrameB=\frac{29.308}{6.78}=4.32kN$
$Vt_{y}@FrameC=\frac{1.965}{1.78}=1.10kN$
$Vt_{y}@FrameD=\frac{25.546}{4.22}=6.05kN$
$Vt_{y}@FrameE=\frac{121.183}{10.22}=11.86kN$

Y-frames	Vd	Vt	Vtot
A	54.44	12.51	66.95
B	29.92	4.32	34.24
C	29.11	1.10	30.21
D	67.31	disregard	67.31
E	54.44	disregard	54.44

B. For X-frames [1-2-3-4]

X-frames	ksx x 10-3	%ksx	Vdx
3	24.00	0.38	88.93
2	19.61	0.31	72.66
1	19.87	0.31	73.62
Σ	63.48	1.00	235.21

$C_{ry}=\frac{\left ( 19.87*0 \right )\left ( 19.61*5 \right )\left ( 24*9 \right )}{63.48}=4.95m$ [from frame 1]

$e=\left | 4.95-\frac{9.0}{2} \right |=0.45m$
$e_{y}=0.45+0.05\left ( 9.0 \right )=0.90m$

Thus:

$M=235.21kN\left ( 0.90m \right )=211.689kN-m$

X-frames	ksx x 10-3	c	c/c1	ksx'	M'	%M	Applied M	Vtx
3	24.00	4.05	0.82	19.636	0.080	0.447	94.635	23.37
2	19.61	0.05	0.01	0.198	0.000	0.000	0.012	0.24
1	19.87	4.95	1.00	19.870	0.098	0.553	117.042	23.64
Σ	63.48				0.178	1.000	211.689

X-frames	Vd	Vt	Vtot
3	88.93	23.37	112.29
2	72.66	disregard	72.66
1	73.62	disregard	73.62
	235.21

PLAIN AND CIVIL

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Friday 2 November 2018

HORIZONTAL DISTRIBUTION: Example 1: REGULAR STRUCTURES

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