PLAIN AND CIVIL: 2018

Wednesday 19 December 2018

HOW TO START WITH THE WIND ANALYSIS OF A BUILDING USING NSCP

The main goal of analyzing the wind load is to convert it into a concentrated force which will be applied to the frames in order for us to start computation of moments and shears using any of the methods discussed earlier:

Portal method
Cantilever method
Q-factor method

However, if you check the NSCP, the procedure with lead us only until $P_{z}$ which in units $\frac{N}{m^{2}}$ . This pressure (distributed load) is still to be converted into a force in order to be applied in the frame, thus $P_{z}$ will then be multiplied with an area,

$Pressure=\frac{Force}{Area}$ . The area of which is the length of the wall being analyzed by the tributary height.

Tuesday 27 November 2018

EXAMPLE 6.1.10. TOTAL HEAD LOSS

Problem: Water flows from the basement to the second floor through the 2cm diameter copper pipe ( a drawn tubing) at a rate of Q= 0.80 liters per second and exits through a faucet of diameter 1.30 cm. Determine the pressure at point 1 if:

Viscous effects are neglected
The only losses included are major losses
All losses are included

With the given Q=0.80 lt/s

Pipe diameter = 2cm

Viscosity = $1.13x10^{^{-6}} \frac{m^{^{2}}}{sn}$

Determine pressure at station 1.

Solution:

$V=\frac{Q}{A}=\frac{0.0008}{\pi \left ( -0.02^{2} \right )}=2.546 m/s$

$Re=\frac{VD}{\nu}=\frac{2.546\left ( 0.02 \right )}{1.13x10_{-6}}=45062\left [ flow is turbulent \right ]$

a. Pressure at station 1 when there is no head loss

Energy equation between (1) and (9):

$\frac{V_{1}^{2}}{2g}+\frac{P_{1}}{\gamma }+z_{1}=\frac{V_{9}^{2}}{2g}+\frac{P_{9}}{\gamma }+z_{9}$

where: $V_{1}=V_{pipe}=2.546\frac{m}{s}$

$V_{2}=V_{faucet}=\frac{Q}{A_{9}}=\frac{0.0008}{\pi \left ( 0.013^{2} \right)}=6.027\frac{m}{s}$
z1 = 0
P9 = 0
z9 = 8.12 m

$P{_{1}}=\gamma \left [ \frac{V_{9}^{2}-V_{1}^{2}}{2g} +z_{9}\right ]$
$P{_{1}}=\9810 \left [ \frac{6.027^{2}-2.546^{2}}{2\left (9.81 \right )} +8.12\right ]=94578.5Pa$

So pressure head is:

$\frac{P_{1}}{\gamma }=9.64m$ compare with z = 8.12m

b. Pressure at station 1 when there is only major head loss

From the same basic BEE equation, add head loss to station 9. After rearranging, you will get:

$P_{1}=\gamma \left [ \frac{V_{9}^{2}-V_{1}^{2}}{2g}+z_{9} +f\frac{L}{D}\frac{V^{2}}{2g}\right ]$

Using Moody diagram, f can be projected using Re=45062 and $\varepsilon =0.00016$

f = 0.0215
L = total length of pipe from station 1 to station 9 = 4.6m+ 6(3m) = 22.60m

$P_{1}=9810\left [ \frac{6.027^{2}-2.546^{2}}{2\left ( 9.81 \right )} +8.12+0.0215\left ( \frac{22.60}{0.02} \frac{2.546^{2}}{2\left ( 9.81 \right )}\right )\right ]=173320Pa$

$\frac{P_{1}}{\gamma }=17.67m$

c. Pressure at station 1 including all head losses

With the same BEE equation, add major head loss (using Darcy Weisbach formula) and minor losses (in terms of velocity head), then you get: $P_{1}=\gamma \left [ \frac{V_{9}^{2}-V_{1}^{2}}{2g}+z_{9}+f\frac{L}{D} \frac{V^{2}}{2g}+\sum K_{i}\frac{V^{2}}{2g}\right ]$

$\sum K{i}=1.5+1.5+0.4+0.4+1.5+10+2 = 17.3$

$P_{1}=9810\left \{ \frac{6.027^{2}-2.546^{2}}{2\left ( 9.81 \right )}+8.12+0.0215\left [ \frac{22.6}{0.02}\left ( \frac{2.546^{2}}{2\left ( 9.81 \right )} \right ) \right ]+17.30\left [ \frac{2.546^{2}}{2\left ( 9.81 \right )} \right ] \right \}$ $P_{1}=229390.30Pa$

$\frac{P_{1}}{\gamma }=23.38m$

Saturday 24 November 2018

FINITE DIFFERENCES

Finite difference method is used for complex differential equations which are difficult to solve by any methods. Finite difference is an approximation method. There are different functions or applications for finite difference method.

Polynomial determination for curve fitting
Interpolation
Differentiation and Integration
Smoothing of data

The methods for each of these applications usually introduce a formula with a certain pattern which could be used after iteration. The iteration process is taking differences between data.

POLYNOMIAL FUNCTION

The number of iteration is taken as the degree of polynomial. If there are 3 iteration steps for constant difference then the formula is:

$f(x)=ax^{3}+bx^{2}+cx+d$

The leading coefficient being a.

INTERPOLATION FUNCTION

There are different methods for solving interpolation using finite difference with difference equation patterns.

Gregory-Newton Forward Interpolation Method
Gregory-Newton Backward Interpolation Method
Gregory-Newton Divided Interpolation Method
Lagrange Interpolation Method
Stirling Formula in Forward Difference
Gauss's Forward Interpolation Formula
Gauss's Backward Interpolation Formula

DIFFERENTIATION AND INTEGRATION FUNCTION

In using finite difference in solving for complex derivatives, three major methods are used:

Backward difference
Forward difference
Central difference

Click here for notes on Finite Difference

Friday 23 November 2018

PAVEMENT DESIGN

With the determination of geometric design and both horizontal and vertical alignments in highway design, the designer can start working on the pavement design.

The two major types of pavement are:

1. Flexible pavement - using asphalt as surface
2. Rigid pavement - using concrete as surface

Formulae for solving thicknesses of pavements on both types are long for manual computation so AASHTO deviced a graphical solution for thickness determination through nomographs.

Click here for pdf notes in Pavement Design

Friday 2 November 2018

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

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