CALCULATOR TECHNIQUES

SEMINAR

BY:

ENGR. JOSE LORENZO D. BUCTON

CALCULATOR TECHNIQUES
-IS THE MAXIMUM UTILIZATION OF A
CALCULATOR TO SOLVE ENGINEERING
PROBLEMS WITH SPEED AND EXACTNESS

-ENGR. ROMEO TOLENTINO

 MODES IN USE
• STAT
• VECTOR
• COMPLEX
• EQUATIONS
• TABLE
• MATRIX
 APPLICATIONS
ALGEBRA LINEAR REGRESSION
TRIGONOMETRY REGRESSION ANALYSIS
ANALYTIC AND SOLID GEOMETRY MATRIX ALGEBRA
CALCULUS STRUCTURAL ANALYSIS
VECTOR ANALYSIS DIFFERENTIAL EQUATIONS
LATERAL EARTH PRESSURE CENTROIDS
HYDROSTATIC FORCE THREE MOMENT EQUATION
BEAM DEFLECTIONS MOMENTS OF INERTIA
SHEAR AND MOMENT DIAGRAMS
DESIGN OF STEEL CONNECTIONS
AND MANY MORE…
 ALGEBRA
AT WHAT TIME BETWEEN 4 AND 5 O’ CLOCK ARE
THE HANDS OF THE CLOCK

A.) OPPOSITE TO EACH OTHER

B.) COINCIDENT
C.) AT RIGHT ANGLES
OPPOSITE TO EACH OTHER
SOLUTION (CONVENTIONAL)

x
180  x  (  120)
12
x
180  x   120
12
3600 60 min s 600
x ( )
11 360 11
x  54.545 min s
time : 4 : 54 : 32.73
COINCIDENT
SOLUTION (CONVENTIONAL)

x
120   x
12
1440 60 min s 240
x ( )
11 360 11
x  21.818 min s
time : 4 : 21 : 49.09
RIGHT ANGLE (1 TIME)
ST
SOLUTION (CONVENTIONAL)

x
120   x  90
12
360 60 min s 60
x ( )
11 360 11
x  5.4545 min s
time : 4 : 05 : 27.27
RIGHT ANGLE (2 TIME)
ND
SOLUTION (CONVENTIONAL)

x
90  x  (120  )
12
2520 60 min s 420
x( )( )
11 360 11
x  38.18 min s
time : 4 : 38 : 10.91
 CALCULATOR TECHNIQUE
INITIAL CONDITION (4:00) FINAL CONDITION (5:00)
 MODE STAT
• A+BX A.)OPPOSITE TO EACH OTHER: C.)RIGHT ANGLE (1ST TIME):

ENTER: 180X’ ENTER: -90X’

DATA INPUT: TIME:4°54’32.73’’ TIME:4°5’27.27’’
X (HOUR) Y (MINUTES) 4:54:32.73 4:05:27.27

4 -120 B.)COINCIDENT: D.)RIGHT ANGLE (2ND TIME):

5 210 ENTER: 0X’ ENTER: 90X’

TIME:4°21’49.09’’ TIME:4°38’10.91’’
4:21:49.09 4:38:10.91
 ALGEBRA
FIND THE TERM IN THE ARITHMETIC
PROGRESSION 4,7,10:
A.)30TH TERM
B.)56TH TERM
 SOLUTION (CONVENTIONAL)
GIVEN: SOLVING FOR d:
a1  4 d  an  a1
a2  7 d  74
a3  10 d 3

an  a1  ( n  1)( d )
a30  4  (30  1)(3)

a56  4  (56  1)(3)

a56  169
 CALCULATOR TECHNIQUE
• MODE STAT:(A+BX)
A.)FOR 30TH TERM:
X Y ENTER: 30Y’
=91
1 4
2 7
3 10 B.)FOR 56TH TERM:
ENTER: 56Y’
=169
 TRIGONOMETRY:

a2  b2  c2
cos A 
COSINE LAW:  2bc
CALTECH VARIATION: a  b  A c;a  c  Ab
 TRIGONOMETRY:

SINE LAW: SINE LAW TECHNIQUE:

MODE 5:1:
a b c
  a b c
sin A sin B sin C Cos(B) Cos(A) c
Sin(B) -Sin(A) 0
 TRIGONOMETRY
FIND THE SIDES OF a AND b:
SOLUTION (CONVENTIONAL)
C  180  A  B
C  180  92  53
C  35

5.1466 a 5.1466 b
 ; 
sin(35) sin(53) sin(35) sin(92)
a  7.166
b  8.967
CALCULATOR TECHNIQUE

SINE LAW TECHNIQUE:

MODE 5:1:

a b c
Cos(92) Cos(53) 5.1466
Sin(92) -Sin(53) 0

a=7.166
b=8.976
 TRIGONOMETRY
FIND SIDE c:
 SOLUTION (CONVENTIONAL)

c 2  a 2  b 2  2ab cos(C )
c  7.26 2  9.03932  2(7.26)(9.0393) cos(35)
c  5.1867
 CALCULATOR TECHNIQUE

c  7.2635  9.0393  5.1867

or
c  9.039335  7.26  5.1867
PRISMATOID TECHNIQUE

h2

v   ( A  Bx  Cx 2 )dx
h1

NOTE: COEFFICIENTS OF A, B AND C

ARE DERIVED FROM STAT MODE:_+Cx^2
 FIND THE VOLUME OF THE SHADED SECTION
INDICATED IN THE FIGURE:

USING
STAT MODE:_+Cx^2
X Y

0 0

1.8214 10.42222

3.6428 0

h2

v  ( A  Bx  Cx A=0; STORE TO A
2
) dx
h1 B=11.44419;STORE TO B
1.8214 C=-3.14159;STORE TO C
v    ) dx  3.83714
2
( A Bx Cx
1.4480
AREA OF CIRCULAR SEGMENT

A   2 Dx  x 2 dx,0, h,0.001

NOTE: PROGRAM CALCULATOR TO

MODE LINE 10
FIND THE AREA OF THE CIRCULAR SEGMENT
INDICATED IN THE FIGURE:

A   2 Dx  x 2 dx,0, h,0.001

A   2 3.6428 x  x 2 dx,0,0.7323,0.001
A  1.49484
 LATERAL EARTH PRESSURE
A RETAINING WALL 8M HIGH SUPPORTS A COHESIONLESS SOIL AS SHOWN IN THE PROFILE
WITH A SHEAR RESISTANCE OF 33. THE SURFACE OF THE SOIL IS HORIZONTAL AND LEVEL
WITH THE TOP OF THE WALL. NEGLECT WALL FRICTION AND USE RANKINE’S FORMULA
FOR ACTIVE EARTH PRESSURE ON A COHESIONLESS SOIL.
A.)DETERMINE THE TOTAL EARTH THRUST ON THE WALL IN Kn/m IF SOIL IS DRY.
B.) DETERMINE THE TOTAL EARTH THRUST ON THE WALL IN Kn/m IF SOIL IS WATERLOGGED
3.5m BELOW SURFACE.
C.)DETERMINE LOCATION OF LATERAL THRUST DURING WATERLOGGED CONDITION.
D.)DETERMINE MOMENT GENERATED FROM LATERAL THRUST.
SOLUTION (CONVENTIONAL)

1  sin  1  sin 33
Ka    0.2948
1  sin  1  sin 33
1 1
Fa  K aH 2  (0.2948)(15.696)(8) 2
2 2
Fa  148.1kN
SOLUTION (CONVENTIONAL)
PRESSURES :
P1  K aH
P1  (0.2948)(15.696)(3.5)
P1  16.195Kpa
P2  K aH
P2  (0.2948)(9.857)(4.5)
P2  13.076 Kpa
P3   H 2O H
P3  (9.81)(4.5)
P3  44.145Kpa
 SOLUTION (CONVENTIONAL)

Y  COORDINATE : FORCES :
3.5 1
Y1  4.5   5.667 m F1  (16.195)(3.5)(1)  28.34 KN
3 2
1 F2  16.195( 4.5)(1)  72.88KN
Y2  ( 4.5)  2.25m 1
2 F3  (13.076)(4.5)(1)  29.421KN
4.5 2
Y3   1.5m 1
3 F4  (9.81)(4.5) 2  99.33KN
4.5 2
Y4   1.5m
3
 SOLUTION (CONVENTIONAL)
VARIGNON ' S  THEOREM :
FT  F1  F2  F3  ...
FT  28.43  72.88  29.421  99.33
FT  229.97 KN
FT Y T  F1Y 1 F2Y 2 F3Y 3...
229.97Y T  (28.43)(5.667)  (72.88)(2.25)  (29.421)(1.5)  (99.33)(1.5)
YT  2.251m
M  517.70928KN .m
CALCULATOR TECHNIQUE
PRESSURES :
P1  K aH
P1  (0.2948)(15.696)(3.5)
P1  16.195Kpa
P2  K aH
P2  (0.2948)(9.857)(4.5)
P2  13.076 Kpa
P3   H 2O H
P3  (9.81)(4.5)
P3  44.145Kpa
 CALCULATOR TECHNIQUE
X(Y-COORDINATE) FREQUENCY (AREA) DATA :
5.667 28.34 N  229.971KN
2.25 72.88 ___

1.5 29.421 X  2.25119 m

1.5 99.33  X  517.70928KN .m
 LIQUID LIMIT
THE FOLLOWING RESULTS WERE OBTAINED
FROM A LIQUID LIMIT TEST ON A CLAY USING
THE CASAGRANDE CUP DEVICE. FIND THE
LIQUID LIMIT:
NUMBER OF BLOWS 6 12 20 28 32
WATER CONTENT % 52.5 47.1 42.3 38.6 37.5
SOLUTION (CONVENTIONAL)
 CALCULATOR TECHNIQUE
STAT MODE: A+BX:
X-(NUMBER OF BLOWS) Y-(WATER CONTENT %)

20 42.3
28 38.6

INPUT:
25Y’
=39.9875%
=40%
 3D TRUSS
IF THE CAPACITY OF EACH LEG IS 15KN, WHAT IS
THE SAFE VALUE OF W?

EQUATIONS OF EQUILIBRIUM:

 FX  0  MX  0
 FY  0  MY  0
 FZ  0  MZ  0
CALCULATOR TECHNIQUE:
COORDINATES:
POINT X Y Z
A 0 0 2.4
B 0.9 1.8 0
C -1.8 0 0
D 0.9 -1.8 0

MEMBER VECTORS:
MEMBER X Y Z
AD 0.9 -1.8 -2.4
AB 0.9 1.8 -2.4
AC -1.8 0 -2.4
 CALCULATOR TECHNIQUE:
 0.9 0.9  1.8 
matA  
  1.8 1.8 0  
MEMBER LENGTH 
  2.4  2.4  2.4
AD 3.132 3.132 0 0
matB  
 0 3.132 0 
AB 3.132

 0 0 3
AC 3 0 
matC   0
 


1 

 0.2873 0.2873  0.6 

matAmatB 1 
  0.574 0.5747 0  

  0.766  0.766  0.8
  0.435
ENTER : matAns 1matC   
  0.435

  0.416

CENTROIDS AND MOMENTS OF
INERTIA

FROM THE GIVEN FIGURE,

DETERMINE THE FOLLOWING:

A.)CENTROID FROM BOTTOM

B.)TOTAL AREA
C.)MOMENT OF INERTIA
PARALLEL TO X-AXIS
X (CENTROID) AREA
50+210+20=280 80x40=3200
50+0.5(210)=155 210x30=6300
0.5(50)=25 50x200=10000
CALCULATOR TECHNIQUE

n  19500  AREA
__
X  108.846  CENTROID
2
n( x )  177461538.5  STORE ( A)

80(40) 3 30(210) 3 200(50) 3

I NA  A  
12 12 12
I NA  203124038.5
 CALCULUS
FIND THE DERIVATIVE OF THE GIVEN FUNCTION:
ex
y x SOLUTION :
x
y  xe
CHOICES :
e x 1 ln( y )  e x ln( x )
A.) x (e  x ln( x ))
x

y' 1
B.)e x x e x 2
(e  ln( x ))
x  e x (  ln( x ))
y x
x
C.) x 2 e (1  e x ln( x )) x 1  x ln( x )
x
1
y'  e x x e ( )
D.) x e e x (1  ln( x x )) x
dy x
 x e 1e x (1  ln( x x ))
dx
 DIFFERENTIAL EQUATIONS
• THE POPULATION OF THE TOWN GROWS AT THE RATE
PROPORTIONAL TO THE POPULATION PRESENT AT ANY TIME
T. THE INITIAL POPULATION OF 500 INCREASED BY 15% IN 10
YEARS. WHAT WILL BE THE POPULATION IN 30 YEARS?
BOUNDARY CONDITIONS: SOLUTION :
A.) p  500; t  0
P=500;T=0 dP
 kp 500  ce 0 k
P=575;T=10 dt
dP c  500
 p  k  dt B.) p  575; t  10
ln( p )  kt  c 575  ce10 k
e ln p  e kt  c k  0.01397
p  ce kt C.) p  500e 0.01397t
@ t  30
p  760.4375
 STRUCTURAL ANALYSIS

KEY EQUATION: KEY EQUATION:

__
x2
x2
ya (3L  a )da
2
6A a ya( L2  a 2 )
  6 EI L
 
x1
L
da
x1
 CANTILEVER BEAM
SOLVE FOR THE DEFLECTION AT C: FOR A+Bx: STATMODE(A+Bx)

X Y

0 40

10 0

x2
ya 2 (3L  a )da
  6 EI 40000
x1

10
( A  Bx) x 2 (30  x)dx 3EI
 
0
6 EI
 CONTINUOUS BEAM
FOR SPAN AB:
FOR THE CONTINUOUS BEAM SHOWN IN THE FIGURE,
DETERMINE THE MOMENT AT B: X Y

0 600

2 600

FOR SPAN CB:

X Y

0 1200

3 0
 THREE MOMENT EQUATION:
MOMENT AREA AB: MOMENT AREA CB:
3
2
( A  Bx)( x)(4  x )dx
2 2 ( A  Bx)( x)(32  x 2 )dx
0 4
 4200 0 3
 3780

900(3)(4  3 )
2 2 3780 store
 D
 4725
4
4200  4725  8925 store
 C

2(4  3)( M B )  (C  D )

M B  907.5 KN .m
 RESULTANT OF FORCES
FIND THE RESULTANT OF THE SYSTEM OF FORCES DESCRIBED IN THE FIGURE
AND THE ANGLE IT MAKES WITH THE HORIZONTAL:

SOLUTION :
R  4000  200150  300240  300300
R  226.7949192  419.6152423i
R  476.9831097  61.6095305
 WELDED CONNECTIONS
FIND THE MAXIMUM MASTER EQUATION:
SHEARING STRESS IN THE
WELDED CONNECTION: T P
R( x, y )   i ( x  yi)( ) 
J L

 2
J  n  x  y
2
 
L3
12
WHERE:
T-TORSIONAL FORCE
J-POLAR MOMENT OF INERTIA
P-AXIAL LOAD
Ө-ANGLE WITH X AXIS
L-LENGTH OF WELD
 WELDED CONNECTIONS
T P
R ( x, y )   i ( x  yi )( ) 
J L


J  n  x  y  
2 2

L3
12
1503
J  4750000  (3)( )  5593750
12
T  90( 233.33)  21000kn.mm
L  n  450mm
P  90kn
  270

21000 90270
R ( x, y )   i ( x  yi )()
5593750 450
R max  755.0237 kn / mm
SOURCES AND ACKNOWLEDGEMENTS:

Mega Review
Review Innovations
Gillesania Review
XU College of Engineering
PICE-XUSC
Tolentino and Associates Review