Solution

Problem Answer
 Conventional
Use of formula

Technology

Problem Concept Answer

May 2015 Board Exam
A 160 m long sag parabolic curve connects a -8% grade to a +2% grade. The grades intersect at Station 1 + 240 with
elevation of 14.87 m. Determine the stationing of the lowest point and its elevation, find the elevation of the first
quarter point of the curve.
L=160m
P.C.

P.T.
Q1
-8% LP +2%

1+240
14.87m
 𝑔1 𝐿
 
𝑥=
𝑔1 − 𝑔2 𝑦1 𝑦2
𝑥 
2
= 2
1 𝑥2
𝐿
𝐻
 = ( 𝑔 − 𝑔1 )
8 2

Problem Answer
A 160 m long sag parabolic curve

Concept
connects a -8% grade to a +2%
grade. The grades intersect at
Station 1 + 240 with elevation of
14.87 m. Determine the stationing
of the lowest point and its
elevation, find the elevation of the
first quarter point of the curve,
determine the stationing of the
lowest point and its elevation.
Geometric
Properties of
Parabolic
Curve
 May 2015 Board Exam
A 160 m long sag parabolic curve connects a -8% grade to a +2% grade. The grades intersect at Station 1 + 240 with
elevation of 14.87 m. Determine the stationing of the lowest point and its elevation, find the elevation of the first
quarter point of the curve.
L=160m
P.C.
80m 80m

P.T.
128m 1+288
  𝑔1 𝐿
𝑥=
𝑔1− 𝑔2 -8% LP +2%

  (− 0.08) ( 160 𝑚 )
𝑥= =128 m
−0.08 −0.02
1+240
𝑆𝑇𝐴
  𝑜𝑓 𝐿𝑃=1+240 − 80+128=1+288
14.87m
 May 2015 Board Exam
A 160 m long sag parabolic curve connects a -8% grade to a +2% grade. The grades intersect at Station 1 + 240 with elevation
of 14.87 m. Determine the stationing of the lowest point and its elevation, find the elevation of the first quarter point of the
curve.
L=160m
P.C. 80m 80m
32m
P.T.
128m
H ¿  𝟐 𝒎 1+288
  = 𝐿 ( 𝑔 2 − 𝑔1 ) -8%
𝐻
8 LP +2% 0.02(32m)=
0.4m

  = 160 ( 0.02−(−0.08) ) y
𝐻
8 H ¿  𝟐 𝒎
1+240 0.96m
0.02(48m)
𝐻=2𝑚
  14.87m
48m
 May 2015 Board Exam
A 160 m long sag parabolic curve connects a -8% grade to a +2% grade. The grades intersect at Station 1 + 240 with elevation
of 14.87 m. Determine the stationing of the lowest point and its elevation, find the elevation of the first quarter point of the
curve.
L=160m
P.C. 80m 80m
32m
P.T.
128m
H ¿  𝟐 𝒎 1+288
-8% LP +2% 0.02(32m)=
0.4m
Solving for “y” using squared
y
property of parabola H ¿  𝟐 𝒎
 𝑦 12 = 𝑦 22 1+240 0.96m
𝑥1 𝑥2 14.87m
48m
  𝑦2 = 22
32 80
𝐸𝑙𝑒𝑣
  𝑜𝑓 𝐿𝑃=14.87+0.96+0.32 ¿  16.15 𝑚
 𝑦=0.32 𝑚
 May 2015 Board Exam
A 160 m long sag parabolic curve connects a -8% grade to a +2% grade. The grades intersect at Station 1 + 240 with elevation
of 14.87 m. Determine the stationing of the lowest point and its elevation, find the elevation of the first quarter point of the
curve.
L=160m
P.C.
80m 80m
40m P.T.

-8% +2%

1+240
14.87m
 L=160m

80m 80m
P.C. P.T.
40m 40m
H ¿  𝟐 𝒎
y
+2%
-8%
0.08(40m)=3.2m H ¿  𝟐 𝒎 Solving for “y” using squared
property of parabola
14.87m
1+240  𝑦 12 = 𝑦 22
𝑥1 𝑥2

  𝑦2= 22
𝐸𝑙𝑒𝑣
  . 𝑜𝑓 𝑓𝑖𝑟𝑠𝑡 𝑞𝑢𝑎𝑟𝑡𝑒𝑟 𝑝𝑜𝑖𝑛𝑡 =14.87+3.2+0.5=18.57 𝑚 40 80

 𝑦=0.5 𝑚
 Tangent Grades
forms a Linear
function with
respect to ‘x”

Problem Answer
A 160 m long sag parabolic curve

Concept
connects a -8% grade to a +2%
grade. The grades intersect at
Station 1 + 240 with elevation of
14.87 m. Determine the stationing
of the lowest point and its
elevation, find the elevation of the
first quarter point of the curve.

Geometric
Properties of
Parabolic Curve
CONCEPT Geometric Property of Parabolic Curve
Tangent Grades forms a Linear function with respect to ‘x”

  𝑏
′
∆  𝑦 ∆ 𝑦 =∫ 𝑓 ( 𝑥 ) 𝑑𝑥  ¿ 𝑦 𝑓 − 𝑦 𝑖
𝑎

  =+
May 2015 Board Exam
A 160 m long sag parabolic curve connects a -8% grade to a +2% grade. The grades intersect at Station 1 + 240 with
elevation of 14.87 m. Determine the stationing of the lowest point and its elevation, find the elevation of the first
quarter point of the curve.
L=160m
P.C.
21.27m

P.T.
Q1
-8% LP +2%
0.08(80m)=6.4m

1+240
80m
14.87m
The grade at PC is -5% and its, elevation is 20m, the elevation of lowest point is 18m. PC
and PT is to be connected by symmetrical parabolic curve, if the grade at PT is +8%,
determine the length of curve from PC to PT

+0.08

-2m

-0.05
208m
L
May 2015 Board Exam
A spiral easement curve has a length of 100 m with 300-m radius central curve. Calculate the spiral angle on the
third-quarter point of the spiral, calculate the offset distance from the tangent to the third-quarter point of the
spiral and determie the corresponding distance along the tangent on that point, determine also the degree of
curve at third quarter point.

x
SC
 𝜃
y

75 𝑚 100
𝑚 𝑅=300 𝑚
𝐿 = =  
𝐿𝑠
 
TS
 2 3
 𝜃= 𝐿  𝑥= 𝐿
2 𝑅 𝑐 𝐿𝑠 6 𝑅 𝑐 𝐿𝑠
 𝑦=𝐿 − 𝐿5
40 𝑅 𝑐2 𝐿𝑠2

Problem Answer
A spiral easement curve has a
length of 100 m with 300-m
radius central curve. Calculate the
Concept
spiral angle on the third-quarter
point of the spiral, calculate the
offset distance from the tangent
to the third-quarter point of the
spiral and determie the
corresponding distance along the
tangent on that point, determine
also the degree of curve at third
Rate of change of
quarter point.
curvature is constant
May 2015 Board Exam
A spiral easement curve has a length of 100 m with 300-m radius central curve. Calculate the spiral angle on the
third-quarter point of the spiral, calculate the offset distance from the tangent to the third-quarter point of the
spiral and determine the corresponding distance along the tangent on that point, determine also the degree of
5
curve at third quarter point.   𝐿 2
  𝐿 3   𝐿
𝜃= 𝑥= 𝑦=𝐿 −
2 𝑅 𝑐 𝐿𝑠 6 𝑅 𝑐 𝐿𝑠 40 𝑅 𝑐2 𝐿𝑠2

x
SC   752   753   75 5
 𝜃 𝜃= 𝑥= 𝑦=75 −
y 2 ( 300 ) ( 100 ) 6 ( 300 ) (100 ) 40 (300¿¿2)(100 ¿¿2)¿¿
= 75 𝑚 00
𝑚 𝑅=300 𝑚
𝐿  =1  
𝐿𝑠
 
𝜃=0.09375
  𝑟𝑎𝑑 𝑥=2.34
  𝑚  𝑦=74.93 𝑚
TS
Curvature forms a
Linear function
with respect to
‘x” distance from
TS

Problem Answer
A spiral easement curve has a
length of 100 m with 300-m
radius central curve. Calculate the
Concept
spiral angle on the third-quarter
point of the spiral, calculate the
offset distance from the tangent
to the third-quarter point of the
spiral and determine the
corresponding distance along the
tangent on that point, determine
also the degree of curve at third
Rate of change of
quarter point.
curvature is constant
  1
(𝑅 )  1
𝑦  (300 )
y x
𝜃
  (𝐿)
 
L
𝐿=75 𝑚 𝐿𝑠=100 𝑚
TS   SC  
𝐿𝑠=100 𝑚 (  𝜃 ) ¿𝐴𝑟𝑒 𝑎𝑜𝑓 𝐶𝑢𝑟𝑣𝑎𝑡𝑢𝑟𝑒𝑑𝑖𝑎𝑔𝑟𝑎𝑚
 
1
  𝐿
𝑦 300  𝑦=   1 ( 𝐿) 𝐿
Integration
 1
=
𝐿 100 30000 𝜃=
2 ( 30000 )
Curvature
(𝑅 )  
𝜃=
𝐿2
60000

Slope (𝜃)
 

Deflection (𝑥)
 
NOV 2009
Figure shows a three-dimensional frame work subjected to a horizontal force of 10 kN applied at point D. A, B
and C are hinged supports.
 𝐴𝐷=⟨ 10 20 −5 ⟩
𝑦   
𝐵𝐷 =⟨ 0 20 5 ⟩
 
𝐶𝐷= ⟨ −1 0 20 −5 ⟩
(0,20,0)
   
2
|  𝐴𝐷|= √ 10 2+20 2+ ( − 5 ) =22.913
2
𝑥  𝐹  2 | 𝐵𝐷|= √ 02+ 202+ ( 5 ) =20.616
𝑧  2
| 𝐶𝐷|=√ ( −10 ) +202 + ( −5 ) =22.913
2

𝐹  1 (0,0
  , −5) 10 20 −5
𝜆  𝐴𝐷 =
⟨ 22.913 22.913 22.913 ⟩
= ⟨ 0 .436 0.873 − 0.218 ⟩

𝐹  3 0 20 5
(−10,0,5)
 
𝜆  𝐵𝐷=
⟨ 20.616 22.616 22.616
=⟨0
⟩ 0.884 0.221 ⟩

−10 20 −5
𝜆 𝐶𝐷 =
⟨ 22.913 22.913 22.913 ⟩
= ⟨ −0.436 0.873 −0.218 ⟩

0.436
  𝐹 1+0 𝐹 2 − 0.436 𝐹3 =−10
0.873
  𝐹1 +0.884 𝐹 2+ 0.873 𝐹 3 =0
−0.218
  𝐹 1+0.221 𝐹 2 − 0.218 𝐹 3 =0
𝐹  1 =−11.47 𝑘𝑁
(10,0,5)
  𝐹  2 =0 𝑘𝑁
𝐹
  3 =11.47 𝑘𝑁
 𝑦 
(0,20,0)
 
 𝐴𝐷=⟨ 10 20 −5 ⟩
𝑥  𝐹  2
𝑧  𝐵𝐷
  =⟨ 0 20 5⟩
𝐹  1 (0,0
  , −5)
𝐶𝐷=
  ⟨ −1 0 20 −5 ⟩
𝐹  3
(−10,0,5)
  2 2 2
 𝐴𝐷=−0.5 √ 10 +20 +5 =− 11.47 kN

2 2 2
  =( 0 ) √ 0 +20 +5 =0 𝑘𝑁
𝐵𝐷

𝐶𝐷=
  ( 0.5 ) √ 102 +202 +52=11.47 𝑘𝑁

(10,0,5)
 
 𝑦  P Determine the value of force P,
(0,20,0)
 
so that the Axial Force on AD
will be zero.
𝑥  𝐹  2
𝑧 
𝐹  1 (0,0
  , −5)
𝐹  3  
𝑷=𝟒𝟎 𝒌𝑵
(−10,0,5)
 

(10,0,5)
 
 𝑦  P =1  𝐴𝐷=⟨ 10 20 −5 ⟩

(0,20,0)
  𝐵𝐷
  =⟨ 0 20 5⟩

𝐶𝐷=
  ⟨ −1 0 20 −5 ⟩
𝑥  𝐹  2
𝑧 
𝐹  1 (0,0
  , −5)
1  0 0 − 10 0 
𝐹  3 20 20 20 1
−5 5 −5 0
(−10,0,5)
 

  1 × 𝐴=0.286 𝐹
80   1 =0.286 𝑃

  1 × 𝐵=0.515 𝐹
40   2 =0.515 𝑃

  1 ×𝐶=0.286
80 𝐹
  3 =0.286 𝑃
(10,0,5)
 
−11.46
  +0.286 𝑃=0
𝑷=𝟒𝟎
  𝒌𝑵
 𝑦 
(0,20,0)
 
 𝐴𝐷=⟨ 10 20 −5 ⟩
𝑥  𝐹  2
𝑧  𝐵𝐷
  =⟨ 0 20 5⟩
𝐹  1 (0,0
  , −5)
𝐶𝐷=
  ⟨ −1 0 20 −5 ⟩
𝐹  3
(−10,0,5)
 

(10,0,5)
 
A trapezoidal gate with lower base width of 1m, upper base width of 0.5 and 1.5m high is immersed at
A
antriangular
angle of 45gate with base
degrees with width of 1m The
the vertical. and vertex
1.5m high is gate
of the immersed at an upward,
is pointing angle of and
45 degrees
the basewith the
is 2.4m
vertical. The vertex
below water surface.of the gate is pointing upward, and the base is 2.4m below water surface.
Evaluate the total force acting on one side of the gate.
Obtain the location of the force from the water
surface.

𝑭=∫ 𝒑𝒅𝑨
2.4m
CONCEPT

 𝐹=𝛾 h́ 𝐴
0.5
m
45 °
    𝐼𝑔
𝑒=
𝐴 ´𝑦
1m

m
1.5
A trapezoidal gate with lower base width of 1m, upper base width of 0.5 and 1.5m high is immersed at
an angle of 45 degrees with the vertical. The vertex of the gate is pointing upward, and the base is 2.4m
below water surface.
Evaluate the total force acting on one side of the gate.
Obtain the location of the force from the water
surface.

2.4m

0.5
m

45 °
 
1m

m
1.5
  𝐴= 1 ( 1+0.5 ) ( 1.5 )=1.125 𝑚 2
Determine First the 2
centroid and Moment
of Inertia. 𝐴 ´𝑦=∑ 𝐴𝑦
 

1
[
( 1.125 ) ´𝑦 =( 0.5 )( 1.5 ) ( 0.75 ) +2 ( 0.25 ) ( 1.5 ) ( 0.5 )
2 ]
0.5m   𝟐
´𝒚 = 𝒎 =𝟎 . 𝟔𝟔𝟕 𝐦
𝟑
  ( 0.5 ) ( 1.53 ) ( 0.5 ) ( 1.5 3 ) 1
𝐼 𝑔=
12
+
36
2
( )
+ ( 0.5 ) ( 1.5 ) ( ( 0.75 −0.667 ) ) +
2
2
( 0.5 ) (1.5 ) ( ( 0.667 − 0.5 ) )
1.5 m

𝑰 𝒈=𝟎 .𝟐𝟎𝟑𝟏 𝒎 𝟒
0.75m  
0.5m 0.5m 0.667m

0.25m 0.5m 0.25m

A trapezoidal gate with lower base width of 1m, upper base width of 0.5 and 1.5m high is immersed at
an angle of 45 degrees with the vertical. The vertex of the gate is pointing upward, and the base is 2.4m
below water surface.
Evaluate the total force acting on one side of the gate.
Obtain the location of the force from the water h́=2.4+0.833
  sin 45=2.989 𝑚
surface.   𝐹=𝛾 h́ 𝐴

𝐹=
  ( 9.81 ) ( 2.989 ) ( 1.125 ) =32.99 𝑘𝑁
2.4m
h́    𝐼𝑔
´𝑦  𝑒=
𝐴 ´
𝑦
e  ´𝑦 = h́ = 2.989 =4.227 𝑚
m sin 𝜃 sin 45
33
0.8
0.5

7m   𝐼𝑔 0.2031
𝑒=
m

6
45 °
  0.6 = =0.043𝑚
𝐴 ´𝑦 ( 1.125 )( 4.227 )
1m

m 𝐿𝑜𝑐 𝑓𝑟𝑜𝑚 𝑊𝑆 =4.227+0.043=4.27 𝑚

1.5  
A trapezoidal gate with lower base width of 1m, upper base width of 0.5 and 1.5m high is immersed at
an angle of 45 degrees with the vertical. The vertex of the gate is pointing upward, and the base is 2.4m
below water surface.
Evaluate the total force acting on one side of the gate.
Obtain the location of the force from the water
 
𝐹=∫ 𝑝𝑑𝐴
surface.
𝑑𝐴=𝑏𝑑𝑥
 
𝑝=𝛾
  h ¿  9.81 ( 𝑥 sin 45 )
2.4m 9 4m 𝑏=𝑓 ( 𝑥)
3. 3  
h b
x
m x y
. 894
4 4.894 3.394 0.5
 
𝐹= 9.81 ( 𝑥 sin ( 45) ) ( 𝐴 + 𝐵𝑥 ) 𝑑𝑥
∫ 4.894 1
3 .394
0.5

𝐹=32.99
  kN 𝑏=
  𝐴+ 𝐵𝑋
dx
m

45 °
 

  ∫ 𝑥 𝑝 𝑑𝐴
b

𝑐𝑝=
1m

m 𝐹
1.5
 𝐹=50.65 𝑘𝑁 𝑐𝑝=4.25𝑚
 

𝐹=∫ 𝑝𝑑𝐴
 

s.g=0.8 2m
𝑑𝐴=𝑏𝑑𝑥
 
 𝑝=𝛾 h+15.696 ¿  9.81 ( 𝑥 sin (45) )+15.696
2.4m 9 4m 𝑏=𝑓 ( 𝑥)
3. 3  
h x b
m
. 894 x y
4 4.894
 
𝐹= (9.81 ( 𝑥 sin (45) )+15.696) ( 𝐴+ 𝐵𝑥 ) 𝑑𝑥 3.394 0.5
∫
3 .394
4.894 1
0.5

𝐹=50.65 kN 𝑏=
dx     𝐴+ 𝐵𝑋
m

45 °
 

  ∫ 𝑥 𝑝 𝑑𝐴
b

𝑐𝑝=
1m

m 𝐹 𝑐𝑝=4.25𝑚
 
1.5