May 2011 Board Exam All Subjects
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Mathematics and Surveying and Transportation Engineering PROBLEM 1: A 1 km x 1 km square lot is divided into 5 equal parts, one square at the middle, while the other four are right triangles. a.) Find the dimensions of the square at the middle. b.) If this square is further divided, find the dimensions of the small square at the middle. c.) Find the percentage ratio of the smallest area to the biggest area.
SOLUTION : a.) Dimensions of the square at the middle.
1 A1 (1000) 2 2 A1 200,000
x12 200,000 x1 4 4 7 . 2 1m b.) Square is further divided, find the dimensions of the small square at the middle.
1 A2 (200,000) 5 A2 40,000 x22 40,000 x2 2 0 0 m c.) Percentage ratio of the smallest area to the biggest area.
40,000 Ratio (100) (1000) 2 Ratio 4% PROBLEM 2: A conical water tank with its vertex down has a height of 20 cm., and a radius of 10 cm. at the top. Water leaks such that the water surface falls at a rate of 0.5 cm/sec. Compute the rate of change of volume when h = 8 cm.
SOLUTION :
V
h 2
V
h
dt
SOLUTION : (2R)2 = (5)2 + (6)2 R = 3.91 m.
PROBLEM 4: Find the distance from point (5, 6, 7) to the origin.
SOLUTION :
d2 = (5-0)2 + (6-0)2 + (7-0)2 d = 10.49
PROBLEM 5: Find the slope of the line that passes through (1,2) if the area is bounded by this line and the coordinates axes in the 1 st quadrant is a minimum. SOLUTION :
A
xy 2
y
x
y2 1
x A
y y2
y y
2 dA 1 y 2 2 y y 1 0 dy 2 y 22 2 y2
2 2 2y 4y y 2 y 4y y4 4
4 2
2
y y1 x x1
40 02
2
PROBLEM 6: Two trains are running in the same direction on two parallel tracks. The trains are 85 m. and 65 m. respectively and running at 120 kph and 240 kph. If the trail end of the first train is 210 m. ahead of the front end of the second train.
2 h
a.) Determine the times required for the front end of the second train to reach the tail end of the first train.
2 dh
12
A rectangular prism with dimensions of 3 x 4 x 6 cm. is inscribed in a sphere, find the radius of the sphere.?
m
3 3
3h
PROBLEM 3:
m
12
dV
dt
x
r 2h V 3 r 10 h 20 h r 2
3 2 5 . 1 3 cm / s ec.
dt b.) Determine the total distance traveled by the second train from the initial position when it has complete overtaken the first train.
dV dt
12
382 0.5
c.) Determine the distance traveled by the first train from the initial position until both ends of the two trains abreast each other.
2
2
4x – 24x + 4y – 8y + 15 = 0
SOLUTION :
a.) Time required for the front end of the second train to reach the tail end of the first train.
a.) Classify the equations above. b.) Determine the points of intersection of the two curve c.) Determine the distance between the two points of intersection. SOLUTION :
a.) Classify the equations above. 2
x – 6x + 3y + 11 = 0 PARABOLA 2 2 4x – 24x + 4y – 8y + 15 = 0 CIRCLE
66.67t = 210 + 85 + 33.33t – 85 33.33t = 360 t = 6.3 sec.
b.) b.) total distance traveled by the second train from the initial position when it has complete overtaken the first train.
Points of intersection of the two
curve.
2
x – 6x + 3y + 11 = 0 2
4x – 24x 2 4x – 24x +
multiply by 4
+ 12y + 44 = 0 2 4y – 8y + 15 = 0
2
4y – 20y – 29 = 0
4y2 – 20y 210 + 85 + 33.33t = 66.67t – 65 33.33t = t = 10.80 sec.
By completing squares 2 4y – 20y + 25 = 29 + 25 2 (2y – 5) = 54
Total distance traveled by second train to overtake the first train. = 66.67 (10.8) = 720 m.
2y – 5 = ± 7.35 y = 6.17 y = 1.174
c.) Distance traveled by the first train from the initial position until both ends of the two trains abreast each other.
= 29
when y = 6.17 2 x – 6x + 3(6.17) + 11 = 0 2 x – 6x + 9 = -29.51 + 9 2 (x - 3) = -20.51 x is complex. when y = -1.174 2 x – 6x + 3(-1.174) + 11 = 0 2 x – 6x + 9 = -7.478 + 9 2 (x - 3) = 1.522 x = 4.233 x = 1.767
210 + 85 + 33.33t = 66.67t t = 8.85 sec. Total distance traveled by the first train Until the two train abreast each other. Distance = 33.33 (8.85) Distance = 295 m. PROBLEM 7: A cylinder is inscribed in a cube. Find the volume of the solid inside the cube but outside the cylinder.
When x = 4.233, y = -1.174 When x = 1.767, y = -1.174 Points of intersection: (4.233, -1.174) and (1.767, -1.174)
c.) Distance between the two points of intersection. Distance = 4.233 – 1.767 Distance = 2.466
SOLUTION :
V 2
2
2 2
1
V 1.72 cu. cm.
PROBLEM 8: A cone is inscribed in a cube such that the base is in one of the face of the cube 1 cm x 1 cm.
PROBLEM 10: A framer owns two square lots of unequal size, together 2 containing 6568 m . If the lots were contiguous. It would require 356 m. of fence to enclose them in a single enclosure of six sides. Find the dimension of the smallest lot.
SOLUTION : 2
SOLUTION : 2 0.5 1 V 3
V = 0.262 cu. cm. PROBLEM 9: Given the following equations of the curve 2 x – 6x + 3y + 11 = 0
2
A=x +y 2 2 x + y = 6568 3x + 3y + y – x = 356 2x + 4y = 356 x + 2y = 178 x = 178 – 2y 2 2 x + y = 6568 2 2 (178 - 2y) + y = 6568 2 2 31684 – 712y + 4y + y = 6568 2 5y – 712y + 25116 = 0
y = 78 m. x = 178 – 2(78) x = 22m. Dimension of smallest lot is 22 m. x 22 m. PROBLEM 11: A sphere has a radius of 15 and its center is at the origin. Which of the following points is inside the sphere. I ( 4, 6, -12) II (-10, -9, 4) III ( 1, -2, 7) a.) b.) c.) d.) 2
2
2
y 13
Since the distance is less than 15, all points are inside the sphere. PROBLEM 12: Francis set out from a certain point at 6 kph. After he had gone 2 hours, Cruz set out to overtake him and went 4 km. rd the first hour, 5 km. the second hour, 6 km. the 3 hour and so on gaining 1 km. per hour. After how many hours were they together.?
c.
a.
(t +2) total time of travel for Francis Distance traveled by Francis = 6(t +2)
t 2
24 t 11
12 t 2 t 8 t 1
12t 24 8t t 2 t 5t 24 0 t = 8 hours
2
t
SOLUTION :
2x 1
x 3
2
x 12 2 2x 6x
x
x 3
lim
0.229
2 1 1 . 1 7 43 6 3
3 2 x 4 x 21x 2 x 10x 21
3 2 ( 3) 4 3 21( 3)
32 10(3) 21
72 0
Do not exist
PROBLEM 15:
In a certain Barangay, 80% have cellphones. If you select 2 persons from this Barangay.
SOLUTION : a.) Probability that both have cellphones. P = P1P2 P = 0.8 (0.8) P = 0.64
2
y –x =4 2yy’ – 2x = 0
2
lim
321.5
8 1.5 1
a. Find the probability that both have cellphones. b. Find the probability that one has a cellphone while other is none. c. Find the probability that both have no cellphones.
PROBLEM 13: What point in the curve y2 – x2 = 4 is nearest to point (6,0).
y'
2a n 1d
6 t2
2
lim
x 1.25
32x
b.
2
x5 2 16x 25
4x
4x 6
t = time in hours Cruz could overtake Francis. a=4 d=1 n=t
x 3
8x 1
c.
2
lim
SOLUTION :
SOLUTION :
n
3 2 x 4 x 21x 2 x 10x 21
2
r = (4 – 0) + (6 - 0) + (-12 – 0) r = 14 < 15 2 2 2 2 r = (-10 - 0) + (-9 - 0) + (4 – 0) r = 14.04 < 15 2 2 2 2 r = (1 - 0) + (-2 - 0) + (7 – 0) r = 7.35 < 15
S
The point is at 3, 13
PROBLEM 14: Evaluate the following limits 2 4x x 5 a. lim x 1.25 16x 2 25 2 x x 12 b. lim x 3 2 x 2 6 x
I and III only II and III only III only I, II and III
SOLUTION :
2x = 6 x=3 2 2 y – (3) = 4 2 y = 13
x y 2
2
d = (x - 6) + (y – 0) 2 2 2 d = (x - 6) + y 2dd’ = 2(x – 6) + 2yy’ = 0
x 6 y' y x x6 y y
b.) Probability that one has a cellphone while the other is none. P = P1P2 P = 0.8 (1 - 0.8) P = 0.16 c.) Probability that both have no cellphones. P = P1P2 P = (1 – 0.8)(1 - 0.8) P = 0.04 PROBLEM 16:
A sphere having a center at the origin has a radius of 15. Which of the points is outside the sphere? A. (1, 2, -7) B. (8, 6, 12) C. (10, 9, 4) D. (-4, 8, -7) SOLUTION : Using (1, 2, -7) 2
2
2
2
r = (1 – 0) + (2 - 0) + (7 – 0) r = 7.34 < 15 (inside the sphere)
Using (8, 6, 12) 2
2
2
Hydraulics and Geotechnical Engineering PROBLEM 1: A pump draws water from reservoir A and lifts it to reservoir B as shown. The loss of head from A to 1 is 3 times the velocity head in the 150 mm pipe and the loss of head from 2 to B is 25 times the velocity head in the 100 mm pipe. When the discharge is 25 liters/ sec.
2
r = (8 – 0) + (6 - 0) + (12 – 0) r = 15.62 > 15 (outside the sphere)
Using (10, 9, 4) 2
2
2
2
r = (10 – 0) + (9 - 0) + (4 – 0) r = 14.04 < 15 (inside the sphere) Using (-4, 8, -7) 2 2 2 2 r = (-4 – 0) + (8 - 0) + (-7 – 0) r = 11.36 < 15 (inside the sphere) therefore point (8, 6, 12) is outside the sphere. PROBLEM 17 : In an organization, there are Civil Engineers, Electrical Engineers, and Mechanical Engineers. The sum of their ages is 2,160; their average age is 36; the average of the C.E. and M.E. is 39; of the M.E. and E.E. 32(8/11); of the C.E. and E.E. 36 (2/3). If each C.E. had been one year, each E.E. 7 years and each M.E. 6 years older, their average age would have been greater by 5 years. Find the average age of the Civil Engineers. Ans. 45 yrs.
1.
Compute the horsepower output of the pump in kilowatts. Compute the pressure head at 1. Compute the pressure head at 2.
2. 3.
SOLUTION : 1. Power output of pump :
V1
0.02
4
0.152
V1 = 1.13 m/s
0.02
V2
4
0.102
V2 = 2.55 m/s
V3
2
2g
P3
Z 3 HA
V4
2
2g
P4
Z 4 HL
0 + HA = 240 + HL 2
PROBLEM 18 : A block of copper has a mass of 50 kg. and a density of 8.91 g. per cubic cm. a.) Find the volume of the copper. b.) If this block is converted to a cylinder with the same volume, find the base area if the length is 1500 cm. c.) Find the diameter of the cylinder.
SOLUTION : a.) Volume of the copper.
V
W
50,000 8.91
V 5 6 1 1 . 6 7 cm
3
b.) Base area if the length is 1500 cm.
V AL
HL
2 31.13
29 . 8 1 HL = 8.48 m.
2
2 252.55 29 . 8 1
HA = 240 + 8.48 HA = 248.48 m Power output = Q E E = H. A. E = 248.48 m. Power output = 0.02 (9.81)(248.48) Power output = 48.75 kW
D V
3V 25V2 HL 1 2g 2g
5611.67 A 1500
V 3 . 7 4 cm
2
2. Pressure head at 1 : 2 2 VA PA V1 P ZA 1 Z1 Hf1 2g 2g 1.132 P1 (20) 31.132 000 2g 2g
c.) Diameter of the cylinder.
D 2 4 2 3.74 D 4 A
P1 19.74m.
D 2 . 1 8 cm .
3. Pressure head at 2 :
2 2 V1 P1 V2 P Z1 HA 2 Z2 2g 2g
1.13
2
2g
19.74 20 248.48
2.55
2
2g
P 2 20
P2 267.95m.
ult f b design
ult FS
ult 3602
ult 720 kN ult f b 720 f
PROBLEM 2: A valve is suddenly closed in a 200 mm pipe. The increase in pressure is 700 kPa. Assuming that the pipe is rigid and the bulk modulus of water is 2.07 x 109 N/m2. a.) Compute the celerity of the pressure wave. b.) Compute the velocity of flow. c.) If the length of the pipe is 650 m. long. Compute the water hammer pressure at the valve if it is closed in 3 sec.
SOLUTION : a.) Celerity of pressure wave. : EB
C
2. Skin friction :
64.74 Q f 655.26 kN 3. Length of pile :
f C. P L
655.26 0.76 55.5 0.36 4 L
L 10.90 m. PROBLEM 4: A confined aquifer has a source of recharge as shown in the figure. The hydraulic conductivity of the aquifer is 40m/day and its porosity is 0.25 . The piezometric head in the two wells 1350 m. apart is 65 m. and 60 m. respectively from a common datum. the average thickness of the aquifer is 25 m. and the average width is 4 km.
2.07 x 109
C
1000 c = 1438.75 m/s
b.) Velocity of the flow. : Increase in pressure Ph = CV 700000 = 1000 (1438.75) V V = 0.486 m/s c.) Water hammer pressure when it is closed in 3 sec. :
2L
t
C
SOLUTION :
2 650
1. Rate of flow : h i L 65 60 i 1350
t
1438.75 t 0.904 sec.
Ph
0.904
700 3 Ph 210.83 kPa. PROBLEM 3:
A 0.36 m. square pre stressed concrete pile is to be driven in a clayey soil as shown in the figure. The design capacity of the pile is 360 kN, with a factor of safety of 2. 1. 2. 3.
Compute the end bearing capacity of the pile. Compute the skin friction expected to develop along the shaft of the pile. Compute the length of the pile if = 0.76.
SOLUTION : 1. End bearing capacity of piles :
C
1. Compute the rate of flow through the aquifer in m3/day. 2. Compute the seepage velocity 3. Compute the time of travel from the head of the aquifer to a point 3 km. downstream in days.
q u 111 2 2
C 55.5 kN/m2 b c NC A tip
h 55.590.360.36 Q b 64.74 kN
i 0.0037 A 254000 A 100,000 m 2 Q Ki A
Q 40 0.0037 100,000 3 Q 1 4 8 0 0 m / d ay
2. Seepage Velocity :
Q A 14800 V 100,000 V 0.148 V VS n 0.148 VS 0.25 VS 0.592 V
3. Time to travel 3km downstream : a.) Compute the lateral pressure in the cell for a failure to occur. b.) Compute the maximum principal stress to cause failure. c.) Compute the normal stress at the point of maximum shear.
Dis tan ce
t
SeepageVelocity
3000 0.592 t 5068 days t
SOLUTION : PROBLEM 5: A tank 12 m. high filled with oil having a unit weight of 9.4 3 kN/m is to be built on a site. The existing soil profile consists of a 3m. sand layer underlain by a 14 m. clay layer. The water table is on the ground surface. Neglecting the weight of the tank.
tan 28
x
65 55.93 3 65
3 13.14 kPa (lateral pressure in the cell) b.) maximum principal stress to cause failure:
1 3 65 65 1 13.14 130
SOLUTION :
1 143.14 kPa c.) normal stress at the point of the maximum shear :
1. Compression index :
PL
N 3 70
LL P L
0.645
31
x 55.93 Sin 28
a.) Compute the compression index of clay. b.) Compute the settlement under the center of the tank. c.) Find the minimum depth in the ground to which the tank must be placed in order to minimize settlement.
P.I.
a.) Lateral pressure in the cell :
N 13.14 65
40 20
N 78.14 kPa
LL 20 LL 51%
PROBLEM 7:
Cc 0.00951 10 Cc 0.009 LL 10
A rectangular channel 5.6 m. wide by 1.2 m. deep is lined with a smooth stone. Well laid and has a hydraulic slope of 0.002. Using n = 0.013.
Cc 0 . 3 6 9
a.) What is the capacity of the channel in m /s. b.) What savings in earth excavation could have been offered by using more favorable proportions but adhering to the same delivery and slope. c.) What savings in lining per meter length by using more favorable proportions but adhering to the same delivery and slope. ?
3
b. Settlement under the center of the tank :
Po 18.88 9.81 3 17.31 9.81 7 Po 79.71
P 9.4 12
P 112.8 kPa
S S
CCH 1 e
log
SOLUTION : a.) Capacity of channel :
P Po
A = 5.6 (1.2) A = 6.72 m2 P = 1.2(2) + 5.6 P = 8 m. A R P
Po
log 112.8 79.71
0.369 1400
1 1.27 S 8 7 . 1 5 mm
79.71
c. Minimum depth in the ground to which the tank must be placed in order to minimize settlement.
R
6.72 8
R = 0.84
A R 2/3 S 1/ 2 Q n 1/ 2 6.72(0.84) 2 / 3 0.002 Q 0.013 3 Q 20.58 m / s PROBLEM 6: A cohesive soil specimen has a shearing resistance equal to 29 and a cohesion of 31 kPa. If the maximum shearing stress of the soil sample is equal to 65 kPa.
2. Savings in earth excavation by using more favorable proportions. : Use most efficient section : b = 2d
R
d 2
b=bd b = 2 d2
Q
AR
2/3 1/ 2 S n
2d 20.58
2 d 2
G S e
S 2/3
0.0011/ 2
S
0.013
d 1.794 b = 2d b = 2 (1.794) b = 3.578 m.
w 1 e 2.68 0.731
1 0.73
3 1.97 g r /c m
S
b. Dry density of the soil :
Savings in excavation : Savings = 5.6 (1.2) – 1.794 (3.587 ) Savings = 0.2849 m2
2.681
dry dr y
1 0.73
3 655 g r /c m
3. Savings in lining per meter length : Lining of old channel = [1.2 (2) + 5.6](1) 2 Lining of old channel = 8 m
c. Water content at a degree of saturation of 92% and 100%.
S
Lining of new channel = [ 1.794(2) + 3.587](1) 2 = 7.175 m
e
0.92
Savings in lining = 8 – 7.175 2 Savings in lining = 0.825 m /m.
0.73
0.25 25%
A conical having a radius of base equal to 35 cm. and a height of 65 cm. has its base at the bottom. a.) If water is poured into the tank, find the total volume to fill up. b.) How much additional water is required to fill the tank if 3 0.025 m of water is poured into the conical tank. 3 c.) Find the height of free surface if 0.025 m of water is poured into a conical tank.
SOLUTION : a.) Total volume to fill up : 2 r h v 3
0.65
0.35
100
0.73
PROBLEM 10: A saturated clay layer has a thickness of 30 m. with a water content of 59% and a specific gravity of 2.76.
a.) Determine the saturated density of the clay. b.) Compute the total stress at the bottom. c.) Compute the effective stress at the bottom.
SOLUTION : a.) Saturated density of the clay :
Gs
S
e
3
3
b.) Additional water required to fill the tank.
V1 = 0.0834 – 0.025 V1 = 0.0584 m3 3. Height of free surface : 3 V
V1
(0.65)
0.0834 0.0584
h13
2.68
27.24%
2
v 0.0834m
2.68
At 100%,
PROBLEM 8:
v
Gs
(0.65)3 h13
h1 = 0.577 m.
PROBLEM 9: A soil has a specific gravity of 2.68 and a void ratio of 0.73 if the degree of saturation is 92%. a.) Compute the total density of the soil. b.) Compute the dry density of the soil. c.) Compute the water content at a degree of saturation of 92% and 100%.
SOLUTION : a.) Total density of the soil :
100
592.76
e
e = 1.628 G e S Sat w 1 e 2.76 1.628 (1) Sat 1 1.628
sa t
3 1 .6 7 g r /c m
b.) Total stress at the bottom : G e S Sat w 1 e 2.76 1.628 (9.81) Sat 1 1.628
3 16.38 kN /m sa t Total stress = 16.38 (30) Total stress = 491.40 kN /m2
c.) effective stress at the bottom: Effective stress = (16.38 – 9.81)(30) Effective stress = 197.10 kN/m2
Structural Design and Construction PROBLEM 1: A 400 mm square column shown is supported by square footing on 5 piles as shown. Dimensions are a = 0.75 m., b = 2 m., effective footing depth = 0.6 m., Ultimate pile capacity = 320 kN Column axial loads : D = 420 kN L = 360 kN E = 210 kN Column moment, ME due to earthquake = 160 kN.m Required strength of the footing is based on.: U = 1.32D + 1.1L + 1.1E a.) Compute the critical beam shear stress at ultimate loads. b.) Compute the ultimate punching shear stress. c.) Compute the maximum design moment.
SOLUTION : a.) Critical beam shear stress at ultimate loads. Pu = 1.32D + 1.1L + 1.1E Pu = 1.32(420) + 1.1(360) + 1.1(210) Pu = 1181.4 kN Mu = 1.32(0) + 1.1(0) + 1.1(160) Mu = 176 kN.m Ultimate reaction of pile at row (1) R
Pu M u C A I 2
I=Ad Assume area of one pile = 1 2 I = 2(1) (2) = 4 C = 1.0
1181.4 176(1) R 5 4 R = 280.28kN < 320 kN (safe) Vu = 280.28(2) Vu = 560.56 kN Vu bd 560560 3500600 0.31 MP a
b.) Ultimate punching shear stress. P MC R A I 1181.4 176(0) R 5 4 Ru = 236.28 kN Vu = 1181.4 – 236.28 Vu = 945.12 kN Vu b d o b
o
41000 4000 mm.
945120 4000600
0.463 MP a
b.) If the compressive stress at the bottom fiber is 12 MPa and the tensile stress at the top fiber is 2 MPa. c. If the compressive stress at the top fiber is 16 MPa and zero at the bottom fibers.
SOLUTION : a.) Value of P and e when the compressive stress of 21 MPa. P f A 21
P 300(600)
P 3780 kN e 0
b.) Value of P and e when compressive stress at the bottom is 12 MPa and tensile stress at the top is 2 MPa. f
P MC A I
P 6M bd bd 2 P 6 Pe f bd bd 2 f
P 6 e 1 stress at the top d bd P 6 e f 1 stress at the bottom b bd d P 1 6 e equation (1) 2 300(600) 600 P 1 6 e equation (2) 12 300(600) 600 f t
Divide equation (1) by (2) 1 600 6e 6 600 6e 600 6e 3600 36e 30e 4200
e 140 mm
P 1 6 140 600 300600 P = 900,000 N P = 900 kN
12
c.) Value of P and e when compressive stress at the top fiber is 16 MPa and zero at the bottom fibers.
P MC A I At the bottom of the beam : P 6 e f 1 b bd d P 1 6 e 0 300600 d 6e At the top of the beam : 1 0 d P 6 e f 1 6e b bd d 1 0 600 P 1 6 100 16 600 6e 0 600 300600 600 P = 1440,000 N e 6 P = 1440 kN e 100 mm f
c.) Maximum design moment. Mu = 2R(0.80) Mu = 2 (280.28)(0.8) Mu = 448.45 kN.m
PROBLEM 2: A beam with width b = 300 mm and depth d = 600 mm is to be prestressed. Considering a 15% prestress loss, compute the value of prestressing force P and eccentricity e.
a.) If the compressive stress of 21 MPa.
PROBLEM 3: A 9 m. high retaining wall is laterally supported at the top and fixed at the base. The wall resists active earth pressure increasing from 0 at the top to 52 kN/m at the base per meter length along the longitudinal axis. a.) Determine the design moment at the base. Apply the 2 2 fixed end moment equation WL at the top and WL at 30 20 the base. EI is constant.
222 6120 4
b.) The lateral support at the top of the wall was removed, determine the design moment at the base.
P
c.) Determine the resulting base shear if the wall is free at the top.
P = 273696 N P = 273.70 kN
SOLUTION : a.Design moment at the base when the top is laterally supported.
WL2 20 2 599 M 20
M
b.) Based on bearing capacity of bolts. P = AbFp P = 16(22)(6)(1.2)(400) P = 1013760 N P = 1013.76 kN c.) Based on block shear strength.
M = 210.6kN.m b. Design moment at the base when the support at the top was removed.
52 91 P 2 P = 234 M = 3P M = 3(234) M = 702 kN.m
c. Base shearing if the wall if the lateral support is removed at the top :
Av = (200 – 25- 25- 12.5)(16) Av = 2200 mm2 At = (100 – 12.5 – 12.5)(16) At = 1200 mm2 P = 0.30 Fv Av + 0.50 Fu At P = 0.30(400)(2200) + 0.50 (400)(1200) P = 504000 N P = 504 kN Second possible failure in block shear :
PV 529 V 2 V 234 kN.
PROBLEM 4: Two plates each with thickness t = 16 mm are bolted together with 6 – 22 mm bolts forming a lap connection. Bolt spacing are as follows : S1 = 40 mm, S2 = 80 mm, S3 = 100 mm Bolt hole diameter = 25 mm Allowable stress : Tensile stress on gross area of the plate = 0.60 Fy Tensile stress on net area of the plate = 0.50 Fu Shear stress of the bolt : Fv = 120 MPa Bearing stress of the bolt : Fp = 1.2 Fu Calculate the permissible tensile load P under the following conditions.
a.) Based on shear capacity of bolts. b.) Based on bearing capacity of bolts. c.) Based on block shear strength.
SOLUTION :
Av = (200 – 25- 25- 12.5)16 Av = 2200 mm2 At = (140 – 12.5 – 25)(16) At = 1640 mm2 P = 0.30 Fu Av + 0.50 Fy At P = 0.30(400)(2200) + 0.5 (400)(1640) P = 592000 N P = 592 kN Use P = 504 kN5.
PROBLEM 5:
A column section shown is reinforced with 8 – 32 mm bars, with a clear concrete cover of 40 mm for the 12 mm ties. Due to reversal of lateral forces. The design axial load due to the reversal effect of DL, LL and WL changes as follows Along the positive x – direction : Mu = -420 kN.m Vu = 370 kN Nu = 1320 kN Along the negative x – direction : Mu = +420 kN.m Vu = 370 kN Nu = 450 kN Use fc’ = 28 MPa and fy = 415 MPa
a.) Based on shear capacity of bolts.
A. ) Allowable shear strength for simplified calculation. 1.) For members subject to shear and flexure only.
Vc
1 f C 'b w d 6
2.) For members subject to an axial compression. 1 Nu Vc 1 f C 'b w d 6 14 Ag B. ) Allowable shear strength for detailed calculation 1.) For members subject to shear and flexure. V d 1 Vc f C ' 120 w u b w d 0.3 f C 'b w d 7 Mu Vu d shall not be greater than 1.0 Mu Where Nu is negative for tension
2.) For members subject to an axial compression. V d 1 Vc f C ' 120 w u b w d 7 Mu Where :
4h d Mm Mu Nu 8 Vu d is not limited to 1.0 Mu Vc = shall not be greater than : 0.3N u Vc 0.3 1 fc ' bw d Ag When Mm is negative : 0.3N u Use Vc 0.3 1 fc ' bw d Ag
3.) For members subject to significant axial tension. 1 0.3N u Vc 1 f c ' b w d 6 Ag Min. area of shear reinforcement : b S AV w 3 fy When factored shear force Vu exceeds the shear strength Vc, shear reinforcement shall be provided and the shear strength Vs shall be computed where shear reinforcement perpendicular to the axis member :
Av f d y S a.) Determine the concrete shear strength for the positive x – direction using simplified calculation. b.) Determine the concrete shear strength for the negative x – direction using simplified calculation. c.) Determine the required spacing of shear reinforcement. Apply provisions on spacing limits of reinforcement when applicable. AS
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