PSAD - Nov.

2019

Problems 6. Compute the value of W (kN) to cause an uplift of 30 kN at

1. A square bar is perfectly fitted between rigid walls. Given A.
the following data: A. 37.5 C. 24.0
Length of bar = 900 mm B. 54.0 D. 13.3
Bar cross-section = 16 mm × 16 mm C
Coefficient of thermal expansion of the bar = 11.7×10-6
m/m-°C
Modulus of elasticity, E = 200,000 MPa
Initial temperature, T1 = 15°C h
What force (kN) is developed in the bar when the
temperature changes to 35°C? A B
A. 9 C. 10.2
B. 12 D. 13.5

2. Given the following data of the cantilever beam shown: a b

Bar diameter, db = 20 mm Situation 2 – The truss shown is acted on a gravity load P = 85
Radius of bend, r = 4db kN and wind loads of H1 = 16 kN, H2 = 27 kN, and H3 = 11
Bar extension = 12db kN
L=2m Given: s = 3 m; h = 1.5 m.
Clear cover = 65 mm 7. Compute the resultant reaction (kN) at A due to wind load
fc’ = 34.5 MPa only.
fy = 413 MPa A. 17.08 C. 36.89
Calculate the total length (mm) of the 20-mm-diameter B. 14.34 D. 40.65
reinforcement. 8. Compute the resultant reaction (kN) at B due to wind load
A. 2388 C. 2834 only.
B. 2588 D. 3004 A. 17.08 C. 36.89
B. 14.34 D. 40.65
9. Compute the axial force in member CF (kN) due to gravity
90° bend load only.
r r A. 101.4 C. 134.4
B. 143.3 D. 114.0
P
L w F
C D
Development of Standard Hooks in Tension h
Development length ldh for deformed bars in tension B
A
terminating in a standard hook shall be the greater of (a) E G
through (c).
𝑓𝑦
a. ( ) 𝑑𝑏 s s s
4.17𝜆√𝑓𝑐 ′
b. 8𝑑𝑏
c. 150 mm
Situation 3 – A solid shaft having a diameter of 50 mm is bent
3. Given the following data of a one-way slab: into a circular arc of radius R = 0.8 m with a central angle θ
Slab thickness, h = 125 mm = 60°. The shaft is fixed at A and supports a vertical load P =
Longitudinal bars = 12 mm 5 kN and end B. Neglect weight of the shaft.
Spacing of bars = 125 mm
Clear concrete cover = 20 mm
Concrete strength, fc’ = 27.5 MPa R
Steel yield strength, fy = 275 MPa
Compute the design strength (kN-m) of the slab. θ
R
A. 26.8 C. 23.3
B. 15.4 D. 20.9

Situation 1 – The girder bars AC and BC are used to lift the load P
W as shown in the figure.
Given: 10. How much is the maximum normal shear stress (MPa) in
a = 3.2 m; b = 4 m; h = 3 m the shaft?
4. As the load W is lifted, which of the following gives the force A. 2.04 C. 0.85
in bar AC? B. 3.40 D. 2.55
A. (13/15)FBC C. (52/15)FBC 11. How much is the maximum torsional shear stress (MPa) in
B. (12/5)FBC D. (13/20)FBC the shaft?
5. How much is the load W relative to the force in bar BC? A. 122 C. 81.5
A. 0.245FBC C. 0.350FBC B. 141.1 D. 21.8
B. 0.733 FBC D. 0.267FBC 12. How much is the maximum flexural stress (MPa) in the
shaft?
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PSAD - Nov. 2019
A. 35 C. 282 CONTINUOUS BEAM – FOUR EQUAL SPANS – THIRD-SPAN UNLOADED
B. 44 D. 163 wl wl wl

Situation 4 – Refer to the semi-circular arch shown: A B

l
C
l
D
l
E

Given:
P1 = 2.7 kN; P2 = 1.35 kN; P3 = 1 kN SHEAR
α = 45°; θ = 30°; r = 2 m
13. Determine the horizontal reaction (kN) at B.
A. 1.63 C. 2.04 MOMENT

B. 2.43 D. 3.04
14. Determine the vertical reaction (kN) at B.
A. 2.04 C. 1.70 CONTINUOUS BEAM – FOUR EQUAL SPANS – LOAD FIRST & THIRD SPAN

B. 1.95 D. 1.37 wl wl

15. Determine the bending moment (kN-m) in the arch at P1. A B C D E

A. 0.92 C. 0.85 l l l

B. 1.70 D. 0.46
SHEAR
P2

MOMENT
P3

P1
CONTINUOUS BEAM – FOUR EQUAL SPANS – ALL SPAN LOADED

wl wl wl wl

A B C D E
l l l

A r r B
s SHEAR

Situation 5 – Steel beams are used to support the deck of a MOMENT

bridge with simple span of 25 m. The live load on each

beam are as follows:
Wheel live loads:
Front wheel = 35.6 kN
Rear wheel = 142.4 kN Situation 7 – The double angle shown in the figure carries a
Wheel base = 4.3 m tensile force P.
16. Compute the maximum support reaction (kN) in the beam. Given:
A. 213 C. 172 Allowable shearing stress of bolt = 150 MPa
B. 268 D. 198 Allowable tensile stress of bolt = 195 MPa
17. Compute the maximum bending moment (kN-m) in the Bolt diameter, d1 = d2 = 25 mm
beam. Slope, V:H = 1:2
A. 1058 C. 1369 22. Find the maximum safe load P (kN) based on tensile and
B. 1288 D. 1037 shearing strength of bolt d2.
18. Determine the maximum shear (kN) at midspan. A. 632 C. 329
A. 82.9 C. 104.1 B. 856 D. 587
B. 96.5 D. 128.5 23. Find the maximum safe load P (kN) based on shearing
strength of bolt d1.
Situation 6 – The beam ABCD is to be analyzed for maximum A. 587 C. 441
forces at ultimate conditions. The beam is simply supported B. 365 D. 220
at A, B, C, D and E. 24. If P = 400 kN, what is the required bolt diameter (mm) d2?
Given: A. 22 C. 25
Span, l = 6 m B. 32 D. 28
Factored loads:
Dead load = 12 kN/m
Live load = 15 kN/m
19. How much is the maximum reaction (kN) at D?
A. 175.8 C. 210.8 bolt, 3d1
V
B. 192.4 D. 136.9 H
20. How much is the maximum shear (kN) at D?
gusset
A. 85.4 C. 99.5
B. 98.3 D. 63.2 plate bolt,4d2
21. How much is the maximum moment (kN-m) at D?
A. 111.3 C. 175.8 t1
B. 154.7 D. 104.1

Situation 8 – A fully restrained beam has a span of 12 m. It

carries a total uniformly distributed load of 30 kN/m. To
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PSAD - Nov. 2019
prevent excessive deflection, a support is added at 32. Compute the maximum bolt hole diameter (mm) based on
midspan. tension on plate when P = 320 kN
25. Compute the reaction (kN) at the added support. A. 65 C. 60
A. 180 C. 90 B. 45 D. 50
B. 270 D. 60 33. Compute the minimum bolt diameter (mm) based on
26. Compute the maximum shear (kN) in the beam. bearing on plate when P = 480 kN.
A. 100 C. 120 A. 20 C. 22
B. 60 D. 90 B. 16 D. 25
27. Compute the maximum bending moment (kN-m) in the s3 s4 s3 s4 s3
beam.
A. 45 C. 90
B. 120 D. 60 s1
Situation 9 – The beam ABC shown is simply supported at B
and C. P s2 P
Given: a = 3 m; b = 9 m; L = 4 m
Properties of W460×106 s1
A = 13,400 mm2 tf = 20.6 mm
d = 470 mm tw = 12.6 mm
bf = 194 mm Sx = 2.08×106 mm3
28. Given: t2
Allowable bending stress of the beam, Fb = 112 MPa. t1
Compute the maximum load w in kN/m. t2
A. 54 C. 29
B. 21 D. 36
29. Given: w = 25 kN/m Situation 11 – A flat steel bar is 5 mm thick by 50 mm wide by
If the beam is connected with two bolts in double shear at C, 300 mm long. Use E = 200 GPa. The bar is bent by moment
determine the minimum bolt diameter (mm) if the applied at its ends.
allowable shearing stress of the bolt is 82 MPa. 34. Determine the radius of curvature (m) of the bar if the
A. 16 C. 25 maximum bending stress in the bar is 124 MPa.
B. 28 D. 20 A. 5.36 C. 3.25
30. Given: w = 25 kN/m B. 4.03 D. 2.89
Compute the bearing stress (MPa) in the plate at B if a 190 35. Determine the moment (N-m) if the maximum bending
mm × 260 mm plate is used. stress in the bar is 164 MPa.
A. 2.4 C. 4.0 A. 69.8 C. 358.9
B. 5.6 D. 3.2 B. 105.4 D. 34.2
36. Determine the deflection (mm) if the applied moment at the
w (kN/m) ends is 0.25 kN-m.
A. 26.7 C. 16.8
B. 12.4 D. 22.4
A
B C Situation 12 – Wooden planks 100-mm thick and 300 mm wide
are used as temporary retaining wall to support a 3 m high
soil.
L Given:
Unit weight of soil = 17.8 kN/m3
Coefficient of active pressure = 1/3
Allowable shear stress of wood = 0.34 MPa
D Allowable bending stress of wood = 10.2 MPa
37. Determine the maximum shearing stress (MPa) in the
plank.
A. 0.4 C. 0.3
a b B. 0.5 D. 0.6
Situation 10 – Given the following data of the butt connection 38. Determine the maximum bending stress (MPa) in the plank.
shown: A. 24 C. 20
Dimensions: B. 22 D. 16
s1 = 50 mm; s2 = 100 mm; s3 = 50 mm; s4 = 90 mm 39. What minimum plank thickness (mm) is required to
t1 = 20 mm; t2 = 12 mm prevent failure?
Allowable stresses: A. 130 C. 140
Shear on bolt = 117 MPa B. 120 D. 150
Tension on net area of plate = 220 MPa
Tension on gross area of plate = 148 MPa Situation 13 – Angle section is considered to be used as a
Bearing on plate = 338 MPa column with unsupported length of 2.3 m. The column is
31. Compute the minimum required bolt diameter (mm) based hinged at both ends. Use the given table for the allowable
on bolt shear when P = 320 kN. stress of compression for yield strength of 248 MPa.
A. 16 C. 25 Properties of L102×102×16
B. 22 D. 20 A = 2974 mm2
Ix = Iy = 2.76×106 mm4
rx = ry = 30.38 mm
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rz = 19.66 mm A. 10 C. 8
x = y = 31 mm B. 12 D. 16
40. Determine the allowable load (kN) if a single angle will be
used.
A. 187 C. 324 bf bf
B. 219 D. 257
41. Determine the allowable load (kN) for double angle welded
together with legs back-to-back.
A. 740 C. 545
B. 325 D. 650
42. Determine the maximum length (m) of a single angle based
d
on the limit of slenderness ratio of a compression member.
A. 3.93 C. 3.21
B. 4.25 D. 2.87
ALLOWABLE STRESS
FOR COMPRESSION MEMBERS OF 248 MPA SPECIFIED YIELD
STRESS

Situation 15 – A 10-m simply supported steel beam is

subjected to uniformly distributed loads. Bending is about
the major axis.
Given:
Superimposed dead load = 16 kN/m
Section W410×100
Area, A= 12,700 mm2
Depth, d = 414 mm
Flange width, bf = 259 mm
Web thickness, tw = 10 mm
Flange thickness, tf = 16.9 mm
Elastic section modulus, Sx = 1.92×106 mm3
Steel yield strength, Fy = 248 MPa
Modulus of elasticity, E = 200 GPa.
46. If the nominal strength of the section is 528 kN-m,
determine the live load (kN/m) based on the design flexural
strength of the beam, Mu;
Given:
Resistance Factor for Flexure, ϕb = 0.90
Factored Load Combination, U = 1.2D + 1.6L
A. 17.6 C. 21.0
B. 11.0 D. 14.7
47. Determine the load live load (kN/m) based on the design
shear strength of the beam, Vu;
Given:
Factored Shear Stress, vu = 0.6Fy
Resistance Factor for Shear, ϕv = 1.0
A. 64.26 C. 35.25
Situation 14 – Two channels are welded tat their tips to form a
B. 52.41 D. 85.42
box column as shown. The column is 9 m long and braced
48. At service load, the allowable midspan deflection due to live
against sidesway in both directions with both ends hinged
load is 50 mm. Determine the service live load (kN/m)
and is laterally supported at mid height about the Y-axis.
based on the allowable deflection.
Use E = 200 GPa and Fy = 248 MPa.
A. 20 C. 30
Channel Properties:
B. 50 D. 40
A = 3923 mm2 tw = 7.2 mm
d = 305 mm Ix = 53.7×106 mm4
Situation 16 – Given the following data of the frame shown
tf = 12.7 mm Iy = 1.61×106 mm4
below.
bf = 75 mm x = 17.7 mm
Given:
43. Determine the effective slenderness ratio with respect to
L1 = L2 = 11.5 m; H1 = 4.2 m; H2 = 3.5 m
the x-axis.
Moment of inertia of members
A. 82 C. 96
Column ADG: W600×82; I = 5.62×108 mm4
B. 74 D. 77
Column BEH: W600×114; I = 8.74×108 mm4
44. Determine the effective slenderness ratio with respect to
Column CFI: W600×82; I = 5.76×108 mm4
the y-axis.
Beam DEF: W600×114; I = 8.74×108 mm4
A. 77 C. 65
Beam GHI: W600×175; I = 14.73×108 mm4
B. 54 D. 74
45. The column is subjected to an axial compressive force of
Use the given alignment charts.
1200 kN. Determine the thickness of 150-mm-wide cover
49. What is the effective length factor of Column DA given that
plates on the top and bottom of the section if the allowable
it is inhibited against sidesway.
compressive stress of the section is 111.5 MPa.
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PSAD - Nov. 2019
A. 0.78 C. 0.88
B. 0.84 D. 0.94
50. What is the effective length factor of Column DG given that
it is inhibited against sidesway.
A. 0.75 C. 0.72
B. 0.80 D. 0.85
51. What is the effective length factor of Column DA given that
it is uninhibited against sidesway.
A. 0.90 C. 1.2
B. 1.9 D. 1.6
G H I

H2
D E F

H1
A B C

L1 L2

Situation 17 – Refer to the cantilever beam shown:

Given:
db = 25 mm h = 600 mm
fc’ = 41.3 MPa L=2m
fy = 413 MPa a = 50 mm

Apply factors for normal weight concrete uncoated

reinforcement. cb = 50 mm, Ktr = 0.

The minimum development length is 300 mm for bars in

tension and 200 mm for bars in compression.
52. How much is the total length L (mm) of bar “A”?
A. 2730 C. 3630
B. 2650 D. 2900
53. How much is the total length L (mm) of bar “B”?
A. 2100 C. 2336
B. 2394 D. 2444
54. How much is the total length L (mm) of bar “A” if 20 mm
bar diameter is used?
A. 2760 C. 2534
B. 2436 D. 2710

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PSAD - Nov. 2019

L1 Lightweight concrete 0.75

In accordance
Lightweight, Lightweight concrete, where 𝑓𝑐𝑡 is
with Section
bar A 𝜆 specified
419.2.4.3
bar B
a Normal-weight concrete 1.0
Reinforcement enclosed within
(1), (2), (3) or (4):
h
1. a spiral
2. a circular continuously wound
tie with 𝑑𝑏 ≥ 6 mm and pitch ≤
100 mm
L Confining 0.75
reinforcement 3. 6 mm  bar or MDI30 wire ties
𝜓𝑟 in accordance with Section
L2 425.7.2 spaced ≤ 100 mm on
center
Development of Deformed Bars and Deformed Wires in 4. hoops in accordance with
Tension Section 425.7.4 spaced ≤ 100
For deformed bars or deformed wires in tension, 𝑙𝑑 shall be mm on center
Other 1.0
calculated by

1 𝑓𝑦 𝜓𝑡 𝜓𝑒 𝜓𝑠
𝑙𝑑 = ( )𝑑 Situation 18 – Refer to the figure shown.
1.1 𝜆√𝑓𝑐 ′ (𝑐𝑏 + 𝐾𝑡𝑟 ) 𝑏 Given:
𝑑𝑏 L1 = 6 m; L2 = 6.8 m
Column dimension = 400 mm × 400 mm
In which the confinement term (𝑐𝑏 + 𝐾𝑡𝑟 )/𝑑𝑏 shall not
exceed 2.5 and Service loads:
40𝐴𝑡𝑟 Dead load = 19 kN/m (all weight included)
𝐾𝑡𝑟 = Live load = 15 kN/m
𝑠𝑛
U = 1.2D + 1.6L
Where n is the number of bars or wires being developed or 55. Determine the moment (kN-m) at B in beam AB.
lap spliced along the plane of splitting. It shall be permitted A. 187 C. 193
to use 𝐾𝑡𝑟 = 0 as a design simplification even if transverse B. 200 D. 213
reinforcement is present. 56. Determine the shear force (kN) at B in beam BC.
A. 224 C. 177
For the calculation of 𝑙𝑑 modification factors shall be in B. 183 D. 193
accordance with Table 425.4.2.4. 57. Determine the maximum positive moment (kN-m) in beam
BC.
Table 425.4.2.4 Modification Factors for Development of A. 196.7 C. 136.9
Deformed Bars and Deformed Wires in Tension B. 174.3 D. 154.6
Modification spandrel beam
Condition Value of factor
factor
A B C
Lightweight concrete 0.75
Masonry wall

Masonry wall
In accordance
COLUMN
Lightweight, Lightweight concrete, where 𝑓𝑐𝑡 is
with Section
𝜆 specified
419.2.4.3
Normal-weight concrete 1.0
Epoxy-coated or zinc and epoxy
dual-coated reinforcement with
1.5
clear cover less than 3db or clear
spacing less than 6db L1 L2
Epoxy[1]
Epoxy-coated or zinc and epoxy
𝜓𝑒
dual-coated reinforcement for all 1.2
other conditions NSCP COEEFICIENTS FOR CONTINUOUS BEAMS AND SLABS
Uncoated or zinc-coated
1.0 Section 408.4 of NSCP states that in lieu of frame analysis, the following
(galvanized) reinforcement
25 mm  and larger bars 1.0 approximate moment and shears are permitted for design of
Size 𝜓𝑠 20 mm , and smaller bars and continuous beams and one-way slabs (slabs reinforced to resist
0.80
deformed wires flexural stresses in only one direction), provided:
More than 300 mm of fresh (a) There are two or more spans,
Casting concrete placed below horizontal 1.3
Position[1] reinforcement (b) Spans are approximately equal, with the larger of two adjacent
𝜓𝑡 spans not greater than the shorter by more than 20 percent,
Other 1.0
(c) Loads are uniformly distributed,
Development of Deformed Bars and Deformed Wires in (d) Unit live does not exceed three times unit dead load, and
Compression (e) Members are prismatic,
Development length, 𝑙𝑑𝑐 shall be the greater of (a) and (b), Positive moment
multiplied by the modification factors of Section 425.4.9.3: End spans
0.24𝑓𝑦 𝜓𝑟 Discontinuous end unrestrained………................𝑤𝑢 𝑙𝑛2 /11
a. ( ) 𝑑𝑏 Discontinuous end integral with support.,,…...𝑤𝑢 𝑙𝑛2 /14
𝜆√𝑓𝑐 ′
Interior spans……………………………...………...…............𝑤𝑢 𝑙𝑛2 /16
b. 0.043𝑓𝑦 𝜓𝑟 𝑑𝑏
Negative moment of exterior face of first interior support
Table 425.4.9.3 Modification factors for Deformed Bars and
Two spans ……………………………………………..…,………𝑤𝑢 𝑙𝑛2 /9
Wires in Compression
Modification More than two spans………………………….………………𝑤𝑢 𝑙𝑛2 /10
Condition Value of factor
factor
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PSAD - Nov. 2019
Clear concrete cover to stirrups = 40 mm
Negative moment at other faces of interior support…..𝑤𝑢 𝑙𝑛2 /11 Concrete strength, fc’ = 27.5 MPa
Maximum size of coarse aggregates = 20 mm
Negative moment at interior face of exterior Steel strength:
support for members built integrally with supports Longitudinal bars, fy = 415 MPa
Where support is a spandrel beam………….……..…..𝑤𝑢 𝑙𝑛2 /24 Lateral ties, fyh = 275 MPa
When support is column…………………………….……...𝑤𝑢 𝑙𝑛2 /16 Reduction factor for shear = 0.75
61. Determine the minimum beam width bw (mm) that will
Shear in end members at face of satisfy the code requirement for bar spacing and clear
first interior support……………………………………………1.15𝑤𝑢 𝑙𝑛 /2 cover.
A. 260 C. 280
Shear at face of all other supports……………………………..𝑤𝑢 𝑙𝑛 /2 B. 270 D. 300
62. Given:
Where 𝑙𝑛 = clear span for positive moment or shear and average of Beam width, bw = 350 mm
adjacent clear spans for negative moment. Spacing of stirrups = 100 mm
Calculate the design shear strength (kN) of the section.
A. 254 C. 230
Situation 19 – Given the following data of a rectangular beam: B. 306 D. 210
Dimensions, b × h = 350 mm × 600 mm 63. Given:
Tension bars = 4-25 mm Spacing of stirrups = 60 mm
Modular ratio, n = 8 Required shear strength, Vu = 310 kN
Concrete strength, fc’ = 34.5 MPa (normal weight) Calculate the required minimum beam width bw in mm.
Concrete rupture strength, fr = 0.62λ√fc ′ A. 300 C. 340
Concrete cover to centroid of tension bars = 65 mm B. 320 D. 360
Beam simple span, L = 6 m
58. Compute the cracking moment (kN-m) of the section. t
A. 62.4 C. 50.7
B. 76.5 D. 42.6
59. Compute the moment of inertia (m4) of the cracked section
transformed to concrete.
A. 0.0266 C. 0.00630 h
B. 0.00266 D. 0.0630 a
60. If Ma = Mcr, compute the midspan deflection in mm.
A. 12.25 C. 3.54
B. 8.47 D. 1.65
bw
According to Section 409.6.2.3 of the NSCP, unless stiffness
values are obtained by a more comprehensive analysis, Situation 21 – Given the following data of a rectangular tied
immediate deflection shall be computed with the modulus column:
of elasticity Ec for concrete as specified in Section 408.6.1 Kn = 0.60; Rn = 0.28
(normal-weight or lightweight concrete) and with the ρ = 0.04; fc’ = 20.7 MPa; fy = 414 MPa
effective moment of inertia as follows, but not greater than Reduction factor, ϕ = 0.65
Ig
𝑀𝑐𝑟 3 𝑀𝑐𝑟 3
𝐼𝑒 = ( ) 𝐼𝑔 + [1 − ( ) ] 𝐼𝑐𝑟
𝑀𝑎 𝑀𝑎
Where:
𝑓𝑟 𝐼𝑔
𝑀𝑐𝑟 =
𝑦𝑡
and for normal-weight concrete, 𝑓𝑟 = 0.62𝜆√𝑓𝑐 ′

𝐼𝑔 = moment of inertia of the gross concrete section

about centroidal axis
𝑦𝑡 = distance from centroidal axis of gross section,
neglecting reinforcement, to extreme fiber in tension.
𝐼𝑐𝑟 = moment of inertia of the cracked section
transformed to concrete, mm4.
𝑀𝑎 = maximum moment in member at stage deflection is
computed.
𝐸𝑐 = 4700√𝑓𝑐 ′ for normal weight concrete

Situation 20 – Refer to the figure shown:

Given:
t = 100 mm; a = 50 mm; h = 400 mm
Tension bars = 8-25 mm 
Compression bars = 4-25 mm 
Stirrups = 10 mm vertical U stirrups

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64. If the column dimension is 600 mm × 600 mm, determine 73. Which of the following comprises the bulk volume of
the design axial strength (kN) of the column. concrete?
A. 3260 C. 2900 A. aggregates C. cement
B. 2640 D. 4470 B. admixture D. water
65. If the column dimension 600 mm × 600 mm, determine the 74. Which of the following tests determines the consistency of
design moment (kN-m) of the column. concrete?
A. 1250 C. 1080 A. trial batch C. cylinder test
B. 920 D. 810 B. arbitrary proportion D. slump test
66. Using a steel ratio ρg = 0.03 and Kn = 0.60, what is the 75. The most important factor in concrete mix
required column depth “h” in mm with the same gross area. A. fineness modulus C. aggregate gradation
A. 730 C. 650 B. water-cement ratio D. none of these
B. 700 D. 780

Situation 22 – Refer to the combined footing shown.

Given:
a = 0.4 m; b = 3.5 m; c = 0.4 m; W = 4 m
Effective depth, d= 600 mm

fc’ = 27.5 MPa; fy = 413 MPa

Net soil pressure at factored loads, qu = 98 kPa

Shear due to factored loads:
e = 156.8 kN; f =610.4 kN; g = 761.6 kN; h = 509.6 kN

Reduction factors:
Shear = 0.75
Moment = 0.90

67. How much is the punching shear stress (MPa) at column B?

A. 0.85 C. 0.64
B. 0.74 D. 0.55
68. How much is the critical wide beam shear stress (MPa)?
A. 0.45 C. 0.29
B. 0.36 D. 0.22
69. Compute the number of 25-mm bars within the length “d”
from the column.
A. 12 C. 22
B. 15 D. 17

Situation 23 – The basic proportioning trial batches for normal

weight concrete with an average compressive strength of
35 MPa at 28 days are as follows:
slump = 75 mm to 100 mm
Water-cement ratio = 0.62
Specific gravity of cement = 3.15
Specific gravity of coarse aggregates = 2.68
Specific gravity of fine aggregates = 2.64
Water (net mixing) = 180 kg/m³
Coarse aggregate = 10.65 kN/m³
Entrapped air = 1%
Unit weight of concrete = 23.4 kN/m3
70. What is the required volume of cement (m3) per cubic
meter of concrete?
A. 0.125 C. 0.065
B. 0.073 D. 0.092
71. Compute the volume of coarse aggregate (m3) per cubic
meter of concrete.
A. 0.587 C. 0.632
B. 0.487 D. 0.405
72. If the combined solid volume of cement, water, coarse
aggregate and entrapped air is 0.58 m3, what is the weight
(kN) of dry sand is required?
A. 10.88 C. 9.12
B. 12.54 D. 8.54

Situation 24 – Answer the following questions.

8
Prepared by:
Engr. Jobert S. De La Cruz