MATHEMATICS, SURVEYING & TRANSPORTATION ENGINIRING
(NOVEMBER 2011)
Situation 1 – The probability of event A happening is 3/5 and the probability of
event B happening is 2/3
1. What is the probability of both A and B happening?
A. 3/5
B. 1/5
C. 2/5
D. 4/5
2. What is the probability of only event A happening i.e. event A happening and
event B not happening?
A. 4/5
B. 1/5
C. 3/5
D. 2/5
3. What is the probability of either A, or B, or A and B happening?
A. 11/15
B. 14/15
C. 3/5
D. 13/15
E.
Situation 2 – Answer the following problems:
4. Six congruent circles are arranged in a circle way that each circle is tangent
to at least two other circles. If the radius of each circle is 2 cm, find the
perimeter of the polygon formed by connecting the centers of each circles.
A. 12 cm
B. 24 cm
C. 30 cm
D. 32 cm
5. Which of the following is/are correct?
I.
sin(-A) = -sin(A)
II.
cos(-A) = -cos(A)
III.
tan(-A) = -tan(A)
A. I only
B. II only
C. I & III only
D. I & II only
6. A solid rectangular block has a volume of 30 cm3. If all side measure are
integers, which of the following is the least possible surface area?
A. 92
B. 82
C. 86
D. 62
E.
Situation 3 – Answer the following problems:
7. What is the distance between the intercepts of the line x + 2y – 6 = 0?
A. 6.231
B. 6.708
C. 8.639
D. 5.449
8. If |x3 – 8| ≤ 5, find the range of values of x.
A. ∛3 ≤ x ≤ ∛13
B. ∛3 ≥ x ≤ ∛13
C. ∛3 ≥ x ≥ ∛13
D. ∛3 ≤ x ≥ ∛13
E.
F.
G. STRUCTURAL ENGINIRING & CONSTRUCTION
(NOVEMBER 2011)
H. Situation 1 – A load of W =
30 kN is lifted through a
boom BCD as shown in the
figure. The boom makes an
angle of 60° with the
vertical. Neglect the weight
of the boom and for this
problem, L1 = L2 = 2m. The
pulley at D is frictionless.
I.
1. Determine the angle .
A. 40°
B. 35°
C. 45°
D. 30°
2. What is the tension in cable AC in kN?
A. 51.96
B. 25.36
C. 34.89
D. 43.21
3. What is the total reaction at B in kN?
A. 54.77
B. 43.21
C. 17.32
D. 51.96
E.
Situation 2 – The strut shown in the figure carries an axial load of P =
148 kN.
F.
4.
4.
4.
4.
4.
4.
Determine the bearing stress between the
pin and the strut:
A. 463 MPa
B. 345 MPa
C. 285 MPa
D. 563 MPa
5. Determine the shearing stress in the pin.
A. 286 MPa
B. 368 MPa
C. 321 MPa
D. 341 MPa
6. Determine the shearing stress in the bolts
A. 159.4 MPa
B. 196.4 MPa
C. 123.9 MPa
D. 167.3 MPa
E.
Situation 3 – The column shown in
the figure is loaded with a vertical
load P = 3 kN and a lateral load H =
0.45 kN. The column is 3 m high
and is made of steel with 300 mm
outer diameter, 6 mm thick and
weighs 150 N/m.
F.
7. What is the maximum stress at the base
due to the load P?
A.
B.
C.
D.
1.78
1.37
2.54
0.87
MPa
MPa
MPa
MPa
8. What is the maximum stress at the base
due to the lateral load?
A. 4.76 MPa
B. 5.28 MPa
C. 3.46 MPa
D. 2.89 MPa
9. If the column is a solid timber with a diameter of 250 mm, what is the
A. 0.089
MPa stress at the base? C. 0.045 MPa
maximum
shearing
B. 0.164 MPa
D. 0.012 MPa
E.
Situation 4 – The frame shown in the figure is acted upon by wind
load pressure of 1.44 kPa. These frames are spaced 6 m apart normal
to paper. Consider the roller support at B and the joint at D as pin.
10.
Determine the horizontal component of the reaction at A.
A. 35.7 kN
B. 26.5 kN
11.
C. 18.3 kN
D. 12.7 kN
Determine the vertical component of the reaction at A.
A. 23.9 kN
B. 20.2 kN
12.
C. 18.5 kN
D. 16.3 kN
Determine the horizontal component of the reaction at B.
A. 26.5 kN
B. 18.3 kN
C. 12.7 kN
D. 35.7 kN
E.
F. Situation 5 – The
sheet pile shown in the
figure is provided with
tension rods spaced 3
meters
apart.
The
wooden stringers has d
= 300 mm and can be
considered
simply
supported
at
each
connection
to
the
tension rod. Allowable
bending and shearing
stresses of the stringer
are 14.7 MPa and 1.48
MPa, respectively.
G.
H.
13.
What is the design moment of the stringer?
A. 54.8 kN-m
B. 74.4 kN-m
14.
C. 42.4 kN-m
D. 63.9 kN-m
What is the value of stringer width “b” based on the bending?
A. 192 mm
B. 249 mm
15.
C. 290 mm
D. 338 mm
What is the value of stringer width “b” based on shear?
A. 321 mm
B. 235 mm
C. 288 mm
D. 254 mm
E.
Situation 6 – The 6 m long prestressed cantilever beam shown in the
figure carries a concentrated live load of 18 kN at the free end and a
uniform dead load due to its own weight. Unit weight of concrete is 20
kN/m3. The strands are 12 mm in diameter with total prestressing
force of 540 kN applied at an eccentricity “e” above the neutral axis of
the cross-section.
16.
What is the maximum stress (MPa) in the bottom fiber of the beam at
the free end when the eccentricity e = 0?
A. -7.86
B. -13.45
17.
C. -2.25
D. -10.35
What is the stress in the top fiber of the beam at the fixed end when
the eccentricity e = 100 mm?
A. +5.4 MPa
B. +6.3 MPa
18.
C. +8.1 MPa
D. +3.6 MPa
What is the required eccentricity e such that the stress in the top fiber
of the beam at the fixed end is zero?
A. 230 mm
B. 160 mm
C. 200 mm
D. 260 mm
E.
F. Situation 7 – Reinforced concrete beams having widths of 400 mm
and overall depths of 600 mm are spaced 3 meters on the centers as
shown in the figure. These beams support a 100 mm thick slab. The
superimposed loads on these beams are as follows:
G.
Dead load (incl. floor finish, ceiling, etc.)……………………….3.2 kPa
Live load ………………………………………………………………….……….3.6 kPa
H. The columns E and H are omitted such that the girder BEHK supports
the beams DEF at E and GHI at H. Assume EI = constant for all
beams. Unit weight of concrete is 24 kN/m 3.
19.
Determine the factored uniform load on beam GHI, in kN/m.
A. 47.71
B. 56.98
20.
Determine the maximum factored shear (in kN) in beam GHI
assuming that G and I are fixed and H is hinge.
A. 143.2
B. 178.9
21.
C. 67.21
D. 41.23
C. 121.1
D. 98.4
Determine the maximum factored positive moment (in kN-m) in beam
GH assuming that G and I are fixed and H is hinge.
A. 213
B. 187
C. 154
D. 112
E.
F. Situation 8 – Channel sections are used as purlin. The top chords of
the truss are sloped 4H to 1V. The trusses are spaced 6 m on center
and the purlins are spaced 1.2 m on centers.
G.
H.
22.
Loads:
Dead load = 720 Pa
Live load = 1000 Pa
Wind load = 1400 Pa
Wind Coefficients:
Windward = + 0.2
Leeward = - 0.6
Properties of C200 x 76
Sx = 6.19 x 104 mm3
Sy = 1.38 x 104 mm3
Weight, w = 79 N/m
Allowable bending stress, Fx = Fy = 207 MPa
Determine the computed bending stress, fbx, due to the combination
of dead and live loads only.
A. 196 MPa
B. 176 MPa
23.
C. 123 MPa
D. 151 MPa
Determine the computed bending stress, fby, due to the combination
of dead and live loads only
A. 169 MPa
B. 123 MPa
C. 143 MPa
D. 103 MPa
24.
Determine the value of the interaction equation using the load
A.
0.87
C. 1.25 side.
combination of 0.75 (D + L +W) at the windward
B. 1.59
D. 1.87
E.
Situation 9 – The column shown in the figure is subjected to shear
force parallel to the 600 mm side. Allowable concrete shear stress for
shear parallel to the 600 mm side is 0.816 MPa. Concrete strength f’ c
= 21 MPa and steel strength for both longitudinal and confining
reinforcements is 415 MPa. The ties are all 12 mm in diameter with
clear cover of 40mm.
25.
Determine the factored shear force Vu that the column can resist if the
nominal shear strength provided by the ties is 375 kN.
A. 378
B. 426
26.
C. 467
D. 532
If the ties are spaced at 225 mm on centers, what is the maximum
value of Vu in kN?
A. 472
B. 421
27.
C. 335
D. 389
If the factored shear force parallel to the 600 mm side is 400 kN,
determine the required spacing of transverse reinforcement in accordance
with the provision for seismic design.
A. 126 mm
B. 164 mm
C. 241 mm
D. 100 mm
E.
F. 5.21.4 Special Provision for Seismic Design
G. 5.21.4.4 Transverse Reinforcement
5.21.4.4.1 Transverse reinforcement as specified below shall be
provided unless a large amount is required by Sec. 5.21.7
H.
(1) The volumetric ratio of spiral or circular hoop reinforcement, ρ s,
shall not be less than that indicated by:
I. ρs = 0.12f’c / fyh
J.
21-2
and shall not be less than
Ag
f 'c
0.45
−1
K. ρs =
Ac
fy
(
)
10-5
L.
(2) The total cross sectional area of rectangular hoop reinforcement
shall not be less than that given by:
M.
N.
0.3
Ash =
Ash =
s hc f ' c Ag
−1
f yh
Ac
0.9
(
s hc f ' c
f yh
)
21-3
21-4
O.
(3) Transverse reinforcement shall be provided by either single or
overlapping hoops. Crossties of the same bar size and spacing as the hoop
may be used. Each end of the crosstie shall engage a peripheral longitudinal
reinforcing bar. Consecutive crossties shall be alternated end for end along
the longitudinal reinforcement.
P.
(4) If the design of the member core satisfies the requirement of the
specified loading combinations including earthquake effect, Eq. (21-3) and
(10-5) need not to be satisfied.
Q.
5.21.4.4.2 Transverse reinforcement shall be spaced at distance not
exceeding (a) one-quarter of the minimum member dimension, (b) six
times the diameter of longitudinal reinforcement, and (c) as defined by
Eq. 21-5
R. sx =
100+
350−h x
3
21-5
S. The value of sx shall not exceed 150 mm and need not be taken less
than 100 mm.
T. Where:
U.
V.
Ach = cross-sectional area of a structural member measured out-to-out
of transverse reinforcement, mm2
Ash = total cross-sectional area of transverse reinforcement (including
crossties) within spacing s and perpendicular to dimension h c
W.
fyh = specified yield strength of transverse reinforcement, MPa
X.
hc = cross-sectional dimension of column core measured center-tocenter of outer legs of the transverse reinforcement comprising area A sh,
mm
Y.
hx = maximum horizontal spacing of hoop of crosstie legs on all faces of
column, mm
Z.
s = spacing of transverse reinforcement measured along the longitudinal
axis of the structural member, mm
AA.
Situation 10 – The girder AB shown in the figure is subjected to
torsional moment from the loads on the cantilever frame. The following
factored forces are computed from this beam:
Factored moment, Mu = 440 kN-m
Factored shear, Vu = 280 kN
Factored torque, Tu = 180 kN-m
AB.
The girder has a width of 400 mm and an overall depth of 500
mm. Concrete cover is 40 mm. The centroid of longitudinal bars of the
girder are placed 65 mm from the extreme concrete fibers. Concrete
strength f’c = 20.7 MPa and steel yield strength for longitudinal bars is
fy = 415 MPa. Use 12 mm U-stirrups with fyt = 275 MPa. Allowable
shear stress in concrete is 0.76 MPa.
28.
Determine the required area of tension reinforcement of the girder, in
mm .
2
A. 4,154
B. 2,732
29.
Determine the spacing of transverse reinforcement due to V u.
A. 137 mm
B. 167 mm
30.
C. 3,873
D. 3,313
C. 98 mm
D. 185 mm
Determine the additional area of longitudinal reinforcement to resist
torsion, in mm2.
A. 3,850
B. 3,420
C. 2,850
D. 4,120
E.
F.
G. Code:
1. Threshold torsion: For Nonprestressed members, it shall be permitted to
neglect torsion effects if the factored torsional moment Tu is less than:
H.
I.
Tu <
Acp 2
1
'
∅ f
12 √ c Pcp
( )
2. Torsional moment strength: The adequacy of solid sections under combined
shear and torsion shall be such that:
J.
√
V u 2 T u Ph 2
(
) +(
)
bw d
1.7 Aoh
≤
∅
(
Vc 2 '
+ √f c
bw d 3
)
3. Where Tu exceeds the threshold torsion, design of cross-section shall be based
on:
K.
L.
∅ Tn ≥ Tu
M.
Tu =
2 Ao At f yt
s
cot θ
N.
O.
Where Ao shall be determined by analysis except that is shall be
permitted to take Ao equal to 0.85Aoh; θ shall not be taken smaller than 30
degrees nor larger than 60 degrees. It shall be permitted to take θ equal to:
P.
(a) 45 degrees for nonprestressed members or members with less prestress than
in (b); or
Q.
(b) 37.5 degrees for prestressed members with an effective prestress force not
less than 40 percent of the tensile strength of the longitudinal reinforcement.
R.
4. The additional area of longitudinal reinforcement to resist torsion, A l, shall not
be less than:
S.
Al =
At
f
ph yt
s
fy
( )
cot2 θ
5. Minimum torsion reinforcement: Where torsional reinforcement is required,
the minimum area of transverse closed stirrups shall be computed by:
T.
U.
Av + 2At = 0.062
√ f 'c
bw s
f yt
V.
but shall not be less than (0.35bws)/fyt
6. Where torsional reinforcement is required, the minimum total area of
longitudinal torsional reinforcement, Al min, shall be computed by:
W.
Al min =
5 √ f ' c Acp A t
f
− ph yt
12 f y
s
fy
( )
7. Spacing of torsion reinforcement: The spacing or transverse torsion
reinforcement shall not exceed the smaller of p h /8 or 300 mm.
X.
The longitudinal reinforcement required for torsion shall be distributed
around the perimeter of the closed stirrups with a maximum spacing of 300
mm. The longitudinal bars or tendons shall be inside the stirrups. There shall
be at least one longitudinal bar or tendon in each corner of the stirrups.
Longitudinal bars shall have a diameter at least 0.042 times the stirrup
spacing, but not less than a No. 10.
Y. Where:
Z.
AA.
Acp
Al
AB.
AC.
torsional
AD.
AE.
At
mm
AF.
AG.
AH.
2
AI.
-area enclosed by outside perimeter of concrete cross section, mm 2
-total area of longitudinal reinforcement to resist torsion, mm 2
Ao
-gross area enclosed by shear flow path, mm2
Aoh
-area enclosed by centerline of the outermost closed transverse
reinforcement, mm2
-area of one leg of a closed stirrup resisting torsion within spacing s,
fyt
-specified yield strength fy of transverse reinforcement, MPa
Pcp
- outside perimeter of concrete cross section, mm
ph
-perimeter of centerline of outermost closed transverse torsional
reinforcement, mm
AJ.
AK.
AL.
AM.
AN.
*** END ***
AO.
Answer Key:
AP.1
D
AV.7
B
BA.
12 C
AQ.
2B
AW.
8C
AR.
3A
AX.
9D
BB.
13
D
AS.
4A
AY.1
0
A
AT. 5
B
AU.
6A
BT.
BU.
BV.
BW.
BX.
BY.
BZ.
CA.
CB.
CC.
CD.
CE.
CF.
CG.
CH.
CI.
CJ.
AZ.
11 B
BC.
14 C
BD.
15 C
BE.
16 C
BF.1
7
D
BJ. 2
1
D
BG.
18
D
BK.
22
D
BH.
19 A
BL.2
3
A
BI. 2
0
B
BO.
26 C
BP.2
7
D
BQ.
28
D
BM.
24 C
BR.
29 A
BN.
25 C
BS.
30 B
CK.
CL.
Solutions
CM.
Situation 1
CN.
CO.
CP.
CQ.
CR.
CS.
CT.
CU.
CV.
CW.
Since pulley is frictionless, the tensions at slack and tight sides are
CX.
T = W = 30 kN
CY.
By inspection,  = 30°
CZ.
a = 2 sec 30° = 2.309 m
DA.
Considering the FBD of the boom:
∑MB = 0
Tc sin60° x a + T x d = T x 4
Tc = 25.359 kN
equal.
d = 4 tan 30° = 2.309 m
DB.
∑FH = 0
BH = Tc cos30° + T
BH = 51.962 kN
DC.
∑FV = 0
BV = T - Tc sin30°
BV = 17.321 kN
DD.
RB =
√B
2
H
+ BV 2
RB =
√(51.962)2+(17.321)2
RB = 54.772 kN
DE.
DF.
DG.
DH.
Situation 2
DI.
P = 148 kN
DJ.
Part 1: Bearing stress between the pin and strut:
Bearing area; Ap = 2 x (Dpin)(tstrut)
Ap = 2 x (16)(10)
Ap = 320 mm2
DK.
P
Ap
fp =
148,000
fp =
320
fp = 462.5 MPa
DL.
Part 2: Shearing stress in pin: (double shear)
Shearing area, AV = 2 x
Shear force,
DM.
π
4
PV = P = 148 kN
PV
AV
fV =
(16)2 = 804.248 mm2
fp =
148,000
402.124
fp = 368.05 MPa
DN.
Part 3: Shearing stress in bolts:
Shearing area, AV = 2 x
Shear force,
DO.
fV =
π
4
(16)2 = 804.248 mm2
PV = P cos 30°
Pv = 148 cos 30°
Pv =128.172 kN
PV
AV
fp =
128,172
804.248
fp =159.4 MPa
DP.
DQ.
DR.
Situation 3
Parts I and II:
Outer diameter, Do = 300 mm
Inner diameter, Di = 288 mm
Area, A =
π
2
2
2
4 (300 – 288 ) = 5,541.77 mm
DS.
Moment of inertia, I =
π
4
4
6
64 (300 – 288 ) = 59.901 x 10
mm4
Moment due to P, Mp = P x e = 3 x 0.1 = 0.3 kN-m
Moment at base due to H, MH = H x L = 0.45 x 3 = 1.35 kN-m
Weight of column, W = w x L = 150 x 3 = 450 N
DT.
Stress due to column weight:
DU.
f1 =
−W
A
f1 =
−450
5541.77
f1 = - 0.081 MPa
DV.
Stress due to P alone:
DW.
f2 =
−P M p c
−
A
I
f2 =
6
−3000 0.3 x 10 (150)
−
5541.77
59.901 x 106
f2 = - 1.293 MPa
DX.
Stress due to lateral load H alone:
DY.f2 =
0.3 x 106 (150)
f2 = 59.901 x 106
−M p c
I
f2 = - 3.381 MPa
DZ.
Part 1:
Maximum base stress due to P:
fmax = -0.081 – 1.293 = -1.374 MPa
EA.
Part 2:
Maximum base stress due to lateral load:
fmax = -0.081 – 3.381 = - 3.462 MPa
EB.
Part 3:
Shear, V = H = 450 N
Diameter, D = 250 mm, r = 125 mm
Shear stress, fV =
4V
3 π r2
fV =
4 (450)
3 π (125)2
fV = 0.012 MPa
EC.
Situation 4
ED.
EE.
Bay, s = 6 m
w=cxpxs
EF.
w1 = 0.08(1.44)(6)
w1 = 6.912 kN/m
EG.
w3 =
0.5(1.44)(6)
w3 = 4.32 kN/m
EH.
θ = arctan (2/6)
θ = 18.435°
EI.
F1 = w1 x 4 =27.648 kN
EJ.
F2 = w2 x 6.325 =
5.464 kN
F2x = F2 sin θ
= 1.728 kN
F2y = F2 cos
θ = 5.184 kN
EK.
F4 = w4 x 4 = 13.824 kN
EL.
∑MA = 0
w2 = 0.1(1.44)(6)
w2 = 0.864 kN/m
w4 = 0.4(1.44)(6)
w4 = 3.456 kN/m
F3 = w3 x 6.325 =
27.322 kN
F3x = F3 sin θ
= 8.64 kN
F3y = F3 cos
θ = 25.92 kN
F1(2) + F4(2) + F3x(5) = BV(12) + F2x(5) + F2y(3)+
F3y(9)
27.648(2) + 13.824(2) + 8.64(5) = 12B V + 1.728(5)
+ 5.184(3)+ 25.92(9)
BV = - 10.944 kN (downward)
EM.
∑FV = 0
AV + BV + F2y + F3y = 0
AV = - 20.16 kN (downward)
EN.
∑MD right = 0
(See figure below)
F3(3.162) + F4(4) + BH(6) + BV(6) = 0
27.322(3.162) + 13.824(4) + B H(6) + (-10.944)(6) = 0
BH = - 12.672 kN (to the left)
EO.
∑FH = 0
(entire frame)
AH + BH + F1 + F4 – F2x + F3x = 0
AH + (-12.672) + 27.648 +13.824 – 1.728 + 8.64 =0
AH = -35.712 kN (to the left)
EP.
EQ.
Situation 5
ER.
ER.
ES.
ET.
F1 =
1
K γ
2 a soil
1
H2 x L
F1 = 2
(1/3)(17.3)(5.4)2(3)
F1 = 252.234 kN
y1 = (2.1 + 3.3)/3 = 1.8 m
EU.
F2 =
1
K γ
2 a water
H2 x L
1
F2 = 2
(9.8)(2.7)2(3)
F2 = 107.136 kN
y2 = (2.7)/3 = 0.9 m
EV.
EW.
EX.
EY.
EZ.
∑Mpin = 0
T(2.1) + F2y2 = F1y1
T = 107.274 kN
From the beam diagram shown:
R = T/2 = 85.137 kN
R = wL/2
w = 2R/L
w = 2(85.137)/3
w = 56.758 kN/M
Mmax =
w L2
8
Mmax =
56.758(3)2
8
Mmax = 63.853 kN-m
Vmax = R = 85.137 kN
FA.
fb =
6M
b d2
6 (63.853 x 106 )
b(300)2
≤ Fb
= 14.7
b = 289.6 m
FB.
fV =
3V
2 bd
≤ FV
3 (85,137)
2b (300)
= 1.48
b = 287.6 m
FC.
Situation 6
FD.
wD =
γ
c
x bh
wD = 20 x (0.4)(0.6)
wD = 4.8 kN/m
FE.
Moment at fixed end;
FF.
M = 18(6) + 4.8(6)(3)
M = 194.4 kN-m
Axial stress due to prestressing force,
fpa =
−Ps
bh
−540,000
fpa = 400(600)
FG.
FH.
Fpa = - 2.25 MPa
Stress due to eccentric position of Ps; fpe = ±
FI.
fpe = ±0.0225e
6M
Stress due to moment, fM = ± b h2
FJ.
FK.
6 Ps e
b h2
(-) for top fiber
(+) for top fiber
Part 1: Stress in bottom fiber at the free end of the beam
when e = 0
FL.
Since M = 0 at the free end, fbot = -2.25 MPa
(uniform in the entire cross
section)
FM.
mm:
Part 2: Stress in the top fiber at fixed end when e = 100
M = 194.4 kN-m
FN.
6M
2
bh
ftop = -2.25 - 0.0225e +
FO.
ftop = -2.25 – 0.0225(100) +
FP.
6 (194.4 x 10 6)
400(600)2
ftop = + 3.6 MPa
FQ.
Part 3: Value of “e” such that the stress in the top fiber
at fixed end is zero:
FR.
ftop = -2.25 – 0.0225e +
6M
b h2
FS.
0 = -2.25 – 0.0225e +
FT.
e = 260 mm
FU.
FV.
6 (194.4 x 10 6)
400(600)2
Situation 7
Unit weight of concrete,
γ
c
= 24 kN/m3
Dead load pressure = 3.2 kPa
Live load Pressure = 3.6 kPa
FW.
Weight of beam:
wb =
γ
c
Ac
wb = 24(0.4)(0.6)
wb = 5.76 kN/m
FX.
Weight of slab:
ps =
γ
c
t
ps = 24(0.1)
ps = 2.4 kPa
FY.
Factored floor pressure:
pu = 1.4(3.2+2.4) + 1.7(3.6)
pu = 13.96 kPa
FZ.
Equivalent load on beam due to factored pressure:
GA.
wu1 =
[ ( )]
( 13.96 )( 3 )
3
3−
6
7.5
[ ( )]
pu s
s
3−
6
L
2
x2
wu1 =
2
x2
wu1 = 39.646 kN/m
GB.
Total factored uniform load (including beam weight)
wu = 1.4(5.76) + 39.646
wu = 47.71 kN/m
→ Part 1
GC.
GD.
Moment at G, MG =
−w u L2
12
MG =
2
−47.71(7.5)
12
MG = - 223.643 kN-m
GE.
Reaction at G, RG =
1
2
wu L
RG =
1
2
(47.71)(7.5)
RG = 178.91 kN
GF.
Maximum factored shear in beam GHI, Vmax = RG =
178.91 kN
GG.
Maximum positive moment in beam GH:
GH.
M=
w u L2
24
GI.
GJ.
M=
47.71(7.5)2
24
M = 111.82 kN-m
Situation 8
GK.
Dead load pressure = 720 Pa
Live load pressure = 1000 Pa
Wind =
1400 Pa
Beam
weight = 79 N/m
Fbx = 207
MPa
Fby = 207
MPa
θ = arctan
(1/4)
θ = 14.036°
GL.
GM.
Wind coefficient:
Windward coefficient = 0.2
Leeward coefficient = -0.6
GN.
Dead load;
wD = 720(1.2) +79
wD = 943 N/m
GO.
Live load;
wL = 1000(1.2)
wL = 1200 N/m
GP.
Wing:
www = 1400(1.2)(0.2)
www = 336 N/m
GQ.
GR.
wlw = 1400(1.2)(-0.6)
wlw = -1008 N/m
Part 1: Due to dead and live load only
GS.
wN = (wD + wL) cos θ
wN = (943 +1200) cos 14.036°
wN = 2079.015 N/m
GT.
wT = (wD + wL) sin θ
wT = (943 +1200) sin 14.036°
wT = 519.754 N/m
GU.
w N L2
8
Mx =
Mx =
2079.015 (6)2
8
Mx = 9.356 kN-m
GV.
fbx =
Mx
Sx
fbx =
9.356 x 106
6.19 x 10 4
fbx = 151.14.MPa
GW.
My =
w T L2
8
519.754 (6)2
8
My =
My = 2.339 kN-m
GX.
fby =
My
Sy
6
fby =
2.339 x 10
4
1.38 x 10
fby = 169.485 MPa
GY.
Part 2: Dead + Live + Wind on windward side
wN2 = 0.75(wN + www)
wN = 0.75(2079.015 +
336)
wN = 1811.262 N/m
GZ.
HA.
wT2 = 0.75(wT)
Mx =
wT = 0.75(519.754)
wT = 389.815 N/m
w N 2 L2
8
2
Mx =
1811.262 (6)
8
Mx = 8.151 kN-m
HB.
fbx =
Mx
Sx
fbx =
8.151 x 10 6
4
6.19 x 10
fbx = 131.675 MPa
HC.
My =
w T 2 L2
8
2
My =
389.815 (6)
8
My = 1.754 kN-m
HD.
fby =
My
Sy
fby =
1.754 x 106
1.38 x 10 4
fby = 127.114 MPa
HE.
HF.
HG.
f ❑bx f by 131.675 127.114
+
=
+
Fbx F by
207
207
= 1.25
Situation 9
bw = 400 mm
fy = 415 MPa
h = 600 mm
Allowable shear stress of concrete, Fvc =
0.816 MPa
f’c = 21 MPa
HH.
Reduction factor, � = 0.085
HI.
HJ.
Effective depth, d = 600 – 40 – 12 – 0.5(0.25)
Effective depth, d = 535.5 mm
HK.
Shear strength provided by concrete, Vc = Fvc bw d
Shear strength provided by concrete, Vc = 0.0816(400)(535.5)
Shear strength provided by concrete, Vc = 174.787 kN
HL.
Part 1:
Vs = 375 kN
HM.
Vn = Vc + Vs
Vn = 174.787 + 375
Vn = 549.787
HN.
Vu = � Vn
Vu = 0.85(549.787)
Vu = 467.319 kN
HO.
HP.
Part 2:
s = 225 mm
Av = 3 x
HQ.
π
4
Vs =
(12)2 = 339.29 mm2
Av f y d
s
Vs =
339.29(415)(535.5)
225
Vs = 335.12 kN
HR.
Vn = Vc + Vs
Vs = 174.787 + 335.12
Vs = 509.906 kN
HS.
Vu = � Vn
Vu = 0.85(509.906)
Vu = 433.42 kN
HT.
Part 3:
Vu = 400 kN
HU.
Vs = Vu – Vc
HV.
Av f y d
Vs
s=
Vs = 400 – 174.787
Vs = 225.213 kN
s=
339.29(415)(535.5)
225.213
HW.
s = 334.8 mm
HX.
Requirements for Seismic Design:
HY.
Ach = (600 – 2 x 40)(400 - 2 x 40) = 166,400 mm 2
Ag = 600 x 400 = 240,000 mm2
Ash = 3 x
π
4
(12)2 = 339.29 mm2
hc = 400 – 2(40) – 12 = 308 mm
HZ.
Ash =
0.3
0.3
s hc f ' c Ag
−1
f yh
Ac
s( 308)(21) 240,000
−1
415
166,400
(
(
)
339.29 =
)
s = 164 mm
IA.
Ash =
0.9
0.9
s hc f ' c
f yh
339.29 =
s( 308)(21)
415
s = 242 mm
IB.
Minimum requirement according to Section 5.21.4.4.2:
a) b/4 = 100 mm
b) 6(25) = 150 mm
c) 100 +
350−h x
3
hx = ½ (600 – 2 x 40) – ½(12) + ½(25) +½(12)
hx = 272.5 mm
100 +
IC.
350−h x
3
= 126 mm
Therefore, uses s = 100 mm
ID.
Situation 10
IE.
Mu = 440 kN-m
Vu = 280 kN
Tu = 180 kN-m
b = 400 mm
h = 500 mm
Allowable shear stress in
IF.
Effective depth, d = 500 – 65 = 435 mm
IG.
Part 1:
Mu = 440 kN-m
IH.
ρb =
cover = 40 mm
f’c = 20.7 MPa
fy = 415 MPa
fyt = 275 MPa
Bar diameter, d = 12 mm
concrete, Fvc = 0.76 MPa
0.85 f 'c β 1 600
f y (600+ f y )
ρb =
0.85(20.7)(0.85)(600)
415(600+ 415)
ρb = 0.0213
II.
ρmax = 0.75 ρb = 0.01598
ρmax f y
IJ.
IK.
ωmax =
'
fc
= 0.3203
Rn max = f’c ω(1 – 0.59ω) Rn max = 20.7(0.3203)[1 –
0.59(0.3203)]
Rn max = 5.378 MPa
IL.
Mn max = Rn max bd2
Mn max = 5.378(400)(435)2
Mn max = 407 kN-m
IM.
� Mn max = 0.90(407) = 366.32
IN.
Since Mu > � Mu max, the beam must be doubly reinforced
IO.
Mu1 = � Mu max = 407 kN-m
IP.
Mu = Mu1 + Mu2
IQ.
Tension steel area, As = As1 + As2
IR.
As1 = ρmax bd
As1 =0.01598(400)(435)
As1 = 2,780 mm2
IS.
Mu2 = T2 (d – d’)
73.678 x 106 = As2 (415)(435 -
440 = 366.32 + Mu2
Mu2 = 73.68 kN-m
65)
As2 = 533 mm2
IT.
IU.
As = 2,780 + 533 = 3, 313 mm2
Part 2:
Vu = 280 kN
Av = 2 x
IV.
IW.
π
4
Vn =
(12)2 = 226.2 mm2
Vu
ϕ
Vn =
280
0.85
Vn = 329.412 kN
IX.
Vc = Fvc bw d
Vc = 0.76(400)(435)
Vc = 132.24 kN
IY.
Vs = Vn – Vc
Vs = 329.24 – 132.24
Vs = 197.17 kN < 1/3
IZ.
S=
A v f yh d
Vs
s=
226.2 ( 275 ) ( 435 )
197,170
s = 137.2 mm
JA.
Maximum spacing (d/2 = 217.5 mm) or 600 mm
JB.
Therefore, s = 137 mm
JC.
Part 3:
Al =
At
f yt
2
ph
cot θ
s
fy
( )
θ = 45°
x = 400 – 46 x 2 = 308 mm
y = 500 – 46 x 2 = 408 mm
Aoh = x y = 308(408) = 125,664 mm2
√f
'
c
bw d
Ao = 0.85Aoh = 106,814
mm
2
ph = 2 (x +y) = 1432
mm
JD.
Tn =
Tu
ϕ
=
180
0.85
=
211.77 kN-m
JE.
Tn =
2 Ao At f yt
cot θ
s
211.77 x 106 =
2 ( 106,814 ) At (275)
cot 45 °
s
JF.
Al =
JG.
At
f yt
2
ph
cot θ
s
fy
( )
At
s
= 3.605 mm
Al = 3.605(1,432)
(
275
)
415
cot2
45°
Al = 3, 420 mm2
JH.
JI.
JJ.
JK.
JL.
JM.
MATHEMATICS, SURVEYING & TRANSPORTATION
ENGINIRING
(MAY 2012)
1. X and Y are inversely proportional with each other. Given that X = 15,000 when
Y = 162,500. Find X when Y = 328,400.
A. 7,422.35
B. 6,567.45
C. 7,849.56
D. 8,956.32
2. The sum of seven consecutive integers is zero. What is the smallest integer?
A. -4
B. -1
C. -3
D. -2
3. The sum and product of three distinct positive integers are 15 and 45,
respectively. What is the largest integer?
A. 5
B. 9
C. 15
D. 7
4. What is the curved surface area of a spherical segment (with two bases) if the
diameters of the bases, which are 25 cm apart, are 100 cm and 140 cm,
respectively.
A. 11,673.43 cm2
B. 10,567.93 cm2
C. 13,783.34 cm2
D. 12,328.75 cm2
5. The area of a park on a map is 500 mm2. If the scale of the map is 1 to 40,000,
determine the true area of the park in hectares (1 hectare = 10 4m2)
A. 40
B. 80
C. 160
D. 12
2π
3
6. Evaluate the interal:
∫ csc x cot x dx
π
3
A. 1
B. 0
C. ½
D. -1
7. Find the general solution of the following differential equation:
8. y” + 3y’ - 4y = 0
A. y = C1 e4x + C2 x e-x
B. y = C1 e-4x + C2 x ex
C. y = C1 e4x + C2 e-x
D. y = C1 e-4x + C2 ex
E.
F.
G.
H.
I. STRUCTURAL ENGINIRING & CONSTRUCTION
(MAY 2012)
1. A vertical load of W is supported by the tripod shown. If the capacity of each leg
is 15 kN, what is the safe value of W?
A.
B.
C.
D.
34.48kN
36 kN
32.12 kN
42 kN
2. Six (6) steel cables are used
to support a circular
moulding having a diameter
of 2 m and weighing 3.6
kN/m. The cables are equally
spaced around the moulding
and attached to a single hook
3 m above the moulding. If
the allowable stress in the
cable is 105 MPa, what is the
required diameter?
A. 8 mm
B. 7 mm
C. 9 mm
D. 10 mm
3. A vertical steel rod is fixed at the top and supports an 8-kN load at the lower
end. The rod is 10mm n diameter and 25 mm long. Unit weight of steel is 77
kN/m3. What is the total elongation of the rod?
A. 12.732 mm
B. 12.853 mm
C. 12.973 mm
D. 12.612 mm
4. A hallow circular tube has an outside diameter of 85 mm and is 5 mm thick. The
tube is fixed (cantilever) at one end and subjected to a torque of 4 kN-m at its
free end. What is the maximum shearing stress in the tube?
A. 76.5 MPa
B. 98.7 MPa
C. 92.3 MPa
D. 84.2 MPa
5. A decorative concrete beam is simply supported over a span of 6 m. The beam
weighs 4 kN/m and the cracking moment is 38 kN-m. What is the safe uniform
load of the beam?
A. 4.44 kN/m
B. 4.84 kN/m
C. 5.24 kN/m
D. 3.84 kN/m
6. A 2.8 m cantilever beam carries a uniformly distributed load of 20 kN/m
throughout its length and a concentrated load of 30 kN at a point 2 meters from
the fixed end. What is the bending moment at the fixed end?
A. 91.3 kN-m
B. 76.7 kN-m
C. 123.9 kN-m
D. 99.2 kN-m
7. A 12 m long beam is simply supported at the left end and at 3 m from the right
end. The beam will be subjected to a uniformly distributed moving load. What
total length of the beam must be subjected to this load to produce maximum
A. moment
9m
C. 7.5 m
negative
at midspan?
B. 3 m
D. 4.5 m
E.
F. Situation 1 – The
hook is subjected to
three forces P, Q
and S as shown. P =
35 kN and Q = 45
kN.
8. Determine the angle 
such that the resultant of
the three forces is 80 kN
acting horizontally to the
right.
A 22.85°
B 21.78°
C 24.98°
D 23.12°
1. If angle  = 60°, find the magnitude of the force S such that the resultant
force is horizontal to the right.
A 48 kN
B 51 kN
C 42 kN
D 45 kN
2. Find the magnitude of the force S such that the three forces are in equilibrium.
A. 43.87 kN
B. 40.93 kN
C. 45.98 kN
D. 38.65 kN
E.
F. Situation 2 –
The horizontal
distance from
A at one end of
the river to
frame C at the
other end is 20
m. The cable
carries a load
of W = 50 kN. The sag “d” of the cable is 1 m.
G.
3. Find the distance x1 such that the tension in segment AB of the cable is equal
to that segment BC.
A. 9 m
B. 10 m
C. 12 m
D. 11 m
4. Calculate the tension in segment BC when x 1 =5 m.
A. 206.56 kN
B. 174.9 kN
C. 165.43 kN
D. 187.92 kN
5. What is the total length of the cable when x1 = 5 m?
A. 20.13 m
B. 20.76 m
C. 21.12 m
D. 19.76 m
E.
F. Situation 3 – The 1.8-mdiameter circular plate
shown is supported by
equally spaced posts along
its circumference. A load P
= 1150 kN is placed at
distance x = 0.45 m from
post A.
6. Neglecting the weight of the
plate, what is the reaction at post
A?
A.
B.
C.
D.
834.2
766.7
191.7
194.6
kN
kN
kN
kN
7. Neglecting the weight of the plate, what is the reaction at post B?
A. 766.7 kN
B. 834.2 kN
C. 194.6 kN
D. 191.7 kN
8. Considering the weight of the plate, what is the reaction at C? the plate is 45
mm thick and the unit weight of steel is 77 kN/m3.
A. 194.6 kN
B. 191.7 kN
C. 834.2 kN
D. 766.7 kN
E. Situation 4 – The
billboard, 3 m high by 4 m
wide, is supported as show
in the figure. The total
weight of the billboard is 30
kN. H = 1.5 m, θ = 60°.
Wind pressure, q =
1.6 kPa
Wind pressure
coefficient, c = 1.0
9. The horizontal component of the reaction at A is nearest to:
A. 19.54 kN
B. 21.89 kN
C. 16.38 kN
D. 12.45 kN
10. What is the axial stress strut BC whose cross sectional dimension is 6 mm x
76 mm?
A. 94.1 MPa
B. 87.3 MPa
C. 76.5 MPa
D. 102.6 MPa
11. If the strut AB were replaced by a 16 mm ∅ steel cable, determine the
normal stress (in MPa) in the cable.
A. 86.5 MPa
B. 90.1 MPa
C. 96.3 MPa
D. 99.1 MPa
E. Situation 5 –
A girder
weighing 18
kN/m is
suspended on
a parabolic
cable by a
series of
vertical
hanger. The length of the beam is 24 m and the sag of the cable is 3
m.
12. What is the vertical component of the reaction at A?
A. 240 kN
B. 250 kN
C. 216 kN
D. 275 kN
13. What is the tension in the cable at the center?
A. 487 kN
B. 412 kN
C. 432 kN
D. 521 kN
14. If the allowable cable tension is 360 kN, what is the minimum sag?
A. 4.5 m
B. 3.5 m
C. 5 m
D. 5.5 m
E. Situation 6 – Steel tank with an outside diameter of 600 mm has a
wall thickness of 8 mm. The tank is used as storage of gas under a
pressure of 2.2 MPa.
15.
Determine the value of the tangential stress in the tank wall.
A. 83.2 MPa
B. 80.3 MPa
16.
C. 89.4 MPa
D. 90.2 MPa
Determine the value of the longitudinal stress in the tank wall.
A. 38.5 MPa
B. 43.1 MPa
C. 34.7 MPa
D. 40.2 MPa
17. If the allowable tensile stress in the wall is 124 MPa, to what value may the
gas pressure be increased?
A. 3.765 MPa
B. 2.873 MPa
C. 4.123 MPa
D. 3.397 MPa
E. Situation 7 – The solid pole
shown in the figure is loaded with
vertical load P = 3kN and lateral
load H = 0.45 kN. The pole is 3 m
high 280 mm diameter and
weighs 22 kN/m3.
18. What is the maximum compressive
stress at the base?
A.
B.
C.
D.
0.75
0.88
0.65
0.52
MPa
MPa
MPa
MPa
19. What is the maximum tensile stress at
the base?
A. 0.88 MPa
B. 0.52 MPa
C. 0.75 MPa
D. 0.65 MPa
20. What is the maximum shearing stress in the pole?
A. 0.0097 MPa
B. 0.0054 MPa
C. 0.0132 MPa
D. 0.0115 MPa
E. Situation 8 –
The barge
shown in the
figure supports
the load w1 and
w2. For this
problem, w1 =
145 kN/m, w2
= 290 kN/m, L1
= 3 m, L2 = 6
m, L3 = 3 m.
21. What is the length of barge “L” so that the upward pressure is uniform?
A. 15 m
B. 12 m
C. 20 m
D. 18 m
22. What is the shear at 3 m from the left end?
A. -162 kN
B. -151 kN
C. -194 kN
D. -174 kN
23. At what distance from the left end will the shear in the barge be zero?
A. 4 m
B. 5.5 m
C. 5 m
D. 4.5 m
E. Situation 9 – A concrete pad supports two distributed loads of 112
kN/m, as shown in the figure. It required to determine the maximum
shear ad moment in the pad due to these loads.
24. What uniform base pressure “q” is induced by these loads?
A. 24 kN/m
B. 32 kN/m
C. 48 kN/m
D. 42 kN/m
25. What is the maximum shear acting on the concrete pad?
A. 24 kN
B. 42 kN
C. 32 kN
D. 48 kN
26. What is the maximum moment on the pad?
A. 42 kN-m
B. 24 kN-m
C. 32 kN-m
D. 48 kN-m
E.
F. Situation 10 – A 10-m long beam is simply supported at the left end
and at 2 m from the right end. The beam will be analyzed for
maximum shear at the midspan that can be induced by a moving load.
27. What is the ordinate of the influence diagram at the midspan?
A. 0.3
B. 0.45
C. 0.25
D. 0.5
28. What is the ordinate of the influence diagram at the free end?
A. 0.3
B. 0.45
C. 0.25
D. 0.5
29. The beam will be subjected to a uniformly distributed moving load. What
total length of this beam must be subjected to this load to produce maximum
shear at the midspan?
A. 4 m
B. 6 m
D. 5 m
C. 3 m
E. Situation 11 –
The trussed
beam shown is
5.4 m long. A
man of weight
“W” is standing
at the middle of the beam. Neglect the weight of the beam.
30. The capacity of the rod is 2kN, what is the safe maximum weight of the man
in kg?
A. 132 kg
B. 129 kg
C. 156 kg
D. 187 kg
31. If the man weighs 85 kg, what is the tensile stress in the rod if its diameter
is 10mm?
A. 12.89 MPa
B. 14.35 MPa
C. 17.87 MPa
D. 16.78 MPa
32. What is the total length of the rod?
A. 6.12 m
B. 5.69 m
C. 5.34 m
D. 7.32 m
E. Situation 12 – The truss shown is made from Guijo 100 mm x 150
mm. The load on the truss is 20 kN. Neglect friction.
F.
G. Allowable stresses for Guijo:
Compression parallel to grain = 11 MPa
Compression perpendicular to grain = 5 MPa
Shear parallel to grain = 1 MPa
Shear longitudinal for joints = 1.45 MPa
33. Determine the minimum value of x.
A. 180 mm
B. 150 mm
C. 160 mm
D. 140 mm
E.
34. Determine the minimum value of y in mm.
A. 34.9
B. 26.8
C. 13.2
D. 19.5
35. What is the axial stress in member AC in MPa?
A. 1.26
B. 1.89
C. 0.67
D. 2.78
E. Situation 13 – The lap joint of a tension member is shown in the
figure. The plate is 252 mm wide and 12 mm thick. The bolts are 20
mm in diameter and the holes are 3 mm larger than the bolt diameter.
Steel is A36 with Fy = 248 MPa and Fu = 400 MPa. It is required to
determine the capacity of the joint based on gross area, net area, and
block shear.
F.
36. Determine the safe value of P based on tension on gross area.
A. 450 kN
B. 420 kN
C. 500 kN
D. 480 kN
37. Determine the safe value of P based on tension on net area.
A. 439 kN
B. 421 kN
C. 453 kN
D. 486 kN
38. Determine the safe value of P based on tension on block shear.
A. 423 kN
B. 469 kN
C. 495 kN
D. 521 kN
E. Situation 14
– A 6-m long
fixed-ended
beam carries a
uniformly
distributed
load of 20
kN/m. Use E =
200 GPa and Ix = 67.5 x 106 mm4.
F.
39. Determine the moment at the fixed end.
A. -60 kN-m
B. -55 kN-m
C. -65 kN-m
D. -50 kN-m
40. What is the maximum shear in the beam?
A. 60 kN
B. 55 kN
C. 65 kN
D. 50 kN
41. Compute the vertical deflection at the midspan.
A. 4 mm
B. 7 mm
C. 5 mm
D. 6 mm
E. Situation 15 – A fixed end beam has a span of 10 m and supports a
super imposed uniformly distributed load of 20 kN/m.
F.
G. Properties of W 450 x 70:
H.
mm2
bf = 150 mm
tf = 15 mm
d = 450 mm
A = 8700
Ix = 274.7 x 106 mm4
Iy = 8.47 x 106 mm4
tw = 10 mm
Wb = 70 kg/m
42. Calculate the maximum bending stress in the beam.
A. 112.56 MPa
B. 132.98 MPa
C. 142.20 MPa
D. 123.87 MPa
43. What is the average shearing stress in the beam?
A. 24.35 MPa
C. 23.15 MPa
B. 26.92 MPa
D. 19.32 MPa
44. Determine the maximum shearing stress in the beam
A. 26.92 MPa
B. 19.32 MPa
C. 2435 MPa
D. 23.15 MPa
E. Situation 16 – A built up
section consisting of W 350
x 90 with two 12 mm
plates welded to form a
box section as shown in
the Figure S01. The section
is used as a column 10
meters long. The column is
fixed at both ends and
braced at midheight about
the weak axis (Y-axis). The
code provision is given in
Figure NSCP-01. Use Fy = 248 MPa.
F.
G. Properties of W350 x 90:
bf = 250
H.
tw = 9.5 mm
Ix = 266 x 106
Iy = 44.54 x 106
A = 11,550 mm2
mm
tf = 16.4
mm
d = 350 mm
45. Determine the effective slenderness ration of the column with respect to
lateral buckling about the x-axis.
A. 42.76
B. 34.89
C. 37.66
D. 35.98
46. Determine the effective slenderness ration of the column with respect to the
lateral buckling about the y-axis.
A. 34.89
B. 35.98
C. 37.66
D. 42.76
47. Determine the axial load capacity of the column in kN.
A. 2435
B. 2895
E. Figure NSCP-01
F. When KL/r < Cc (short column)
C. 3219
D. 2663
G.
Fa =
[ ]
KL 2
Fy
r
1−
2
2 Cc F . S .
( )
F.S. =
KL 3
KL
)
r
5
r
+
−
3 8 Cc
8 C3c
3(
( )
H. When KL/r > Cc (long column)
I.
J.
Fa =
12 π 2 E
KL 2
23
r
( )
K = effective length factor
K = 0.5 for both ends fixed
K = 1 for both ends pin
K = 0.7 for one end fixed and other end pin
K. Situation 17 – A box column is formed by welding two channel
sections at the tip of their flanges. The column has an unsupported
length of 4 m and is hinged at both ends (K=1.0).
L. The property of each channel section is as follows:
bf = 90 mm
tf = 10 mm
d = 250 mm
M.
tw = 12 mm
Ix = 38.1 x 106 mm4
Iy = 2.91 x 106 mm4
x́
= 21 mm
A = 4560 mm2
N.
48. What is the compressive stress in the column due to an axial load of 900
kN?
A. 98.7 MPa
B. 91.2 MPa
C. 89.4 MPa
D. 102.5 MPa
49. What is the maximum bending stress in the column due to a moment of 70
kN-m, about the x-axis of the section?
A. 114.8 MPa
B. 123.9 MPa
C. 96.5 MPa
D. 1181.1 MPa
50. What is the critical (maximum) effective slenderness ratio of the column?
A. 48.2
B. 76.1
C. 54.4
D. 65.2
E. Situation 18 – The deck of a bridge consists of a ribbed metal deck
with 100 mm concrete slab on top. The superstructure supporting the
deck is made of wide flange steel beams strengthened by a cover plate
16 mm x 260 mm one at the top and one at the bottom, and is spaced
1.2 m on centers. The beams are simply supported over a span of 25
m. The loads on each beam are as follows:
F.
G. Dead load = 12 kN/m (including beam weight and deck)
Wheel live loads:
Front wheel = 18 kN
Rear wheel = 72 kN
Wheel base = 4.3 m
H.
Impact factor =
15
L+37
≤ 30%, where L = length in m.
I. Properties of W 850x185:
J.
A = 23,750 mm2
d = 850 mm
bf = 290 mm
tf = 20 mm
tw = 15 mm
Ix = 2662 x 106 mm4
Iy = 81.52 x 106 mm
51. Calculate the maximum bending stress in the beam due to dead load.
A. 123 MPa
B. 107 MPa
C. 92 MPa
D. 98 MPa
52. Calculate the maximum bending stress in the beam due to live load plus
impact.
A. 79 MPa
B. 62 MPa
C. 68 MPa
D. 56 MPa
53. Calculate the maximum average web shear stress in the beam due to live
load plus impact.
A. 7.6 MPa
B. 8.5 MPa
C. 9.1 MPa
D. 12.4 MPa
E. Situation 19 – The W450x86 beam is supported by a concrete wall
and a 130-mm- wide bearing plate as shown. The beam reaction is
250 kN. All steel are A36 steel with Fy = 248 MPa. Concrete strength f’ c
= 27.5 MPa.
F. Properties of W450x86 are as follows:
G.
d = 450 mm
bf = 190 mm
H.
tf = 18
mm
mm
tw = 10
I. k = 38 m
J. Allowable bearing stress of concrete, Fp = 0.35 f’c
Allowable bending stress of weak axis of plate, Fb = 0.75 Fy
54. What is the required width of the bearing plate “W”?
A. 220 mm
B. 240 mm
C. 180 mm
D. 200 mm
55. Using the width in Part 1, wat is the required plate thickness? Assume that
the critical section in bending for bearing plate is distance “k” from the axis of
the beam.
A. 28.4 mm
B. 32.1 mm
C. 24.5 mm
D. 21.2 mm
56. Determine the web yielding stress at the web toe of fillet.
A. 102 MPa
B. 85 MPa
C. 127 MPa
D. 111 MPa
E.
F. Situation 20 – The floor framing plan of a reinforced concrete
structure is shown in the figure. The beams are 280 mm wide and 520
mm deep and the slab is 110 mm thick. Other than concrete weight,
the floor is subjected to additional (superimposed) dead load of 3 kPa
and live load of 5.2 kPa. Unit weight of concrete is 23.5 kN/m 3.
G. Due to space consideration, the columns E and H are to be removed.
This will make girder BEHK support the beams DEF at E and GHI at H.
H. Use tributary area method.
57. Determine the uniform service dead load on beam DEF.
A. 19.87 kN/m
B. 21.34 kN/m
C. 17.38 kN/m
D. 16.21 kN/m
58. Determine the uniform service live load on beam DEF.
A. 13 kN/m
B. 14 kN/m
C. 11 kN/m
D. 12 kN/m
59. Determine the factored concentrated load at E due to loads on beam DEF.
A. 287.9 kN
B. 145.8 kN
C. 254.5 kN
D. 321.2 kN
E.
F.
G.
H.
I.
J.
K.
L.
M.
Situation
21 – The
floor
framing plan
of a
reinforced
concrete
structure is shown in the figure. Then the columns E and H are
deleted, girder BEHK carries the reaction of BEF at E and GHK at H.
this girder maybe considered fixed at B and K. the uniform load on this
girder is 5 kN/m and the concentrated load at E and H are each 270
kN.
60. Calculate the maximum shear at B due to uniform and concentrated loads.
A. 321 kN
B. 289 kN
C. 265 kN
D. 253 kN
61. Calculate the maximum shear at E due to concentrated load only.
A. 300 kN
B. 280 kN
C. 290 kN
D. 270 kN
62. Calculate the maximum positive moment in the girder due to uniform load
only.
A. 11.72 kN-m
B. 13.21 kN-m
C. 9.65 kN-m
D. 10.12 kN-m
E.
F.
G.
H.
I.
J.
K.
L. Situation 22 – The rectangular footing shown is subjected to axial
load of P = 1200 kN and a moment of M = 360 kN-m. it is required to
determine the safe gross bearing capacity of the soil to support the
given loads. The unit weights of concrete and soil are 23.5 kN/m 3 and
18 kN/m3, respectively.
M.
63. What is the maximum foundation pressure in kPa?
A. 256 kPa
B. 274 kPa
C. 287 kPa
D. 321 kPa
64. What is the minimum foundation pressure in kPa?
A. 64 kPa
B. 69 kPa
C. 82 kPa
D. 54 kPa
65. What is the minimum required gross allowable soil bearing capacity to carry
the given loads?
A. 310 kPa
B. 280 kPa
C. 290 kPa
D. 300 kPa
E. Situation 23 – The Tbeam shown resulted
from monolithic
construction of the
beam and slab. The
effective flange width
is 1100 mm and the
uniform slab thickness
is 120 mm. width of
beam is 340 mm, total
depth of the T-section
is 590 mm. The
centroid of steel is 70
mm from extreme
concrete fiber. Concrete strength f’ c = 21 MPa and streel strength fy =
415 MPa. Use strength design method.
F.
66. Calculate the nominal strength of the beam for positive moment neglecting
the contribution of the top reinforcement.
A. 567.2 kN-m
B. 503.2 kN-m
C. 456.1 kN-m
D. 526.5 kN-m
67. Calculate the nominal strength of the beam for negative moment.
A. 289.88 kN-m
B. 321.98 kN-m
C. 432.12 kN-m
D. 238.43 kN-m
68. Calculate the required nominal shear strength of the beam if it is subjected
to a factored shear of 220 kN.
A. 289.4 kN
B. 269.5 kN
C. 258.8 kN
D. 231.9 kN
E. Situation 24 – A reinforced concrete beam has a width of 300 mm
and an overall depth of 480 mm. The beam is simply supported over a
span of 5 m. Steel strength fy = 415 MPa and concrete strength f’ c =
28 MPa. Concrete cover is over 70mm from the centroid of the steel
area. Unit weight of concrete is 23.5 kN/m 3. Other than the weight of
the beam, the beam carries a superimposed dead load of 18 kN/m and
live load of 14 kN/m. Use the strength design method.
69. Determine the maximum factored moment on the beam.
A. 135 kN-m
B. 121 kN-m
C. 168 kN-m
D. 183 kN-m
70. If the design ultimate moment capacity of the beam is 280kN-m, determine
the required number of 20 mm tension bars.
A. 8
B. 6
C. 9
D. 7
71. If the beam will carry an additional factored load of 240 kN at midspan,
determine the required number of 20 mm tension bars.
A. 14
B. 9
C. 10
D. 12
E. Situation 25 – The
section of a column
is shown in the
figure. For this
problem, b1 = 300
mm, b2 = 180 mm,
d1 = 250 mm, d2 =
350 mm. f’c = 28
MPa, fy = 414 MPa.
72. Determine the location of
the gross concrete area
measured from y-axis.
A. 281 mm
B. 262 mm
C. 274 mm
D. 253 mm
73. Determine the location of the plastic neutral axis of the column measured
from the y-axis. Neglect the area of concrete occupied by the steel.
A. 272 mm
B. 302 mm
C. 282 mm
D. 292 mm
74. Determine the factored moment Mu due to factored load Pu = 3200 applied
400 mm from the y-axis. Assume that the column is reinforced such that
plastic neutral axis is 290 mm from the y-axis.
A. 352 kN-m
B. 387 kN-m
C. 326 kN-m
D. 376 kN-m
E. Situation 26 – The column
shown in the figure is
subjected to shear force
parallel to the 600 mm side.
Allowable concrete shear
stress fir shear parallel to the
600 mm side is 0.816 MPa.
Concrete strength f’c = 21 MPa
and steel strength for both
longitudinal and
reinforcements is 415 MPa.
The ties are all 12 mm in diameter with clear cover of 40 mm.
75. Determine the factored shear force Vu that the column can resist if the
nominal shear strength provided by the ties is 375 kN.
A. 421
B. 514
C. 486
D. 452
76. If the ties are spaced at 230 mm o centers, what is the maximum value of
Vu, in kN?
A. 446
B. 521
C. 389
D. 416
77. If the factored shear force parallel to the 600 mm side is 400 kN, determine
the required spacing of transverse reinforcement in accordance with the
provisions for seismic design.
A. 154.8 mm
B. 112.5 mm
C. 125.8 mm
D. 208.1 mm
E. Situation 27 – A prestressed concrete beam ha a width of 300 mm
and an overall depth of 600 mm. the prestressing tendons are placed
at a distance “e” below neutral axis of the beam and the applied
prestressing force is P = 1500 kN. There is 15% loss of prestress.
78. Determine the compressive stress in concrete when P is applies at the
centroid of the beam.
A. 6.43 MPa
B. 8.21 MPa
C. 7.08 MPa
D. 7.54 MPa
79. What is the maximum compressive stress in the beam when e = 120 mm?
A. 14.32 MPa
B. 18.72 MPa
C. 15.58 MPa
D. 16.92 MPa
80. Determine the value of eccentricity “e” such that the resulting stress at the
top fiber of the beam is zero.
A. 100 mm
B. 120 mm
C. 200 mm
D. 150 mm
E. Situation 28 – The section of a prestressed double-tee concrete floor
joist is shown in the figure. The prestressing force in each tee is 750
kN. Unit weight of concrete is 23.5 kN/m3.
F.
G. The properties of the double tee section are:
Area = 220,000 mm2
I = 1890 x 106 mm4
y1 = 90 mm
y2 = 270 mm
y3 = 75 mm
Simple span, L = 8 m
H. Service load on floor:
Dead load = 2.5 kPa
Live load = 6 kPa
81. Determine the initial stress at the bottom fibers due to prestressing force
along?
A. -42.3 MPa
B. -48.6 MPa
C. -52.8 MPa
D. -37.6 MPa
82. Determine the stress at the bottom fibers due to service load and
prestressing force. Assume that there is a loss of prestress of 20% at service
loads.
A. -8.9 MPa
B. -9.87 MPa
C. -12.32 MPa
D. -6.56 MPa
83. Calculate the additional load can the floor carry so that the stress at the
bottom fibers at the midspan is zero.
A. 5.43 kPa
B. 7.98 kPa
C. 4.89 kPa
D. 3.04 kPa
E. Situation 29 – A square footing is shown in the figure. The footing is
to support a 350 mm x 400 mm column that carried an axial dead load
of 740 kN and an axial live load of 460 kN. Use f’ c = 20.7 MPa and fy =
275 MPa. Main bar diameter is 20 mm, concrete cover from center of
main bars = 90 mm.
84. Calculate the factored shear on footing at critical section for wide-beam
action.
A. 435 kN
B. 612 kN
C. 504 kN
D. 587 kN
85. Calculate the factored shear on footing at critical section for two-way action.
A. 1432 kN
B. 1873 kN
C. 1648 kN
D. 1256 kN
86. Determine the required number of 20-mm bars.
A. 15
B. 17
C. 13
D. 11
E.
F.
G.
H.
I.
Situation 30 –
Answer the
following questions:
87. Which of the following deals with forces at rest?
A. Impact
B. Kinetic
C. Static
D. Dynamic
88. Which of the following forces determines whether a body will be at rest or in
motion?
A. Resultant
B. Equilibrant
C. Work
D. Momentum
89. Energy by virtue of velocity
A. Potential
B. Kinetic
C. Work
D. momentum
E.
F. Situation 31 – Answer the following questions on axial deformation of
rigid bodies:
90. Within proportional limit, the stress is directly proportional to strain.
A. Elastic limit
B. Young’s Modulus
C. Poisson’s Ratio
D. Hooke’s Law
91. The ratio of lateral strain to longitudinal strain.
A. Hooke’s Law
B. Poisson’s Ratio
C. Young’s Modulus
D. Elastic Limit
92. Within elastic range, the slope of the straight line portion of the stress-strain
curve.
A. Young’s Modulus
B. Elastic Limit
E.
F.
G.
C. Hooke’s Law
D. Poisson’s Ratio
H.
I.
J.
K.
L.
M.
N.
O. ANSWER KEY:
P.
Z.
AI.
AR.
26
Q.
AA.
11
AB.
12
S.
AC.
13
AS.
27
AK.
20
W.
AF.
AG.
16
Y.
BI.
BA.
35
BQ.
49
BR.
50
AM.
22
AN.
23
AO.
24
BB.
BC.
36
AU.
29
X.
AH.
17
AT.
AL.
AD.
14
AE.
15
AZ.
34
BY.
BJ.
T.
U.
BH.
41
AJ.
R.
V.
AQ.
BD.
37
BS.
51
BK.
44
AW.
31
AX.
32
AP.
BF.
BG.
40
BU.
53
BM.
BO.
47
AY.
BP.
CA.
58
CB.
59
CE.
62
BW.
55
BX.
CJ.
CK.
67
CF.
CP.7
2
A
CQ.
73
D
CR.
74
B
CS.
75
A
CT.
CL.
CD.
61
BV.
BN.
46
CH.
65
CI.
CC.
60
BL.
AV.
BE.
38
BT.
BZ.
57
CG.
64
CM.
69
CU.
76
C
CV.7
7
C
CN.
70
CW.
78
D
CO.
71
CX.
79
D
CY.
DA.
82
DB.
83
CZ.
81
DC.
84
DD.
85
DF.
DH.
88
DI.
DG.
87
DJ.
DW.
DX.
DY.
DZ.
EA.
EB.
EC.
ED.
EE.
EF.
EG.
EH.
EI.
EJ.
Solutions
DK.
91
DL.
92
DE.
DV.
DM.
93
DN.
94
DO.
95
DP.
DQ.
96
DR.
97
B
DS.
98
DU.
10
0
DT.
A
EK.
1
EL.
LAD =
√ 1.8 + 2.42 +0.92 ππ
2
LAD =3.1321 m
LAB = LAD =
3.1321 m
LAC =
√ 2.42 +1.82
= 3m
EM.
FACy =
2.4
3
FAC = 0.8 FAC
2.4
FABy = 3.1321
EN.
FAB = 0.7662 FAB
EO.
By symmetry, FAB = FAD
EP.
∑MBD = 0
FACy(2.7) = W(0.9)
EQ.
ER.
Set FAC = 15 kN
∑MCE = 0
2(FABy)(2.7) = W(1.8)
ES.
ET.
0.8FAC(2.7) = (0.9)
FAC = 0.4167 W
W = 36 kN
5.4(0.7662FAB)= W(1.8)
FAB = 0.43501 W = FAD
Set FAB = 15 Kn
W = 34.482 kN (governs)
2
EU.
EV.
θ = arctan(3/1) = 71.565°
EW.
Total weight, W = 3.6 x
EX.
∑FV = 0
EY.
F t = T x Ac
π (2) = 22.619 kN
6 x T sin θ = 22.619
T = 3.974 kN
105 = 3,974 x
π
4
(dc) 2
dc = 6.9 say 7 mm
EZ.
FA.
3
D
= 10 mm
L=
25 m
P=8
kN
γ
s
= 77
kN/m3
FB.
Area, A =
π
2
2
4 (10) = 78.54 mm
E = 200
GPa (for
steel)
FC.
Weight of rod,
γ
W=
s
Vs =
77,000[78.54/10002](25)
W = 154.189 N
FD.
Elongation due to concentrated load P:
δ
FE.
1
PL
AE
=
δ
1
=
8,000(25,000)
78.54 (200,000)
δ
FF.
1
= 12.732 mm
Elongation due to own weight:
δ
FG.
2
1
WL
2
AE
=
δ
2
=
1
(151,189)(25,000)
2
78.54 (200,000)
δ
FH.
Total deformation,
FI.
δ
2
= 0.1203 mm
=
δ
1
+ δ
Outside diameter, D =
Inside diameter, d = D – 2t =
Torque, T =
FK.
Maximum shearing stress:
τ
FL.
=
16 TD
π ( D 4−d 4)
τ
τ
FN.
= 12.853 mm
4
FJ.
FM.
2
=
16 (4 x 106 )(85)
π (854 −754 )
= 84.22 MPa
5
Cracking moment, M = 38 kN-m
Weight of beam = 4kN/m
FO.
M=
w L2
8
2
38 =
w(6)
8
w = 8.444 kN/m
FP.
Safe uniform load = 8.444 – 4 = 4.444 kN/m
FQ.
FR.
6
MA = 30(2)+
10(2.8)(1.4)
MA = 99.2 kNm
FS.
FT.
FU.
FV.
7
FW.
FX.
FY.
Moment = w x Area under the influence diagram
FZ.
Maximum negative moment at B will occur when the uniform load is
within CD only. Total length = 3 m
GA.
Situation 1
GB.
Given:
P = -35i
Q = (45 cos 60°)i –
(45 sin 60°)j
Q = 22.5i – 38.97j
GC.
Part 1:
GD.
80i
GE.
Resultant, R =
R=P+Q+S
80i = -35i + (22.5i-38.9j) + S
S = 92.5i + 38.97j
Sx = 92.5 kN , Sy = 38.97 kN
α
GF.
= arctan
Sy
Sx
α
= arctan
α
= 22.85°
Part 2:
GG.
horizontal
Resultant is
to the right with
α
=
60°
GH.
Rxi + 0j
R=P+Q+S=
Rxi = -35i + (22.5i – 38.97j)
+ S(cos 60° i + sin 60° j)
Rxi + 0j = (-12.5 0.5S)I +
(-38.97+0.866S)j
GI.
0 = -38.97 + 0.866S
S = 45 kN
GJ.
Part 3:
38.91
92.5
GK.
0
P+Q+S=
-35i + (22.5i –
38.97j) + S = 0
S = 12.5i + 38.97j
Sx = 12.5
kN
Sy = 38.97
kN
GL.
S=
√ 12.52+38.97 2
S = 40.927 kN
GM.
GN.
GO.
Situation 2
Part 1:
The tensions in the cables are equal when their angles of
inclination
are equal. Since A and C are on the same elevation, therefore x 1
= x2 = 10.
GP.
Parts 2 & 3:
GQ.
θ
GR.
= arctan (5/1) = 78.69°
α
= arctan (15/1) = 86.19°
β
= 180 ° - α
– β
= 15.12°
GS.
polygon:
From the force
T1
sin α
GT.
=
T2
sinθ
=
W
sin β
GU.
T1 =
50
sin 86.19 °
sin15.12 °
= 191.21
kN → Part 2
GV.
T2 =
50
sin 78.69 °
sin15.12 °
= 187.92
kN
GW.
Length of cable:
L = x1 sec θ + x2 sec
α
L = 5 sec 78.69° + 15 sec 86.19°
L = 20.13 m → Part 3
GX.
GY.
Situation 3
Parts 1 & 2:
Neglecting the weight of the plate:
GZ.
∑MA = 0
2RB (1.35) = 1150 (0.45)
RB = 191.667 kN = RC
HA.
∑FV = 0
RA = 1150 – 2 (191.667)
RA = 766.67 kN
HB.
HC.
Part 3
Considering the weight of the plate:
γ
W=
W = 77 x
π
4
W = 8.817 kN
s
Vs
(1.8)2(0.45)
∑MA = 0
2RB (1.35) = 1150 (0.45) + 8.817(0.9)
RB = 194.61 kN = RC
HD.
HE.
Situation 4
HF.
HG.
HH.
a = 1.5 cot 30°
|∑MC =0|
HI.
HJ.
HK.
HL.
a = 2.598 m
AH(3) + 19.2(1.5) = 30(2.598)
AH = 16.38 kN → Part 1
At joint A:
∑FH = 0
FAB cos 30° = 16.38
FAB = 18.915 kN
Stress in member AB, fAB =
18.915 x 10
π
2
( 16 )
4
3
Stress in member AB, fAB = 94.08 MPa → Part 3
At joint B:
∑FH = 0
FBC cos 30° = 19.2 + 18.915 cos 30°
FBC = 41.085 kN
41.085 x 103
=
6(76)
HM.
Stress in member BC, fBC
HN.
Stress in member BC, fBC = 90.1 MPa → Part 2
HO.
HP.
Situation 5
W1 = 18(12)
W1 = 216 kN
HQ.
θ=
arctan(3/6)
θ = 26.565°
HR.
polygon:
From the force
HS.
T = W1 csc θ
T = 216 csc
26.565°
T = 483 kN
HT.
AV = T sin θ = 216 kN
HU.
H = W1 cot θ
HV.
Part 3:
When T = 360 kN
H = 216 cot 26.565°
H = 432 kN
W
T
HW.
θ = arcsin
= arcsin
HX.
Sag = 6 tan θ = 4.5 m
216
360
HY.
HZ.
IA.
IB.
Situation 6
Given:
Outside diameter, Do = 600 mm
Thickness, t = 8 mm
Pressure inside, pi = 2.2 MPa
IC.
Inside diameter, D = Do – 2t = 584 MPa
ID.
Part 1:
σ
IE.
t
=
pD
2t
σ
σ
IF.
Part 2:
t
t
=
2.2(584)
2(8)
= 80.3 MPa
= 36.87°
σ
IG.
l
pD
4t
=
σ
σ
l
l
2.2(584)
4 (8)
=
= 40.15 MPa
IH.
Part 3: Note: in thin walled cylindrical tanks, tangential stress twice as
critical as longitudinal stress.
II.
σ
IJ.
σ
t allow
t
= 124 MPa
=
pD
2t
124 =
p( 584)
2(8)
p = 3.397 MPa
IK.
IL.
Situation 7
A=
π
4
I=
π
64
(280)2 = 61,575.2 mm2
(280)2 = 301.719 x 106 mm4
Total vertical load at the base:
Pt = P +
γ Vol = 3 + 22 +
π
4
(0.28)2(3) = 7.064 kN
IM.
Total moment at base:
M = H x L +P x e = 0.45(3) + 3(0.1) = 1.65 kN-m
IN.
Parts 1 & 2: Maximum normal stress at the base:
IO.
f=
Pt M C
±
A
I ; c = D/2 = 140 mm
-
6
f=-
1.65 x 10 (140)
3000
±
61,575.2 301.718 x 106
f = - 0.1147 ± 0.7656
IP.
Maximum compressive stress, fc = - 0.1147 - 0.7656
Maximum compressive stress, fc = -0.88 MPa
IQ.
Maximum tensile stress, ft = - 0.1147 + 0.7656
Maximum tensile stress, ft = 0.651 MPa
IR.
Part 3:
IS.
Shear, V = H = 450 N
Diameter, D =280 mm, r = 140mm
4V
3 πr
Shear stress, fv =
4(450)
fv = 3 π ( 140 )2
fv = 0.0097 MPa
IT.
Situation 8
IU.
IV.
IW.
W = 145(3) + 290(3) = 1305 kN
Location of W:
Wx = 145(3)(1.5) + 290(3)(10.5)
x = 7.5 m
IX.
For the uniform pressure at the bottom of the barge, x = L/2.
L = 2(7.5)
L = 15 m → Part 1
IY.
Upward pressure, q =
IZ.
Shear at a point 3 m from the left end (@B):
VB = q(3) – w1(3) VB = 87(3) - 145(3)
VB = -174 kN → Part 2
JA.
Point of zero shear:
q(x1) = w1(3)
87(x1) = 145(3)
x1 = 5m → Part 3
W
L
=
1305
15
= 87 kN/m
JB.
Situation 9
JC.
JD.
JE.
JF.
Part 1:
JG.
q=
Force
L
q=
2 x 112 (1.5)
7
q = 48 kN/m
JH.
Part 2: Maximum shear:
VB = VA + 48(1) = 48 kN
VC = VB – 112(1.5) + 48(1.5) = -48 kN
Thus, Vmax = 48 kN
JI.
Part 3: Maximum moment:
MG = 48(1.75)(1.75/2) – 112(0.75)(0.75/2) = 42 kN-m
MH = 48(3.5)(3.5/2) – 112(1.5)(1.75) = 0
JJ.
JK.
Maximum moment = MG = 42 kN-m
JL.
Situation 10
JM.
JN.
Shear = w x Area under the influence diagram.
JO.
From the influence diagram, the uniform load must be within AB and
CD to produce maximum area. The total length is 4 + 2 = 6 m
JP.
JQ.
Situation 11
The beam is assumed hinged at B.
The force in the strut is W.
JR.
θ = arctan(0.9/2.7)
θ = 18.435°
JS.
JT.
∑FV = 0
2T sin θ = W
JU.
Part 1: T = 2kN
W = 2(2) sin 18.435°
W = 1.265 kN = 1265 N
JV.
Mass, M =
W
g
M=
JW.
JX.
JY.
1265
9.81
M = 128.99 kg
Part 2:
M = 85 kg
By ratio and proportion from the previous question:
T
85 kg
JZ.
=
2kN
128.99 kg
T = 1.318 kN
KA.
Stress, ft =
T
Ar
ft =
ft = 16.78 MPa
KB.
Part 3:
Length of rod = 2
√ 2.72 +0.92
Length of rod = 5.692 m
KC.
KD.
Situation 12
1,318
π
( 10 )2
4
KE.
KF.
KG.
KH.
KI.
KJ.
KK.
KL.
= 11 MPa
q = 5 MPa
fV = 1 MPa
p
α
KM.
=
arctan(0.75/1.2)
α
KN.
= 32°
At joint C:
2F sin α
=
20
F = 18.868 kN
KO.
Axial stress on member AC =
F
100 (150)
= 1.258
MPa
KP.
R1 = F sin α
KQ.
Considering R2:
KR.
On surface ab: θ =
KS.
Fab =
KT.
R2 = fab x Aab
KU.
Shear: Fv = 1 MPa
R 2 = Fv + A v
KV.
Situation 13
R2 = F cos α
= 10kN
α
pxq
p sin θ+q cos 2 θ
2
= 16 kN
= 32°
= 8.227 MPa
16,000 = 8.227 x y x 100
y = 19.45 mm
16,000 = 1 x (100x)
x = 160 mm
KW.
Part 1: Tension on gross area:
Ag = 252(12) = 3024 mm2
Allowable tensile stress, Ft = 0.6Fy = 0.6(248) = 148.8
MPa
KX.
P = Ft x A
Part 2: Tension on net area.
Allowable tensile stress, Ft = 0.5Fu = 200 MPa
Net area, An = (252 – 3 x 23)(12) = 2196 mm2
KY.
KZ.
P = Ft x An
LA.
LB.
P = 148.8(3024)
P = 450 kN
P = 200(2196)
P =439.2 kN
Part 3: Block shear:
Path 1:
Tension:
At =
(63x2- 2x23)(12)
At = 960
2
mm
LC.
AV =
2[63x2 +38
–
2.5x23](12)
AV =2556
mm
LD.
LE.
2
Ft = 0.5Fu = 200 MPa
FV = 0.3Fu = 120 MPa
P = Ft x At + FV x AV
P = 498.72 kN
LF.
LG.
P = 200(960) + 120(2556)
Path 2:
Tension:
At = (63x3 - 2.5x23)(12)
At = 1578 mm2
AV = [63x2 + 38
– 2.5x23](12)
AV = 1278 mm2
LH.
Ft =
0.5Fu = 200 MPa
FV = 0.3Fu
= 120 MPa
LI.
P = Ft x At + FV x
AV
P = 200(1578) +
120(1278)
P = 468.96 kN
LJ.
LK.
Situation 14
Part 1: Moment
at fixed end:
LL.
MA = -
2
w L2
12
MA = -
20 (6)
12
MA = -60 kN-m
LM.
Part 2: Maximum shear:
LN.
Vmax = RA
Vmax =
20 (6)
2
Vmax =
LO.
Part 3: Midspan deflection:
LP.
w L4
384 EI
δ
mid
=
LR.
LS.
δ
δ
LQ.
wL
2
= 60 kN
20(6)(1000)4
200,000 (67.5 x 106 )
mid
=
mid
= 5 mm
Situation 15
Total load:
w = 20 +
70 (9.81)
1000
= 20.834 kN/m
LT.
Maximum moment, Mmax = MA =
w L2
12
MB = LU.
Mmax = -
20.834 (10)2
12
LV.
Mmax = -173.613 kN-m
LW.
= RB
Maximum shear, Vmax = RA
LX.
Vmax =
LY.
Vmax = 104.168 kN
LZ.
wL
2
Part 1: Maximum bending stress:
MC
Ix
fb max =
6
173.613 x 10 (
max
=
fb
450
)
2
274.7 x 106
fb max =
142.2 MPa
MA.
stress:
Part 2: Average shearing
V
dtw
fv ave =
fv ave =
104.168 x 10
450(10)
3
fv ave = 23.15 MPa
MB.
Part 3: Maximum shearing stress:
fv max =
MC.
MD.
VQ
Ix t
Q = ∑Ay
Q = 150(15)(210+7.5) + 210(10)(105)
Q = 709.875 x 103 mm3
t = 10 m
ME.
104.168 x 10 3 (709.875 x 103)
274.7 x 106 (10)
fv max =
fv max = 26.92 MPa
MF.
Situation 16
MG.
Cc =
√
√
2π2E
Fy
=
2 π 2 (200,000)
248
Cc = 126.17
MH.
section:
Properties of built-up
A = 11,550 + 2 x
(350)(12)
A = 19,950 mm2
MI.
Ix = 266 x 10 + 2 x
6
12 ( 350 )
12
3
Ix = 351.75 x 106 mm4
rx =
MJ.
√
Ix
A
= 132.784 mm
Iy = 44.54 x 106 + 2
[
350 ( 12 )3
250 122
+( 350)(12)(
+
)]
12
2
2
Iy = 188.79 x 106 mm4
ry =
MK.
LRx =
LRy =
ML.
K y Ly
ry
√
Iy
A
K X LX
rx
=
Maximum
=
= 97.28 mm
0.5(10,000)
132.784
0.7(5,000)
97.28
KL
r
= 37.66
= 35.98
= 37.66 < Cc short column
→ Part 1
→ Part 2
α
MM.
KL
r
Cc
=
α
α
MN.
(
1−
MO.
Fa =
MP.
P = Fa x A
MQ.
2
α Fy
2 FS
)
37.66
126.17
= 0.2984
5 3
α3
+ α−
3 8
8
FS =
=
FS = 1.775
Fa = 133.476 MPa
P = 133.476(19,950)
P = 2662.8 kN
→ Part 3
Situation 17
MR.
A = 2A1
A = 2(4560)
A = 9,120 mm2
MS.
Ix = 2Ix1
Ix = 2(38.1 x
106)
Ix = 76.2 x 106
MT.
Iy = 2(Iy1 + A x12)
Iy = 2 [2.91x106
+ 4560(69)2]
Iy = 49.24 x 106
4
mm
MU.
rx =
√
√
Ix
A
rx =
Iy
A
MV.
ry =
MW.
Part 1:
Axial load = 900 kN
Axial compressive stress:
MX.
ry =
fa =
P
A
√
√
76.2 x 106
9,120
49.24 x 106
9,120
fa =
fa = 98.68 MPa
= 91.41 mm
= 73.48 mm
900,000
9,120
MY.
Part 2:
Moment about x-asis, Mx = 70 kN-m
MZ.
Bending stress:
NA.
fb =
250
)
2
76.2 x 10 6
70 x 10 6 (
M xc
Ix
fb =
fb = 114.83 MPa
NB.
Part 3: Critical slenderness ratio
( KLr )
NC.
( KLr )
y
=
x
=
1( 4000)
91.407
1( 4000)
74.48
= 43.76
= 54.44 ← Critical
ND.
Situation 18
NE.
NF.
NG.
Moment of inertia
of the beam with cover
plate:
Ix = 2662 x
NH.
10 + 2 x
6
3
260 ( 16 )
+260 ( 16 ) ( 433 )2
12
¿
Ix = 4,222 x 106
NI.
NJ.
Part 1: Bending stress due to dead load
MD =
w d L2
8
MD =
12 ( 25 )
8
MD = 937.5 kN-m
2
NK.
fb =
6
Mc
Ix
fb =
c = 441 mm
NL.
937.5 x 10 (441)
4,222 x 106
fb = 97.925 MPa
Part 2: Bending stress due to live load plus impact
Maximum moment in the beam due to two moving loads:
NM.
( PL−Pd )2
4 PL
Mmax =
NN.
base = 4.3 m
P = total load = 90 kN
Ps = smaller load = 18 kN
d = wheel
L = beam length =
25 m
NO.
Mmax =
NP.
(90 ( 25 )−18 ( 4.3 ))2
4 (90)(25)
Impact factor =
15
L+37
=
= 524.466 kN-m
15
25+ 37
= 0.2419 < 0.3
(ok)
NQ.
Maximum moment with impact:
M = Mmax(1 + Impact factor)
M = 524.466(1 + 0.2419) = 651.33 kN-m
NR.
fb max =
Mc
Ix
fb max =
651.33 x 10 6 (441)
4,222 x 106
c = 441 mm
fb max = 68 MPa
NS.
NT.
Part 3: Maximum average shearing stress.
NU.
Maximum shear occurs at the reaction where the heaviest
load is nearest.
NV.
NW.
∑MR1 = 0
NX.
Maximum shear including impact:
Vmax = 86.904 x (1 + Impact factor)
Vmax = 89.904(1+0.2419)
Vmax =107.93 kN
NY.
25 R2 = 18(20.7) + 72(25)
R2 = 86.904 kN
fv ave =
107.93 x 10
850 (15)
V
dtw
fv ave =
3
fv ave = 8.465 MPa
NZ.
OA.
Situation 19
Part 1:
Load, P = 250 kN
Allowable bearing stress of concrete, Fp = 0.35f’c = 9.625
MPa
OB.
130
P = Fp A
250,000 = 0.625 x W x
W = 199.8 mm say 200 mm
OC.
Part 2:
Actual
bearing pressure:
OD.
fp =
P
130 (200)
=
9.615 MPa
OE.
x=
100 – k = 62 MPa
OF.
t=
t=
√
√
3 f p x2
Fb
3 ( 9.61 ) ( 62 )
0.75(248)
2
t = 24.4 mm
Part 3: Web yielding stress at toe of fillet (end reaction):
OG.
fa =
P
( N +2.5 k ) t w
fa =
OH.
250,000
[ 130+2.5 ( 38 ) ] 10
fa = 111.11 MPa
OI.
Situation 20
OJ.
OJ.
OJ.
OJ.
OK.
OL.
Loads:
Dead load:
wd = beam weight + slab weight + dead load pressure
wd =
γ
c
Ab +
γ
c
Ac + pd x S
wd = 23.5(0.52)(0.28) + 23.5(2.5)(0.11) + 3(2.5)
wd = 17.384 kN/m
→ Part 1
OM.
Live load:
w1 =pl x S
w1 = 5.2(2.5)
w1 = 13 kN/m
→ Part 2
ON.
Part 3: Factored concentrated load at E:
Factored load:
wu = 1.4 wd + 1.7 wl
wu = 1.4(17.384) + 1.7(13)
wu = 46.438 kN/m
OO.
Factored concentrated load at E:
RE = wu(6.2)
RE = 46.438(6.2)
RE = 287.9 kN
OP.
OQ.
OR.
OS.
OT.
Situation 21
OU.
OV.
OW.
OX. Part 1: Shear at B due to concentrated and uniform loads:
OY.
OZ.
VB = RB1 + RB2
VB = ½(5)(7.5) + ½(270 + 270)
PA.
VB = 288.75 kN
PB.
PC.
PD.
Part 2: Maximum shear at E due to concentrated load
PE.
PF.
In Figure (2):
PG.
VE1 = 270 kN
PH.
VE2 = 270 – 270 = 0
PI.
VEmax = 270 kN
PJ.
PK.
Situation 22
PL.
PM.
m
PN.
PO.
Given:
PP.
Eccentricity, e =
PQ.
PR.
PS.
PT.
P = 1200 kN
M = 360 kN-m
L = 2.5 m
Thickness of concrete, hC = 0.7
B = 3m
Thickness of soil, hS = 1.5 m
M
P
e=
360
1200
e = 0.3 m < (B/6 = 0.5 m) OK
Net foundation pressure:
PU.
q= -
P
L(B)
±
6M
L B2
q=-
1200
2.5(3)
±
6(360)
2.5 (3)2
PV.
PW.
PX.
Part 1
PY.
Part 2
PZ.
QA.
q= -256 kPa and -64 kPa
Maximum ne foundation pressure = 256 kPa

Minimum net foundation pressure = 64 kPa

Part 3: Gross allowable soil pressure, qa:
qe = qa - γchc - γshs
QB.
QC.
QD.
QE.
QF.
QG.
QH.
QI.
qe = effective (net) soil bearing capacity = 256 kPa
256 = qa – 23.5(0.7) – 18(1.5)
qa = 299.45 kPa
Situation 23
Top bar, As1 = 3 x
π
2
2
4 (25) = 1473 mm
π
2
2
4 (25) = 2454 mm
QJ.
Bottom bar, As2 = 5 x
QK.
Flange area, Af = 1100(120) = 132,000 mm2
QL.
Effective Depth, d = 590 -70 = 520 mm
600d
600 + f y
QM.
Balance, cbalance =
= 307 mm
QN.
β1 = 0.85
QO.
top bar.
QP.
Part 1: Strengh of beam for positive momen, neglecting
since f’c < 30 MPa
As = 2454 mm2
QQ.
QR.
Assuming fs = fy:
QS.
QT.
Asfy = 0.85 f’c Ac
2454(415) = 0.86(21)Ac
QU.
Ac = 57,062 mm2
< Af
QV.
QW.
Ac = bf x a
57,602 = 1100 x a
QX.
a= 51.9 mm
QY.
QZ.
c = a / βf
c = 61 mm <
cbalance (fs = fy)
RA.
RB.
Mn = T(d - a/2)
Mn = Asfy(d - a/2)
RC.
Mn = 2454(415)
(520 – 51.9/2)
RD.
Mn = 503.2 kN∙m
RE.
RF.
Part 2: Negative moment
RG.
RH.
RI.
RJ.
RK.
RL.
RM.
RN.
RO.
Assuming fs = fy and f’s < fy:
RP.
T = Cc + C’s
RQ.
RR.
RS.
RT.
RU.
RV.
Asfy = 0.85 f’c a b + A’s f’s
RW.
fs = 600
c -d '
c
a = β1c
RX.
RY.
1473(415) = 0.85(21)(0.85c)(340) + 2454 x 600
RZ.
SA.
c = 80.68 mm < cbalance (fs = fy)
SB.
f’s = 600
SC.
SD.
a= β1c = 68.6mm
c - 70
c
SE.
SF.
SG.
= 79.425 MPa < fy (OK)
Mn = Cc(d – a/2) + C’s(d – d’)
Mn = 0.85 f’c a b (d – a/2) + A’s f’s (d – d’)
SH. Mn = 0.85(21)(68.6)(340)(520 – 68.6/2)
SI.
+ 2454(79.425)(520 – 70)
Mn = 289..88 kN∙m
SJ.
SK.
SL.
SM.
SN.
Part 3:
Factored shear, Vu = 220 kN
SO.
Nominal shear strength, Vn =
258.82 kN
SP.
SQ.
SR.
SS.
ST.
SU.
SV.
SW.
SX.
SY.
SZ.
TA.
TB.
TC.
TD.
80.68 - 70
80.68
Vu
Φ
=
220
0.85
=
Situation 24
Given: b = 300 mm
d = 480 – 70 = 410 mm
fy = 415 MPa
Bar diameter, db = 20 mm
f’c = 28 MPa
β1 = 0.85
ρmin = 1.4/fy = 0.00337
Weight of beam, wb = γcAb = 23.5(0.3 x 0.48) = 3.384 kN/m
Part 1:
Factored load, wu = 1.4(3.384 + 18) + 1.7(14)
Factored load, wu = 53.738 kN/m
Maximum factored moment:
TE.
Mu =
w u L2
8
53.738(5 )2
8
Mu =
TF.
Mu = 167.93 kN∙m
TG.
TH.
Part 2:
TI.
Mu = 280 kN∙m
TJ.
TK.
Solve for the Mumax to determine whether compression
steel is needed
TL.
TM.
TN.
TO.
TP.
TQ.
0.85 f'c β 1 600
f y (600+ f y )
ρb =
0.85(28)(0.85)(600)
415(600 + 415)
ρb =
ρb = 0.02881
ρmax = 0.75 ρb
ρb = 0.02161
ρ max f y
f' c
TR.
ωmax =
TS.
TT.
TU.
TV.
TW.
TX.
TY.
TZ.
Ru max = f’c ωmax(1 – 0.59 ωmax) = 7.274
Mu max = Φ Ru max b d2 = 330.14 kN∙m
UA.
Required Mu = 280 kN∙m < Mu max
Mu = Φ Ru max b d2
ρ=
UB.
UC.
UD.
UE.
UF.
UG.
UH.
UI.
UJ.
UK.
UL.
UM.
UN.
UO.
ωmax = 0.03203
280 x 106 = 0.90Ru(300)(410)2
Ru = 6.169 MPa
[√
0.85 f'c
2R
1- 1- u
fy
0.85 f'c
ρ=
π
4
]
[√
0.85 (28)
2(6.169)
1- 1415
0.85(28)
As = ρ b d
As =
(singly reinforced)
]
= 0.01755 > ρmin
As = 0.01755(300)(410)
As = 2159 mm2
db2 N
2159 =
π
4
(20)2 N
N = 6.9 say 7 bars
Part 3:
Part 3:
Pu = 240 kN at midspan
Mua = 167.93 kN∙m
(From part 1)
UP.
Mu =
Pu L
4
+ 167.93 = 467.93 kN∙m > Mu max
(doubly)
UQ.
Mu1 = Mu max = 330.14 kN∙m
UR.
As1 = As max = 2,658 mm2
US.
UT.
Mu2 = Mu – Mu1 = 137.79 kN∙m
UU.
UV.
Mu2 = ΦT2(d – d’)
137.79 x 106 = 0.90 As2(415)
(410-70)
UW.
As2 = 1,085 mm2
UX.
UY.
As = As1 + As2
As = 2,658 + 1,085
UZ.
As = 3,743 mm2
VA.
VB.
VC.
As =
π
4
db N
2
VD.
N = 11.9 say 12 bars
VE.
VF.
Situation 25
VG. Part 1:
VH. Geometric centroid:
VI.
A1 = 300(25) = 75,000
2
mm
VJ.
x1 = 125 mm
VK.
VL.
A2 = 180(350) = 63, 000
2
mm
VM.
x2 = 250 + 350/2
VN.
x2 = 425 mm
VO.
VP.
A = A1 + A2 = 138,000 mm2
VQ.
VR.
VS.
VT.
A x́
x́
= A1x1 + A2x2
=
75,000(125) + 63,000(425)
138,000
VU.
VV.
VW.
VX.
3,743 =
π
4 (20)2 N
x́
= 262 mm
Part 2: Plastic Centroid
VY.
The plastic centroid of a column cross section is the point through
which the resultant column load must pass to produce uniform strain in
failure. It represents he location of h resultant force produced by the steel
and concrete.
VZ.
WA.
WB.
WC.
WD.
WE.
WF.
WG.
WH.
Cc1 = 0.85 f’c A1
xc1 = 125 mm
Cc1 = 0.85(28)(75,000)
Cc1 = 1785 kN
Cc2 = 0.85 f’c A2
xc2 = 425 mm
Cc1 = 0.85(28)(63,000)
Cc1 = 1499.4 kN
Cs1 = As1 fy
Cs1 = 6 x
WI.
WJ.
xcs1 = 125 mm
π
4
(20)2(414)
Cs1 = 780.37 kN
Cs1 = 4 x
π
4
WK.
Cs2 = As2 fy
(28)2(414)
WL.
WM.
WN.
WO.
WP.
xcs1 = 516 mm
Cs1 = 1019.69 kN
Resultant Force, C = Cc1 + Cc2 + Cs1 + Cs2
Resultant Force, C = 5084.46 kN
Location of C from x-axis:
WQ.
WR.
C x́
= Cc1xc1 + Cc2xc2 + Cs1xcs1 + Cs2xcs2
x́
1785(125) + 1499.4(425) + 780.37(125) + 1019.69(516)
5084.46
=
WS.
x́
= 291.9 mm
WT.
WU. Part 3:
WV. The eccentricity of a column
load is the distance from the load to
the plastic centroid of the column
WW.
WX.
Mu = Pu x e
WY.
WZ. Mu = 3200 x 0.11
XA. Mu = 352 kN∙m
XB.
XC.
Situation 26
XD.
XE.
0.816
XF.
XG.
XH.
XI.
XJ.
XK.
XL.
XM.
XN.
XO.
XP.
XQ.
XR.
XS.
XT.
XU.
XV.
bw = 450 mm
h = 600 mm
MPa
f’c = 21 MPa
fy = 415 MPa
Allowable shear stress of concrete, Fvc =
Reduction factor, Φ = 0.85
Effective depth, d = 600 – 40 – 12 – 0.5(25)
Effective depth, d = 535.5 mm
Shear strength provided by concrete, Vc = Fvc bw d
Shear strength provided by concrete, Vc = 0.816(450)(535.5)
Shear strength provided by concrete, Vc = 196.64 kN
Part 1:
Vs = 375 kN
Vn = Vc + Vs
Vn = 196.64 + 375
Vn = 571.64 kN
XW.
XX.
XY.
XZ.
YA.
Vu = ΦVn
Part 2:
s = 230 mm
YB.
Vs =
Av fy d
s
Vs = 327.83 kN
Vn = 196.64 + 327.83
Vn = 524.47 kN
Vu = ΦVn
Vu = 0.85(524.47)
Vu = 445.8 kN
Part 3:
Vu = 450 kN
Vs =
YO.
YP.
Vu
Φ
– Vc
Vs =
s=
Av fy d
Vs
s=
– 196.64
339.29(415)(535.5)
332.78
s = 226.6 mm
Requirements for Seismic Design:
Ach = (600 -2 x 40)(450 – 2 x 40) = 192,400 mm 2
Ag = 600 x 450 = 270,000 mm2
π
2
2
4 (12) = 339.29 mm
YX.
Ash = 3 x
YY.
YZ.
hc = 450 – 2(40) – 12 = 358 mm
sh c f' c A g
-1
Ash = 0.3 f yh
A ch
(
s(358)(21) 270,000
-1
415
192,400
(
ZB.
450
0.85
Vs = 332.78 kN
YQ.
ZA.
339.29(415)(535.5)
230
Vs =
Vn = Vc + Vs
YN.
YR.
YS.
YT.
YU.
YV.
YW.
π
2
2
4 (12) = 339.29 mm
Av = 3 x
YC.
YD.
YE.
YF.
YG.
YH.
YI.
YJ.
YK.
YL.
YM.
Vu = 0.85(571.64)
Vu = 485.89 kN
)
339.29 = 0.3
)
s = 155 mm
sh c f' c
Ash = 0.09 f yh
ZC.
339.29 = 0.09
s(358)(21)
415
ZD.
s = 208 mm
ZE.
Minimum requirement according to Section 5.21.4.4.2:
a) b/4 = 112.5 mm
b) 6(25) = 150 mm
c) 100 +
350 - h x
3
ZF.
ZG.
hx = ½(600 – 2 x 40) – ½(12) + ½(25) + ½(12)
hx = 272.5 mm
ZH.
100 +
ZI.
350 - h x
3
= 126 mm
Therefore, use s = 112.5 mm
ZJ.
ZK.
Situation 27
ZL.
ZM.
The compressive stress at the top and bottom of the beam due to P is
given by the formula:
ZN.
fc = -
Pe
bh
6Pe e
±
bh
2
(+) for top fiber, (-) for
bottom fiber
ZO.
ZP.
Effective prestressing force, Pe = P – 15%P
Effecting prestressing force, Pe = 0.85(1500) = 1275 kN
ZQ.
ZR.
ZS.
Part 1:
When e = 0;
ZT.
fc =
ZU.
ZV.
ZW.
P
bh
fc =
1275 x 10
300(600)
3
fc = -7.08 MPa
Part 2:
ZX.
fc top = -
6Pe e
P
bh
+
bh
1275 x 103
300(600)
fc top = -
2
+
6(1275 x 103 )(120)
300(600 ) 2
ZY.
ZZ.
fc top = 1.417 MPa
AAA.
fc bot = -
P
bh
6Pe e
-
bh
fc bot = -
2
1275 x 103
300(600)
-
6(1275 x 103 )(120)
300(600 ) 2
AAB.
fc bot = -15.583 MPa
AAC.
AAD.
Part 3:
AAE.
Since the stress at the top is zero, P acts at h/3 from the bottom
of the beam, or e = h/2 – h/3 = h/6
AAF.
e = 600/6
AAG.
e = 100 mm
AAH.
AAI.
AAJ.
Situation 28
Loads:
wd = pd x b = 2.5(2.5) = 6.25 kN/m
wl = pl x b = 6(2.25) = 15 kN/m
wb =
AAK.
AAL.
γ
c
A = 23.5 x (220,000/10002) = 5.17 kN/m
Total service load, w = wd + wl + wb = 26.42 kN/m
Moment at midspan, M =
w L2
8
kN-m
AAM.
Stress due to initial presses:
e = y2 – y3 = 270 - 75 = 195 mm
=
26.42 ( 8 )
8
2
= 211.36
AAN.
ftop = -
P Pec
+
A
I
ftop = -
2( 750,000) (220,00 x 195)(90)
+
220,000
1890 x 106
ftop = 7.11 MPa
AAO.
fbot = -
P Pec
−
A
I
fbot = -
2( 750,000) (220,00 x 195)(270)
+
220,000
1890 x 106
fbot = - 48.604 MPa
AAP.
→ Part 1
Stress at midspan due to loads:
ftop = -
Mc
I
211.36 x 106 ( 90)
1890 x 106
ftop = -
ftop = - 10.065 MPa
AAQ.
fbot =
6
Mc
I
fbot =
211.36 x 10 (270)
1890 x 106
fbot = 30.194 MPa
AAR.
prestress:
Part 2: Stress at bottom, fibers due to service loads and
Note: There is a loss of prestress of 20% at service loads.
AAS.
fbot = 30.194 – 48.609(1 – 0.20)
fbot = -8.689 MPa
AAT. Part 3: Additional service loads to “zero” the stress at the bottom at
midspan
AAU. The additional load must induce a stress of 8.689 MPa at the
bottom fibers.
AAV. fbot =
Mac
I
8.689 =
M ( 270)
1890 x 10 6
M = 60.822 kN-m
M=
w a L2
8
60.822 =
w = 7.603 kN/m
wa (8)2
8
AAW. pa =
w
b
7.603
2.5
pa =
pa = 3.04 kPa
AAX.
AAY.
Situation 29
Dead load, PD = 740 kN
Live load, PL = 460 kN
Factored load, PU = 1.4 PD + 1.7 PL = 1,818 kN
AAZ. Factored base pressure, qU =
Pu
A ftg
=
1,818
2.4 (2.4)
= 315.625 kPa
ABA. Effective depth, d = 450 - 90 = 360 mm
ABB. Parts 1 & 2: Factored shear on footing, Vu:
ABC.
ABD.
d = 0.36 m
ABE.
Wide beam shear:
x = ½ (2.4 – 0.35) – d = 0.665 m
ABF.
Vu = qu x Area
ABG.
Vu = 315.625 x (2.4)(0.665)
Vu = 503.74 kN → Part 1
Punching shear:
x1 = 0.4 + d = 0.76 m
x2 = 0.35 + d = 0.71 m
ABH.
Vu = qu x Area
0.76(0.71)]
Vu = 1647.7 kN → Part 2
Vu = 315.625 x [2.42 –
ABI. Part 3:
ABJ.
x = ½ (2.4-0.35) = 1.025 m
ABK.
(x/2)
Mu = qu x 2.4(x)
Mu = 315.625 (2.4)
(1.025)2/2
Mu = 397.924 kN-m
ABL.
Mu = φ Ru b d2
ABM.
Ru =
397.924 x 10 6
0.90 ( 2400 )( 360 )2
ABN.
ABO. ρ =
Ru = 1.421 MPa
[ √
]
[ √
0.85 f 'c
2 Ru
1 − 1−
fy
0.85 f 'c
ρ=
0.85(20.7)
2(1.421)
1− 1−
275
0.85 (20.7)
]
ρ = 0.0054 > (ρmin = 1.4/fy = 0.00509)
ABP.
As = ρ b d
As = 0.0054(2400)(360)
As = 4663 mm2
ABQ. N =
A¯¿
As
¿
N=
N=
4663
π
( 20 )2
4
14.8 sa 15 bars
ABR.
ABS.
Situation
30
ABT. Answers:
ABU.
Part 1: Static
ABV.
Part 2: Resultant
ABW. Part 3: Kinetic
ABX.
ABY.
31:
ABZ.
Situation
Answers:
Part 1: Hooke’s
Law
ACA.
Part 2: Poisson Ratio
ACB.
Part 3: Young’s
Modulus
ACC.
ACD.
ACE.
ACF.
ACG.
ACH.
ACI.