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.
fV =
π 4
(16)2 = 804.248 mm2
PV = P = 148 kN
PV AV
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. IJ.
IK.
ρmax = 0.75 ρb = 0.01598 ωmax =
ρmax f y f 'c
= 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
´x
= 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
260 ( 16 )3 +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 x´
x´
= A1x1 + A2x2 =
75,000(125) + 63,000(425) 138,000 VU. VV. VW. VX.
3,743 =
π 4 (20)2 N
x´
= 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 x´
= Cc1xc1 + Cc2xc2 + Cs1xcs1 + Cs2xcs2
x´
1785(125) + 1499.4(425) + 780.37(125) + 1019.69(516) 5084.46
=
WS.
x´
= 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 ±
b h2
(+) 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.
