2017-ce_may.pdf

CE BOARD May 2017

PROBLEM 1:

A 120 kg man walks on a 4 m. plank resting on rough surfaces at A and B. Given: Coefficient of friction at A = 0.40 Coefficient of friction at B = 0.20 ➀ At what distance x will the plank start to slide? ➁ What is the friction force (N) at A? ➂ Find the vertical force (N) at B?

Solution: ➀ Distance x when the plank start to slide ΣMB = 0 R A Cos 8.2˚ (4) = 120(9.81)x R A = 297.43x Using Sine Law: RB 120(9.81) = Sin8.2˚ Sin160.49˚ R B = 502.75 N RA 120(9.81) = Sin11.31˚ Sin160.49˚ R A = 691.28 N R A = 297.34x 691.28 = 297.35x x = 2.32 m.

➁ Friction force at A FA = 691.28 Sin 21.8˚ FA = 256.72 N

➂ Vertical force at B NB = RB Cos 11.31˚ NB = 502.75 Cos 11.31˚ NB = 493 N

CE BOARD May 2017

PROBLEM 2:

Pre stressed hollow core slabs with typical section shown in the figure are used for the flooring of a library. Properties of the slab are as follows: A = 1.4 x 105 mm2 St = Sb = 6.8 x 106 mm3 a = 1.2 m. b = 200 mm

b

The slab is pre stressed with 820 kN force at an eccentricity e = 63 mm below the neutral axis of the section. Slab weight = 2.7 kPa Super imposed (DL) = 2.0 kPa Live load = 2.9 kPa

a

The slab is simply supported on a span of 8 m. Allowable stresses at service loads are 2.0 MPa in tension and 15.5 MPa in compression. Consider 15% loss of pre stressed at service loads. ➀ ➁ ➂

➀

Compute the stress at the top fibers of the slab at the ends due to the initial pre stress force. Determine the resulting stress at the top fibers of the slab at midspan due to the loads and pre stress force. Determine the maximum total load (kN/m) including its own weight, that the slab can be subjected to if the allowable stress at service loads are not to be exceeded.

Solution:

ft = ft = -

P M + A S 820000 1.4 x 105

1.2 m

+

820000(63) 6.8 x 106

ft = +1.74 MPa (tension) ➁

200 mm

Stress at the top fibers of the slab at the ends due to the initial pre stress force

N.A.

P

M=P e

Resulting stress at the top fibers of the slab at midspan due to the loads and pre stress force LL = 2.9(1.2) = 3.48 kN/m Slab weight = 2.7(1.2) = 3.24 kN/m Super imposed load = 2(1.2) = 2.4 kN/m Total load = 3.48 + 3.24 + 2.4 = 9.12 kN/m WL2 9.12(8)2 M= = = 72.96 x 106 N.mm 8 8

CE BOARD May 2017

PROBLEM 2:

Continuation

P Pe C MC + A I I P Pe M ft = - + A St St

ft = -

P = 820(0.85)

ft = - 9.25 MPa (compression)

ft = -

72.96 x 106

697000(63)

+ 1.4 x 105 6.8 x 106 6.8 x 106 ft = - 4.98 + 6.46 - 10.73

P = 697 kN (after loss of stress)

➂

697000

Maximum total load (kN/m) including its own weight, that the slab can be subjected to if the allowable stress at service loads are not to be exceeded - 4.98 + 6.46 - ft = - 15.5

ft = 16.98 MPa - 4.98 - 6.46 + fb = 2.0

-15.5

+

+

fb = 13.44 MPa

-f t

+Pe/S b

-P/A

=

When fb = 13.44 MPa fb =

M Sb

13.44 =

-P/A M

6.8 x 106 M = 91.39 kN.m When ft = - 16.98 ft =

-f t

+6.46

-4.98

M St

16.98 =

+2.0

+f b

-Pe/S b

+

-15.5

+

=

M

6.8 x 106 M = 115.46 kN.m Safe moment = 91.39 kN/m WL2 M= 8 W(8)2 91.39 = 8 W = 11.42 kN / m

-4.98

-6.46

+f b

+2.0

CE BOARD May 2017

PROBLEM 3:

A hollow circular steel column is supported on a steel base plate and a concrete pedestal. Column ends are hinged and sidesway is prevented. Given: Column axial load = 780 kN Column inside diameter = 250 mm ➀ ➁ ➂

Allowable column compressive stress = 55 MPa Allowable concrete bearing stress = 10 MPa

Calculate the critical slenderness ratio of the column if its thickness is 10 mm and a height of 3 m. Find the minimum required thickness (mm) of the column. What is the safe diameter (mm) of the base plate?

Solution: ➀

Critical slenderness ratio of the column if its thickness is 10 mm π ⎡(270)4 - (250)4 ⎤ = 69.12 x 106 I= ⎦ 64 ⎣

A=

π ⎡(270)2 - (250)2 ⎤ = 8168.14 ⎦ 4 ⎣

r=

I 69.12 x 106 = = 91.99 A 8168.14

10 250 10

KL (1)(3000) = = 32.61 mm r 91.99 ➁

Minimum required thickness of the column

σ= 55 =

P A 780,000 π ⎡(250 + 2t)2 - (250)2 ⎤ ⎦ 4 ⎣

(250 + 2t)2 = 80,556.85 250 + 2t = 283.83 t = 16.92 say 18 mm

➂

t 250

t

Safe diameter of the base plate P σ= A

780,000 π 2 d 4 d = 315.14 say 320 mm

10 =

CE BOARD May 2017

PROBLEM 4: Given:

P = 40 kN W = 8 kN/m L1 = 2.5 m L2 = 0.3 m

w P 3

0. L2= L

1=2

.5

➀ Which of the following gives the bending moment at C? ➁ Which of the following gives the twisting moment at C? ➂ Which of the following gives the shear at C?

Solution:

➀ Bending moment at C Moment at C : 8(25)2 M= + 40(2.5) 2 M = 125 kN.m.

➁ Twisting moment at C MC = 40(0.3) MC = 12 kN.m. ➂ Shear at C VC = 40 + 8(25) VC = 60 kN

CE BOARD May 2017

PROBLEM 5: A 250 mm thick concrete wall is to be centrally located on a footing. ➀ How much is the required least resisting moment so that there will be no tensile stress in the footing if the loads on the footing are as follows. Total vertical load = 432 KN Overturning moment = 288 KN.m Footing width = 4 m. ➁ Find the smallest footing width that will prevent uplift. Total vertical load = 360 KN Resisting moment = 756 KN.m Overturning moment = 252 KN.m ➂ Which of the following gives the resulting maximum soil pressure in MPa if the load on the footing are as follows. Total vertical load = 250 KN Resisting moment = 470 KN.m P=432 kN Overturning moment = 245 KN.m Footing width = 3.0 m. OM=288

RM

Solution:

➀ Least resisting moment so that there will be no tensile stress in the footing. 4m

R y x = RM - OM ⎛ 4⎞ 432 ⎜ ⎟ = RM - 288 ⎝ 3⎠ RM = 864 KN.m

2m

O

B/3

e

Ry

x=4/3

2m B/3

B/3

CE BOARD May 2017

PROBLEM 5:

Continuation

➁ Smallest footing width that will prevent uplift. R y x = RM - OM

P = 360 kN OM = 252 kN.m

RM = 756 kN.m

⎛ B⎞ 360 ⎜ ⎟ = 756 - 252 ⎝ 2⎠ B = 2.8 m. Ry

O

x =B/2

B

➂ Resulting maximum soil pressure: P=250 kN

R y x = RM - OM 250 x = 470 - 245 x = 0.90 Ry = 250 =

OM=254 kN.m

RM=470 kN.m

( )

2.7 fmax 2

( )

2.7 fmax 2

3m 3(0.9)=2.7

fmax

fmax = 185 kPa

0.90

Ry =350

CE BOARD May 2017

PROBLEM 6:

At ultimate condition, the uniformly distributed loads for beams KLMNO and PQRST are as follows. Dead Load,WDL = 15KN/m (beam weight included) Live Load, WLL = 19 KN/m. Given L = 6 m. s = 2.5 m. t = 100 mm bxh = 300x450 mm For maximum stress, apply full live load. ➀ How much is the maximum negative moment in beam KLMNO? ➁ Find the maximum shear in beam KLMNO. ➂ What is the maximum positive moment in beam MN?

Solution: ➀ Maximum negative moment in beam KLMNO L = 6 – 0.3 L = 5.7 Wu = 15 + 19 Wu = 34 KN/m Wu L2 Mu = 10

34(5.7)2 Mu = = 110.5 kN.m. 10 ➁ Maximum shear 1.15WuL u Vmax = 2

Vmax =

1.15(34)(5.7 = 111.4 kN 2

➂ Maximum positive moment

Wu L2 Mu = + 14 34(5.7)2 Mu = + = 78.9 kN.m. 14

CE BOARD May 2017

PROBLEM 7:

The line of action of the force P coincides with the axis of the angle which is at distance “a” from the back of the connected leg. Do not include end turns. P Given: Properties of vertical member gusset plate 2 agles 75 mm x 75 mm x 8 mm thick. A = 21 mm. Area of 2 angles = 2290 mm2 Yield stress, Fy = 248 MPa. L2 L1 Allowable weld shear stress: Fvw = 93 MPa. Weld thickness = 5 mm ➀ Which of the following gives the maximum tensile stress a=25 capacity of the angles? end returns ➁ Given: L1 = 50 mm L2 = 130 mm. Which of the following gives the allowable load P based on the capacity of the welded connection? ➂ Given: P = 270 KN Which of the following gives the required length of weld L2?

Solution:

➀ Maximum tensile stress capacity of the angles P = 0.60 FyAg P = 0.60 (248)(2290) = 340.75 kN ➁ Allowable load based on capacity of welded connections P = 0.707 tLFv 2 P = 0.707(5)(50+130)(93) 2 P = 118352 N P = 118.35 kN ➂ Length of weld L2 if P = 270 kN ΣM A = 0

135000(54) = P2 (75) P2 = 97200 P2 = 0.707tL 2Fv 97200 = 0.707(5)L 2 (93) L 2 = 295.66 mm say 300 mm

CE BOARD May 2017

PROBLEM 8:

To prevent the ladder from sliding down, the man exerts a horizontal force at C. Given: L1 = 1.5 m. L2 = 4.5 m. Ø = 30˚

➀ Find the horizontal force at C, which the man should exert if the weight of the ladder is 500 N. Assume that surface A and B are frictionless. ➁ If the weight of the ladder is 500 N, what is the reaction at A? Surface A and B are frictionless. ➂ Find the maximum weight of the ladder that can be held from sliding if the man exerts force C = 200 N. Coefficient of friction at A is 0.30 and at B is 0.20.

Solution: ➀ Horizontal force at C that the maximum should exert to prevent the ladder from sliding down ∑Fv = 0 RA = 500 F = RB ∑MA = 0 6 Sin 30˚ RB = 500(3)Cos 30˚ + F Cos 60˚ (1.5) 3 F = 1300 + 0.75 F 2.25 F = 1300 F = 577.7 N ➁ Reaction at A: if A and B are frictionless RA = 500 N

CE BOARD May 2017

PROBLEM 8:

Continuation

➂ Maximum weight of the ladder that can be held from sliding if the man exerts a force C = 200 N. Coefficient of friction at A is 0.30 and at B is 0.20. tan θ = 0.30 tan ß = 0.20 θ = 16.70˚ ß = 11.31˚ ΣFh = 0 R A Sin 16.70 + 200 = R B Cos 11.31˚ R A + 696 = 3.41 R B R A = 3.41 R B - 696 ΣFv = 0 R A Cos 16.7˚ + R B Sin 11.31˚ = w (3.41 R B - 696)Cos16.7˚ + R BSin 11.31˚ = w 327 R B - 666.64 + 0.196 R B = w 3466 R B = w + 666.64 ΣM A = 0 200(1.5) Sin 30˚ + w (3) Cos 30˚ = R B Sin 11.31˚(6) Cos 30˚ + R B Cos 11.31˚(6) Sin 30˚ 150 + 2.6 w = 1.02 R B + 2.94R B 150 + 2.6 w = 3.96R B 3.96R B = 150 + 2.6 w

by 3.466

3.466R B = 666.64 + w

by 3.96

13.725R B = 520 + 9 w 3.466R B = 2640 + 3.96 w 520 + 9w = 2640 + 3.96 w 5.04 w = 2120 w = 420.6 say 421 N

CE BOARD May 2017

PROBLEM 9:

The water supply pipe shown is suspended from a cable using a series of close and equal spaced hangers. The length of the pipe is supported by the cable is 60 m. The total weight of the pipe filled with water is 6.5 KN/m. ➀ What is the maximum sag at the lowest point of the cable which occurs at mid length if the allowable tensile load in the cable is 2550 KN? ➁ The sag of the cable at mid length is 3 m. Find the maximum axial stress (MPa) if the cable diameter is 75 mm. ③ The sag of the cable at mid length is 2 m. If the allowable tensile load in the cable is 2000 KN, how much additional load can the cable carry?

Solution:

➀ Maximum sag at the lowest point of the cable if the allowable tensile load in the cable is 2550 KN H2 = (2550)2 – (195)2 H2 = 2542.53 KN

ΣMA = 0 2542.53(y) = 195(15) y = 1.15 m. (maximum sag) ➁ Maximum axial stress if the sag at mid length is 3 m. and the diameter of cable is 25 mm ΣMA = 0 3H = 195(15) H = 875 T2 = (145)2 + (975)2 T = 994.31 KN T = A s fs π

2 994310 = 4 (75) fs fs = 225 MPa

③ Additional load that the cable can carry if the sag is 2m and the allowable tensile load in the cable is 2000 KN ΣMA = 0 H = 225 w (2000)2 = (30w)2 + (225 w)2 w = 8.81 KN.m Additional load = 8.81 – 6.5 = 2.31 KN/m

CE BOARD May 2017

PROBLEM 10: Given:

L=8m S1 = 2.5 m S2 = 3 m Total dead load = 4.6 kPa Live load = 4.8 kPa U = 1.2D + 1.6L Beam KLMNO is to be analyzed for the maximum forces at ultimate condition.

S1=2.5 m

A

B

C

D

E

K

L

M

N

O

F

G

H

I

J

S2=3 m

wL

➁ Which of the following gives the span moment when all spans are fully loaded? ➂ Which of the following gives the max. shear at span NO?

wL M

L

L

K

wL L

L

L

L

R B=1.143wL

R C=0.928wL

R D=1.143wL

0.393wL

0.536wL

0.464wL

0.607wL

L R A=0.38wL

L R B=1.223wL

L

L

R C=0.357wL

R D=0.598wL

0.603wL

0.38wL

O

R E=0.442wL

0.558wL

0.464wL

0.607wL +0.0772wL 2

+0.0364wL

2

0.397wL

0.62wL +0.072wL

2

+0.0364wL 2

0.393L

-0.0714wL 2

-0.1071wL 2

0.38L

-0.0179wL -0.1205wL 2 0.603L

2

K

L

M

L R A=0.446wL R B=0.572wL

O

L

L

R C=0.464wL

R D=0.572wL

0.482wL

0.518wL

+0.0805wL

2

MOMENT -0.0536wL 2

-0.0357wL

R E=0.054wL

0.054wL

0.016wL

+0.0996wL 2

0.446L

0.393L

N

L

SHEAR

0.442L

-0.1071wL 2

wL

0.554wL

-0.058wL 2

+0.0772wL 2

0.536L

0.442wL

+0.0977wL 2

+0.0611wL 2

MOMENT

0.040wL

0.393wL

0536wL

MOMENT

0.446wL

SHEAR

R E=0.393wL

SHEAR

wL

N

O

R A=0.393wL

wL M

wL N

0.536L

wL

L4=8

L3=8

wL

K

➀ Which of the following gives the maximum shear at span LM?

L2=8

L1=8

2

-0.0536wL 2 0.518L

CE BOARD May 2017

PROBLEM 10:

Solution:

Continuation

➀ Max. shear at span LM Using Live Load Patterns table. WDL = 4.6(2.75)(1.2) WDL = 12.65 kN/m WLL = 4.8(2.75)(1.6) WLL = 13.2 kN/m Dead load reactions: VDL = 0.536 WDL (L) VDL = 0.536 (12.65)(8) VDL = 54.24 KN VLL = 0.603 WL(L) VLL = 0.603(13.2)(8) VLL = 63.68 kN Max. shear at panel LM. Vmax = 1.2 VDL + 1.6 VLL Vmax = 54.24(1.2) + 63.68(1.6) Vmax = 166.97 kN say 167 kN

wL K

wL

wL

wL

M

L

L

N

L

O

L

L

R A=0.393wL

R B=1.143wL

R C=0.928wL

R D=1.143wL

0.393wL

0.536wL

0.464wL

0.607wL

R E=0.393wL

SHEAR 0.464wL

0.607wL +0.0772wL 2

+0.0364wL

2

0.393wL

0536wL

+0.0364wL 2

+0.0772wL 2

MOMENT 0.393L

-0.0714wL 2

-0.1071wL 2

0.393L

-0.1071wL 2

0.536L

0.536L

WDL=12.65 kN/m K

L

M

8 RA=0.393wL

N

8 RB=1.143wL

8 RC=0.928wL

8 RD=1.143wL

➁ Max. span moment when all spans are fully loaded wL wL Consider exterior span KL. Dead Load: K L M WDL = 12.65 kN/m3 L L L + M = 0.0772 w L2 R =0.38wL R =1.223wL R =0.357wL + M = 0.0772(12.65)(8)2 0.603wL 0.38wL + M = 62.50 kN.m SHEAR Live Load: 0.397wL WLL = 13.2 kN/m 0.62wL + M = 0.0996 w L2 +0.072wL +0.0611wL + M = 0.0996(13.2)(8)2 MOMENT + M = 84.14 -0.0179wL 0.38L -0.1205wL Max. M = 1.2 MDL + 1.6 MLL 0.603L Max. M = 1.2(62.50)+1.6(84.14) Max. M = 75 + 134.63 Max. M = 209.63 kN.m A

B

C

2

2

2

2

O

RE=0.393wL

wL N

O

L R D=0.598wL

R E=0.442wL

0.558wL

0.040wL

0.442wL

+0.0977wL 2

-0.058wL 2

0.442L

CE BOARD May 2017

PROBLEM 10:

Continuation

➂ Max. shear at span NO

wL

WDL = 12.65 kN/m VDL = 0.607 wDL L

K

L

VDL = 0.607(12.65)(8) VDL = 61.42 KN Live Load: WLL = 21.12 kN/m VLL = 0.62 WL(L)

VNO = 1.2VDL+1.6VLL VNO = 1.2(61.42)+1.6(65.47) VNO = 73.71 + 104. 76

M

L

N

L

R A=0.446wL R B=0.572wL

O

L

L

R C=0.464wL

R D=0.572wL

0.482wL

0.446wL

0.518wL

0.554wL +0.0996wL 2

R E=0.054wL

0.054wL

0.016wL

SHEAR

VLL = 0.62(13.2)(8) VLL = 65.47 kN

wL

+0.0805wL

2

MOMENT 0.446L

-0.0536wL 2

-0.0357wL

2

-0.0536wL 2 0.518L

VNO = 178.5 KN Note: if span MNO was loaded and KL was loaded, the same value of shear at L and N could be used.

CE BOARD May 2017

PROBLEM 11:

A box column is built up by welding two channels at their flanges. Given: Properties of each channel. Depth = 250 mm Web thickness = 17 mm. Flange width = 77 mm. Flange Thickness = 11 mm. Area = 5690 mm2 Moment of inertia: Ix = 43x106 mm4 Iy = 1.6x106 mm4 Distance from the centroid y-axis to the back of the web = 16.5 mm Modulus of elasticity E = 200 MPa. Yield stress, Fy = 248 MPa. Column height = 6 m. The column is pin connected at the ends and side way is prevented K = 1.0 ➀ Calculate the maximum load that the column can support without buckling. ➁ What is the maximum axial load so that the yield stress will not be exceed? ➂ Find the largest allowable axial load that the column can support before it either begins to buckle or the material yields

Solution : ➀ Maximum load that the column can support without buckling IX = 43X106 (2) IX = 86X106 mm4 Iy = [1.6x106 + 5690(60.5)2 ] 2 Iy = 44.85x106 mm4 π 2EI π 2 (200000)(44.85)106 Pcr = = = 2459 KN (KL u )2 (1)(6000)2

➁ Maximum axial load so that the yield stress will not be exceeded P = A Fy P = 5690(2)(248) P = 2822240 N P = 2822.24 kN

➂ Largest allowable axial load that the column can support it either begins to buckle or the material yields. P = 2459 kN (smaller of “a” and “b”)

CE BOARD May 2017

PROBLEM 12:

A simply supported girder spans 10 m. The girder carries 3 concentrated loads, each 30 KN at 2.5 m. spacing. The girder also carries a uniformly distributed load of 15 KN/m throughout its length. Given: Girder properties. Section: = 2 460 x 97 kg/m Flange width, bf = 193 mm 2 Area, A = 12320 mm Flange thickness = 19mm Depth, d = 465 mm Web thickness = 11 mm Moment of inertia: Ix = 445x106 mm4 Iy = 23x106 mm4 Yield strength, Fy = 248 MPa. Rt (radius of gyration of the flange components plus 1/3 of the web area in compression) = 50 mm. ➀ Which of the following gives the maximum bending stress. ➁ What is the maximum web shear stress. ➂ Calculate the maximum shear stress at the neutral axis of the section.

Solution:

➀ Maximum bending stress 97(9.81) W = 15+ = 15.95 KN/m 1000 2 R1 = 30(3)+15.95(10) R1 = 124.75 KN 124.75+84.875) 54.875+15) (2.5) + (2.5) 2 2 M = 349.375 KN.m M=

MI 349.375(232.5)(10)6 fb = = = 182.5 MPa C 445x106

➁ Maximum web shear stress V 124750 fv = = = 24.4 MPa. dt w 465(11) ➂ Maximum shear stress at the neutral axis of the section Q = 193(19)(223)+213.5(11)(106.75) Q = 1068443 τ=

VQ 124750(1068443) = = 27.2 MPa. Ib 445x106 (11)

CE BOARD May 2017

PROBLEM 13:

The truss shown is on a roller support at A and on hinged at support B. It is subjected to a lateral load H = 12 KN. ➀ Determine the reaction at A. ➁ Determine the force in member AE. ➂ Determine the reaction at B. Solution: ➀ Reaction at A. ΣMB = 0

7 R A = 12(3.5) R A = 6 KN ➁ Force in member AE.

3.5 5.5 θ = 32.47˚

AC

tanθ =

3.5 tan ß = 1.5 ß = 66.80˚ ΣFh = 0

66.80˚ A

AE 32.47˚

6

AE Cos 32.47˚ = AC Cos 66.80˚

➂ Reaction at B

AC = 2.142 EA

R Bx = 12

ΣFv = 0

R By = 6

AC Sin 66.80 = AE Sin 32.47˚+ 6

R B = (12)2 + (6)2

2.142 EA Sin 66.80˚ = AE 32.47˚+ 6

R B = 13.4 KN

AE = 4.2 KN (tension)

CE BOARD May 2017

PROBLEM 14:

A column 800 mm. in diameter is reinforced with 12 – 25 mmø bars and 12 mm diameter spirals spaced at 72 mm on centers. Use ø = 0.75 fc’ = 30 MPa fy = 413 MPa for main bars, fyt = 275 MPa for spirals. Allowable shear stress fv = 0.90 MPa. ➀ What is the nominal shear strength provided by the concrete? ➁ What is the nominal shear strength provided by the shear reinforcement? ➂ Find the shear stress in the column if Vu = 800 kN.

Solution: ➀ Nominal shear strength provided by concrete Vc = 0.90 bwd Vc = 0.90(800)(0.8)(800) Vc = 460800 N Vc = 461 kN

➁ Nominal shear strength provided by shear reinforcement π A v = (12)2 = 113 4 A fy d S= v t Vs

d = 0.80 D d = 0.80(800) d = 640 72 =

113(275)(640) Vs

Vs = 276 kN

25 mm ø

12 mm ø

D=800

➂ Shear stress in the column if Vu = 800 kN Vu fv = ø bw d

fv =

800000 0.75(800)(0.80)(800)

fv = 2.08 MPa

CE BOARD May 2017

PROBLEM 15:

A 15 m. long precast pile is to be lifted at two points from casting bed. ➀ At what equal distance from the ends should the pile be lifted so that the maximum bending stress is the least possible? ➁ At what equal distance from the ends should the pile be lifted so that the resulting shear is smallest? ➂ As the pile is being lifted, its left end is laid on the ground while it is supported at 2.5 m. from the right end. At this instance, what is the resulting maximum positive moment due to its weight of 8.5 KN/m. Solution: ➀ Equal distance from the ends where the pile could be lifted for minimum bending stress 2R = 15 w R = 7.5 w For minimum bending stress to occur M1 = M2 −wx(x) 2 −wx 2 M1 = 2 (w)(7.5 − x)(15 − 2x) M 2 = M1 + 2(2) M1 =

−wx 2 (w)(7.5 − x)(15 − 2x) M2 = + 2 4 M1 = M 2 (do not include the sign of M1 ,use absolute values only)

2wx 2 w = (7.5 − x)(15 − 2x) 2 4 2 4x = (7.5-x)(15-2x) 4x 2 = 112.5 - 15x - 15 x + 2x 2 2x 2 = 112.5 - 30x x 2 + 15x - 56.25 = 0 x = 3.10 m.

CE BOARD May 2017

PROBLEM 15:

Continuation

➁ Equal distance from the ends where the piles will be lifted so that the resulting shear is the smallest wx = 7.5w – wx x = 7.5 – x x = 3.75 ➂ Resulting positive moment if the pile is lifted at 2.5 m. from the right end due to its weight of 8.5 KN/m

ΣMR = 0 1

12.5 R 2 = 8.5(15)(7.5) R 2 = 76.5 R1 + R 2 = 8.5(15) R1 = 51 8.5x = 51 x=6 Positive moment M=

51(6) = 153 KN.m. 2

CE BOARD May 2017

PROBLEM 16:

Given: a = 2 m., b = 5 m., c = 5 m., d = 2 m. A

B

a=2

C

D

b=5

c=5

E

d=2 V=12

V=4 V=4 V=12





➀ Which of the following gives the reaction of D? ➁ Find the load at C. ➂ How much is the maximum moment. Solution: ➀ Reaction at D RD = 4 + 12 = 16 kN

P

w=6

A

B

w=6

C

a=2

RB=16

b=5

D

c=5

d=2

RD=16 12

➁ Load at C P = 4 + 4 = 8 kN ➂ Maximum moment M = - 12 kN.m.

4 -4 -12 8

-12

E

-12

CE BOARD May 2017

PROBLEM 17:

A wooden log is to be used as a foot bridge to span a 2m. gap. The log is to support a concentrated load of 27 kN. Allowable shear stress = 1.2 MPa. Allowable bending stress = 6.3 MPa. ➀ What is the diameter of the log needed for a maximum shear? ➁ Find the diameter of the log needed so that the allowable bending stress is not exceeded. ➂ A 150 wide rectangular log is to be used instead of the circular log. What is the required minimum depth of the log so as not to exceed the allowable bending stress?

Solution: ➀ Diameter of log needed for a maximum shear. For maximum shear, the 27 KN should be placed at the end of the log. ΣMR 2 = 0 2R1 = 27 KN(2) R1 = 27 kN 4V 3A 4(27000) 1.2 = 3A A = 30000 fv =

A=

π 2 D 4

π 2 D 4 D = 195.4 mm say 200 mm. 30000 =

➁ For a maximum moment placed the load at the center PL M= 4 27 M = (2) 4 M = 13.5 kN.m. MC f= I 13.5 x 10 6 (D / 2) 6.3 = πD4 64 D = 280 mm.

➂ Minimum depth of rectangular log

6M b d2 6(13.5)10 6 6.3 = 150 d2 d = 300 mm. f=

CE BOARD May 2017

PROBLEM 18: Given: As = 8 – 28 mm∅ As’ = 4 – 28 mm∅ ds = 12 mm diam. ties h1 = 125 mm h2 = 475 mm a = 55 mm fc’ = 28 MPa fyb = 415 MPa (longitudinal bars) fyv = 275 MPa (ties) Shear strength reduction factor = 0.75 Clear concrete cover = 40 mm Specified maximum aggregates size in the concrete mixture = 20 mm ➀ Find the minimum width of beam “b” required to satisfy on cover requirements ➁ Find the minimum width of beam “b” adequate for a factor shear force Vu = 600 kN if the spacing of 12 mm diameter ties is 50 mm. ➂ If Vu = 450 kN and spacing of 12 mm diameter ties is 70 mm what is the required minimum width of the beam “b” mm?

Solution: ➀ Minimum width “b” required for cover requirements 12 mm ø

Value of x : should be the larger value of the following. ➊ bar diameter = 28 ➋ 25 mm. 4 1 ➌ 1 3 max. aggregate size = 3 (20) = 26.67

use x = 28 mm. b = 40(2) + 12(2) + 28(4) + 3(28) = 300 mm.

28

40

x

28

x b

x

40

CE BOARD May 2017

PROBLEM 18:

Continuation

➁ Value of b if Vu = 600 KN and spacing of tie wires = 50 mm 28 55 d = 600 - 40 - 12 2 2 d = 506.5 mm Vu = Vs + Vc ø A v fyt d S= Vs

π (12)2 (2)(275)(506.5) 50 = 4 Vs Vs = 630121 Vu - Vs ø 600000 Vc = - 630121 = 169879 0.75 Vc = 0.17λ fc 'b w d Vc =

169879 = 0.17(1) 28b(506.5) b = 373 say 370 mm

➂ Value of b if Vu = 450 kN and spacing of tie wires is 70 mm

A v fyt d

S=

Vs

π (12)2 (2)(275)(506.5) 70 = 4 Vs Vs = 450087 Vu - Vs ø 450000 Vc = - 450087 0.75 Vc = 149912 Vc =

Vc = 0.17λ fc 'b w d 149912 = 0.17(1) 28b(506.5) b = 329.03 say 330 mm.

CE BOARD May 2017

PROBLEM 19:

A cantilever beam 300 mm wide x 400 mm deep and 3 m. long is designed with tension reinforcement only. Superimposed dead load = 12 kN/m Live load at free end = 20 kN Concrete unit weight = 23.5 kN/m3 Concrete: fc’ = 30 MPa Steel: fy = 415 MPa Assume 70 mm concrete cover to the centroid of the tension reinforcement. ➀ Calculate the maximum factored design moment (Mu). U = 1.4 DL + 1.7LL ➁ What is the maximum factored design shear (Vu)? ➂ Calculate the nominal bending strength (kN.m.) of the section if the tension reinforcement consists of 4 – 25 mm∅. 20 kN DL=12 kN/m Solution: ➀ Max. factored design moment (Mu) Total dead load = 12 + 0.3(0.4)(23.5) 3m Total dead load = 14.82 kN/m Wu L2 Mu = + PL 70 4 - 25 mm ø 2 1.4(14.82)(3)2 Mu = + 1.7(20)(3) 2 400 d=330 M u = 195.4 kN.m. ➁ Max. factored design shear (Vu) Vu = 1.4(14.82)(3) + 1.7(20) Vu = 96.24 kN

300 70

➂ Nominal bending strength T = A s fy π (25)2 (4)(415) 4 T = 814,850 N.mm T=

C=T

400

d=330

⎛ a⎞ Mn = T ⎜ d - ⎟ 2⎠ ⎝

Mn

300

0.85 fc ' ab = A s fy 0.85(30)(a)(300) = 814,850

M n = 225.5 kN.m.

(d-a/2) C

a

⎛ 106.52 ⎞ M n = 814850 ⎜ 330 2 ⎟⎠ ⎝

a = 106.52 mm

T

4 - 25 mm ø

0.85fc’

CE BOARD May 2017

PROBLEM 20:

A simply supported steel beam spans 9 m. It carries a uniformly distributed load of 10 kN/m, beam weight already included. Beam Properties mm2

Area = 8530 Depth = 306 mm Flange width = 204 mm Flange thickness = 8.5 mm Moment of inertia, Ix = 145 x 106 mm4 Modulus of elasticity, E = 200 GPa ➀ What is the maximum flexural stress in the beam? ➁ To prevent excessive deflection, the beam is propped at midspan using a pipe column. Find the resulting axial stress in the column. Outside diam. = 200 mm. Thickness = 10 mm. Height = 4 m. ➂ How much is the maximum bending stress in the propped beam?



Solution:

➀ Max. flexural stress in the beam

w=10 kN/m

2

wL 8 10(9)2 M= 8 M = 101.25 kN.m. M=

fb =

MC I

101.25 x 10 6 (153) fb = 145 x 10 6 fb = 107 kPa

9m

85 153 N.A.

306

153 85 204

CE BOARD May 2017

PROBLEM 20:

Continuation

➁ Resulting axial stress in the column 5wL4 δ1 = 384EI RL3 δ2 = 48EI δ1 = δ 2 4

3

w=10 kN/m

1=2

5wL RL = 384EI 48EI 5wL =R 8 5(10)(9) R= = 56.25 kN 8 R 56250 σ= = = 9.4 MPa A π 2 2 ⎡(200) - (180) ⎤ ⎦ 4 ⎣

R

➂ Max. bending stress in the propped beam ∑Fv = 0 2R1 + 56.25 = 10(9)

w=10 kN/m

R1 = 16.875 kN 10(4.5)2 MB = - 16.875(4.5) 2 MB = 25.31 kN.m. fb =

MB C I

25.31 x 10 6 (153) fb = 145 x 10 6 fb = 26.7 MPa

A

R1

4.5

B

56.25

4.5

C

R1

CE BOARD May 2017

PROBLEM 21:

A cylindrical tank is filled with water to a depth of 2.5 m. on its vertical position. The tank dimensions are: Height = 3 m. Diameter = 2 m. Thickness = 6 mm Weight = 18 kN ➀ Find the maximum circumferential stress in the tank in MPa. ➁ The tank is supported only at the top. What is the max. longitudinal stress (MPa) in the tank? ➂ The tank is supported at the top and at the base. Determine the max. longitudinal stress in MPa. Solution: ➀ Max. circumferential stress in the tank F = 9810(2.5)(2)(0.001) F = 49.050 N

2.5

2T = F

0.001m

49.050 T= 2 T = 24.525 N

2m

Circumferential stress : σ=

T

T 24.525 = = 4.09 MPa A 1(16)

1mm

F 6 mm T

σ = 4.1 MPa

2000 mm

➁ Max. longitudinal stress when the tank is supported only at the top P 18000 σ1 = = = 0.48 MPa A π(2000)(6)

6 mm

support

P = 9810(2.5) P = 24525 N/m2 PD 4t 24525(2) σL = 4(0.006) σL =

Total longitudinal stress = 0.48 + 2.04

2.5

= 2.52 MPa

σ L = 2043750 N/m2 σ L = 2.04 MPa ➂ Max. longitudinal stress when the tank is supported at the top and at the base Max. longitudinal stress = 0 MPa

open

CE BOARD May 2017

PROBLEM 22:

a=1.92 kN/m

A pole fixed on the ground is subjected to the load as shown. Given: a = 1.92 kN/m b = 1.44 kN/m H = 15 m. Pole outside diameter = 180 m Thickness of pole = 15 mm

H=15 m

➀ Find the maximum shear force (kN) in the pole. ➁ What is the maximum moment in the pole? ➂ If the allowable tensile stress in the pole is 138 MPa, what is the allowable uniformly distributed load in kN/m? Solution: ➀ Max. shear force

Vmax =

b=1.44

kN/m

0.48

1.44

(1.92 + 1.44) (15) = 25.2 kN 2 15 m

➁ Max. moment in the pole ⎛ 0.48 + 0 ⎞ ⎛ 2⎞ M = 1.44(15)(7.5) + ⎜ (15) ⎜⎝ 3 ⎟⎠ (15) ⎝ 2 ⎟⎠

M = 198 kN.m. 1.44 kN/m

➂ Allowable uniformly distributed load if the allowable tensile stress in the pole is 138 MPa M = w(15)(7.5) kN.m. MC f= I π ⎡(180)4 - (150)4 ⎤ I= ⎦ 64 ⎣ I = 26.68 x 10 6 C = 90 180 15 m w 150 MC f= I M(90) 138 = 26.68 x 10 6 M = 40.91 kN.m. 40.91 = 15(7.5)w w = 0.364 kN / m

CE BOARD May 2017

PROBLEM 23:

A 25 m. long girder of a bridge is simply supported at the right end and at 3m. from the left end. It is subjected to a highway laneload consisting of a uniformly distributed load and a concentrated load as follows. Uniformly distributed load = 9.53 kN/m Concentrated load for moment = 80 kN Concentrated load for shear = 115.7 kN ➀ What is the maximum span positive moment? ➁ Calculate the maximum support reaction. ➂ What is the maximum negative moment? Solution: ➀ Max. span positive moment 11(11) y= = 5.5 22 Max. positive moment = area of positive influence diagram times uniform load plus concentrated load times the maximum ordinate of the positive influence diagram. Max. positive moment =

80 kN

w=9.35 kN/m

h 3m

22 m

9.35(22)(5.5) + 80(5.5) = 1006 kN.m 2 y

➁ Max. support reaction Max. support reaction: y 1 = 25 22 y = 1.136 9.35(25)(1.136) R1 = 115.7(1.136) + 2 R1 = 264 kN ➂ Max. negative moment

y=5.5

1.0

3m R1

1.136

80 kN 5.5

w=9.35 kN/m

h 3

R2

115.7 kN 9.35 kN/m

3m

Max. negative moment = 80(3) + 9.35(3)(1.5) = 282 kN.m

22 m

11

11

R1

22 m

R2

CE BOARD May 2017

PROBLEM 24:

Given: D = 2.0 m Width of cylinder = 3.2 KN.

H1 = 2.4 m. H2 = 3.0 m.

The coefficient of friction between the cylinder and the surface is 0.20. The cylinder is filled with water to a depth of 2.4 m. Find the maximum value of the force F (KN) that will satisfy the following conditions. ➀ Translation of the cylinder is prevented. ➁ Tipping of cylinder is prevented. ➂ The cylinder remains in equilibrium. Solution: ➀ Value of F when the translation of cylinder is prevented π(2)2 N2 = (2.4)(9.81) 4 W2 = 74. KN W1 + W2 = 3.2 + 74 = 77.2 KN F = MN F = 0.20(77.2) F = 15.4 KN

➁ Value F so that tipping of cylinder is prevented ΣM A = 0

F(2.4) = 77.2(1) F = 32.17 kN ➂ Value of F so that the cylinder remains horizontal F = MN

F = 0.20(77.2) F = 15.4 kN