Reinforced Concrete Design 2 - Test 1 19/20 PDF

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SKAA 4333 1

School of Civil Engineering

TEST 1 SEMESTER II, SESSION 2019/2020 COURSE CODE

:

SKAA 4333

COURSE

:

REINFORCED CONCRETE DESIGN 2

PROGRAMME

:

SKAW

DURATION

:

2 HOURS

DATE

:

MARCH 2020

INSTRUCTION TO CANDIDATES:

1.

Answer ALL questions.

2.

All solutions should be in accordance with EUROCODE 2 : MS EN 1992-1-1: 2010

3.

Reinforcement detailing and related drawings/sketches should be accompanied for corresponding solutions, if any.

WARNING!

SKAA 4333 2

Students caught copying/cheating during the examination will be liable for disciplinary actions and the faculty may recommend the student to be expelled from the study. This examination question consists of ( 5 ) printed pages only.

Q1.

The longitudinal cross-section of a reinforced concrete staircase connecting the ground and the first floor levels of a commercial building is shown in FIGURE Q1. The supports are provided by the ground beam and the first floor beam which are cast monolithically with the stair slab. The staircase, with a width of a riser of

and

the landing is

has a going and

respectively, while the overall depth of the waist and

. Other relevant dimensions of the structure are also given in the

figure. The following data are to be used in the design of the structure. Characteristic variable action

= 4.0 kN/m2.

Characteristic permanent action from finishes and handrails

= 1.0 kN/m2

Characteristic strength of concrete

= 30 N/mm2

Characteristic strength of reinforcement

= 500 N/mm2

Concrete cover to main reinforcement

= 25 mm

Unit weight of concrete

= 25 kN/m3

(a)

Design all reinforcement required in the staircase.

(b)

Check your design satisfaction in shear, deflection and cracking

(c)

Show your proposed arrangement of reinforcement in a neat sketch of the longitudinal section of the structure.

250

270

170

8@270 = 2160

250

1200

FIGURE Q1

All dimensions are in mm

250

12@270 = 3240

250

SKAA 4333 3

SKAA 4333 4

Q2.

(a)

FIGURE Q2(a) shows a four-span continuous beam with their respective total characteristic permanent (Gk) and variable (Qk) actions. Sketch the various action arrangements for the beam based on mentioned in MS EN 1992 : 2010, showing their respective values. Gk (kN/m) Qk (kN/m)

28 20

24 17

24 17

28 20

8m

6m

6m

8m

FIGURE Q2(a)

(b)

Part of the floor plan of a five storey reinforced concrete building, and the front elevation of its substitute frame in the transverse plane, are shown in FIGURE Q2(b). Shear walls are to be provided at suitable locations in the building to resist lateral forces from the wind. The in the substitute frame have a cross-section of and are of and . The with a cross-section , are high for th , and for the . The overall thickness of the . Other relevant dimensions are also shown in FIGURE Q2(b). The building is to be designed to carry the following vertical actions : Characteristic variable action (including partition) Characteristic permanent action from finishes, services etc. (excluding self-weight)

= 4.50 kN/m2 = 1.25 kN/m2

(i) Determine the bending moments in using method, with loading arrangements suggested in MS EN 1992: 2010, and draw the bending moment envelope, showing all the important values. (The maximum span moments in the beams should be determined from shear force diagrams). (ii) Using One-Free Joint Sub-Frame method determine the bending moments in Column B/5 resulted from the most critical arrangement of ultimate actions on Beam 5/A-C, Level 1, and draw the bending moment diagram.

SKAA 4333 5

A

B

C

4 4000 250 x 700

250 x 700

5 4000 6

9000

All dimensions are in mm

7000

5

4

3 4 @ 3600 = 14400 2

400 1 250 x 700

250 x 700 400

9000

7000

4200

Level 0

A

B

FIGURE Q2(b)

C

TEST 1 : S2 201920 : Q1

page 1/4

Ref.

Calculations

Output

(a)

Characteristic Actions: Permanent, g k Variable, q k

= =

Materials: Characteristic strength of concrete, f ck Characteristic strength of steel, f yk Unit weight of reinforced concrete Assumed: Cover, C = 25 mm Dimension: Riser, R = Waist, h =

2

1.0 kN/m (Excluding selfweight) 2 4.0 kN/m

170 mm 250 mm

2

30 N/mm 2 500 N/mm

= =

f bar = Going, G =

AVERAGE THICKNESS 2 2 1/2 y = h . [ (G + R ) /G] 1/2 2 2 = 250 [ (270 + 170 ) / 270] Average thickness t = [y + (y + R )] / 2 = [ 295 + 295 + 170] / 2 = ACTIONS Landing Slab selfweight = 0.250 x 25 Permanent load (Excluding selfweight) Characteristic permanent action, g k Characteristic variable action, q k Design action, n d = 1.35 g k + 1.5q k

25 kN/m3 16 mm

=

=

270 mm

= 295 mm

y

R

t

y 380 mm

6.25 kN/m2 = 1.00 kN/m2 2 = 7.25 kN/m 4 2 = 4.00 kN/m 2 = 15.79 kN/m Table A1.2B : EN 1990

msy '11

TEST 1 : S2 201617 : Q1

page 2/4

Ref.

Calculations

Output

Flight Slab selfweight = 0.380 x 25 Permanent load (Excluding selfweight) Characteristic permanent action, g k Characteristic variable action, q k Design action, n d = 1.35 g k + 1.5 q k ANALYSIS Consider 1m width, 20.19 kN/m 15.79 kN/m 2.29

1.20 6.85

=

9.51 kN/m2 = 1.00 kN/m2 2 = 10.51 kN/m 3 2 = 4.00 kN/m 2 = 20.19 kN/m Table A1.2B : EN 1990

20.19 kN/m 3.37 M

6.1

M Total action, F = (20.2 x 2.29) + (15.8 x 1.2) + (20.2 x 3.37) = 133.0 kN/m Bending Moment, M = FL /10 = 133.02 x 6.85 / 10 = 91.1 kNm/m MAIN REINFORCEMENT

6

Effective depth, d = h - C - 0.5 f bar = 250 - 25 - (0.5 x 16) = 217 mm Design moment, M Ed = 91.1 kNm/m 2 K = M / bd f ck 2 6 = 91.1 x 10 / (1000 x 217 x 30) = 0.064 < K bal = 0.167 Compression reinforcement is not required z = d [ 0.5 + 0.25 - K /1.134) ] = 0.94 d ≤ 0.95d A s = M / 0.87 f yk z 6 = 91.1 x 10 / (0.87 x 500 x 2 = 1027 mm /m 9.2.1.1

Main bar : 0.94 x 217 )

H16 - 175 2 (1149 mm /m)

Minimum and maximum reinforcement area, Secondary bar : A s,min = 0.26(f ctm/f yk) bd = 0.26 x (2.90 / 500) x bd H10 - 200 2 2 (393 mm /m) = 0.0013 bd = 0.0013 x1000 x 217 = 282 mm /m 2 A s,max = 0.04Ac = 0.04 x1000 x 250 =10000 mm /m

msy '09

TEST 1 : S2 201617 : Q1

page 3/4

Ref.

Calculations SHEAR 20.19 kN/m

Output 91.1 kNm

15.79 kN/m

20.19 kN/m

1.20 6.85

3.37

2.29 Va

Vb

SM @ b = 0 6.85 V a (20.2 x 2.3 x 5.71) (15.8 x 1.2 x 4.0) - (20.2 x 3.4 x 1.7) + 91.1 = 0 V a = ( 263.3 + 75.1 + 114.3 - 91.1 ) / 6.9 = 52.8 kN Vb = (46.1 + 18.9 + 67.9) - 52.8 = 80.2 kN Design shear force, V Ed = 80.2 kN/m 6.2.2

Design shear resistance, V Rd,c = [ 0.12 k (100r 1 f ck)1/3 ] bd 1/2

≤ 2.0 k = 1 + (200/d ) 1/2 = 1+(200 / 217) = 1.96 r 1 = A sl /bd ≤ 0.02

≤ 2.0

= 1149 / (1000 x 217) = 0.0053 ≤ 0.02 1/3 V Rd,c = 0.12 x 2.0 x (100 x 0.0053 x 30) x 1000 x 217 = 128304 N/m = 128.3 kN/m 3/2 V min = [ 0.035k f ck1/2 ] bd 3/ 2

1/2

= 0.035 x 2.0 x 30 x 1000 x 217 = 114152 N/m = 114.2 kN/m So, V Rd,c = 128.3 kN/m > V Ed 7.4

Table 7.4N

Ok !

DEFLECTION Percentage of required tension reinforcement, r = A s,req / bd 1027 / 1000 x 217 Reference reinforcement ratio, 1/2 -3 r o = (f ck) 1/2 x 10 -3 = (30) x10 = Factor for structural system, K = 1.3  l  K 11  1.5 d 

fck

ro  3.2 r

r  fck  o  1   r 

3/ 2

0.0047 0.0055

  

= 1.3 (11 + 9.5 + 1.09) = 28.1 Modification factor for spspan less than 7 m = 1.00 Modification factor for steel area provided, 1149 / 1027 = 1.12 < 1.5 = A s,prov/A s,req = Therefore allowable span-effective depth ratio, (l /d )allowble = 28.1 x 1.00 x 1.12 = 31.4 Actual span-effective depth (l /d )actual = 6850 / 217

= 31.6

>

(l /d )allowble

Fail ! msy '09

page 4/4 Ref.

7.3.3

Table 7.3N

Calculations

Output

CRACKING h = 250 mm > 200 mm Steel stress under the action of quasi-permanent loadin f s = [(G k + 0.3Q k)/(1.35G k + 1.5Q k)](A s.req/A s.prov)(f yk/1.15) G k + 0.3Q k = 11 + (0.30 x 4) = 11.7 kN (1.35G k + 1.5Q k) = 20.2 kN f s = (11.7 / 20.2) (1027 /1149) (500 / 1.15) = 0.58 x 0.89 x 435 = 226 N/mm2 For design crack width = 0.3 mm Max. allowable bar spacing = Max. bar spacing, Main bar = Sec. bar =

175 mm 200 mm

200 mm < <

200 mm 200 mm

Ok ! Ok !

DETAILING 0.3L = 2055

. . . . H16-175 0.2L = 1370

H16-175 H10-200

6850

TEST 1 : S1 201920 : Q2

page 1/6

Ref. 2(a)

Calculations

Output

Action Pattern for a four-span continuous beam Gk = 28 kN/m Gk = 20 kN/m 1.35G k = 1.35 x 28 = 37.8 kN/m 1.35G k+1.5 Q k = (1.35 x 28) + (1.50 x 20) = 67.8 kN/m Gk = 24 kN/m Gk = 17 kN/m 1.35G k = 1.35 x 24 = 32.4 kN/m 1.35G k+1.5 Q k = (1.35 x 24) + (1.50 x 17) = 57.9 kN/m Case 1

67.8

57.9

32.4

37.8

Case 2

37.8

57.9

57.9

37.8

Case 3

37.8

32.4

57.9

67.8

Case 4

67.8

32.4

57.9

37.8

Case 5

37.8

57.9

32.4

67.8

SPECIFICATION Loading: Finishes, ceiling, services etc. Density of concrete Imposed load Dimension: Slab thickness, h Beam size, b x h = Column size, b x h =

= = =

1.25 kN/m2 25 kN/m3 5.00 kN/m2

= 150 mm 250 x 700 mm 300 x 400 mm

msy '11

TEST 1 : S1 201920 : Q2

page 2/6

Ref. 2(b)

Calculations

Output

ACTIONS 2 Loads on slab, n kN/m : Slab selfweight = 0.15 x 25 Finishes, ceiling etc. = Characteristic permanent load, g k = Characteristic variable load, q k = Loading distribution:

=

3.75 kN/m2 2 1.25 kN/m 2 5.00 kN/m 2 4.50 kN/m

1

4.0m 0.5 0.5

0.5 0.5

4.0m

2 B

9.0m

A

7.0m

C

Loads on beam, w kN/m : w

w

A

9.0 m

B

7.0 m

C

Span A-B : w Perm. load from slab = 1.00 x 5.00 x 4.0 = 20.00 kN/m Beam selfweight = x x = 3.44 kN/m Characteristic permanent load, g k = 23.44 kN/m Variable load fr.slab = 1.00 x 4.50 x 4.0 = 18.00 kN/m Chacracteristic variable load, q k = 18.00 kN/m Span B-C : w Perm. load from slab = 1.00 x 5.00 x 4.0 = 20.00 kN/m Beam selfweight = 0.55 x 0.25 x 25 = 3.44 kN/m Characteristic permanent load, g k = 23.44 kN/m Variable load fr.slab = 1.00 x 4.50 x 4.0 = 18.00 kN/m ChaA acteristic variable load, q k = 18.00 kN/m Analysis Stiffeness, I /L = (bh 3/12)/L 3 7.9 x 105 mm3 Beam : K AB = [ 250 x 700 /12 ]/ 9000 = 3 KBC = [ 250 x 700 /12 ]/ 7000 = 10.2 x 105 mm3 Distribution factor : F = K / SK Joint A: F AB = K AB/(K AB) = 1.00 Joint B: F BA = K AB/(K AB+K BC) = 0.44 F BC = K BC/(K AB+K BC) = 0.56 Joint C: F CB = K BC/(K BC) = 1.00 msy '11

page 3/6 Ref.

Calculations Case 1 : Max, Max w = 58.6 kN/m

w =

9.0 m

A

Output

58.6 kN/m 7.0 m

B

Fixed End Moment, 2 -M AB = M BA = 58.6 x 9.0 /12 2 -M BC = M CB = 58.6 x 7.0 /12

C

= =

395.8 kNm 239.4 kNm

A 1.00

B 0.44 0.56

C 1.00

-395.8 395.8

395.8 -239.4 -68.4 -88.0

239.4 -239.4

0.0 0.0 0.0

197.9 -119.7 -34.2 -44.0 491.1 -491.1

0.0 0.0 0.0

Shear force : 58.6

491.1 491.1

9.0 V AB

58.6 7.0

V BA

V BC

V CB

SM @ B = 0 9.0 V AB (58.6 x 9.0 x 4.5) + 491.1 V AB = ( 2374.9 - 491.1 ) / 9.0 = V BA = ( 58.6 x 9.0) 209.3 =

= 0 209.3 kN 318.5 kN

SM @ C = 0 7.0 V BC (58.6 x 7.0 x 3.5) + V BC = ( 1436.7 +491.1 ) / 7.0 V CB = ( 58.6 x 7.0) 275.4

- 491.1 = 0 = 275.4 kN = 135.1 kN

msy '11

page 4/6 Ref.

Calculations Case 2 : Max, Min w = 58.6 kN/m A

w =

9.0 m

Output

31.6 kN/m 7.0 m

B

Fixed End Moment, 2 -M AB = M BA = 58.6 x 9.0 /12 2 -M BC = M CB = 31.6 x 7.0 /12

C

= =

395.8 kNm 129.2 kNm

A 1.00 -395.8

B 0.44 0.56 395.8 -129.2

C 1.00 129.2

395.8 0.0 0.0 0.0

-116.6 -150.0 197.9 -64.6 -58.3 -75.0 418.8 -418.8

-129.2 0.0 0.0 0.0

Shear force : 58.6

418.8 418.8

9.0 V AB

31.6 7.0

V BA

V BC

V CB

SM @ B = 0 9.0 V AB (58.6 x 9.0 x 4.5) + 418.8 V AB = ( 2374.9 - 418.8 ) / 9.0 = V BA = ( 58.6 x 9.0) 217.4 = SM @ C = 0 7.0 V BC (31.6 x 7.0 x 3.5) + V BC = ( 775.2 +418.8 ) / 7.0 V CB = ( 31.6 x 7.0) 170.6

= 0 217.4 kN 310.4 kN

- 418.8 = 0 = 170.6 kN = 50.9 kN

msy '11

page 5/6 Ref.

Calculations Case 3 : Min, Max w = 31.6 kN/m A

w =

9.0 m

Output

58.6 kN/m 7.0 m

B

Fixed End Moment, 2 -M AB = M BA = 31.6 x 9.0 /12 2 -M BC = M CB = 58.6 x 7.0 /12 A 1.00 -213.6 213.6 0.0 0.0 0.0

B 0.44 213.6 11.3 106.8 5.7 337.3

C

= =

213.6 kNm 239.4 kNm C 1.00 239.4 -239.4 0.0 0.0 0.0

0.56 -239.4 14.6 -119.7 7.3 -337.3

Shear force : 31.6

337.3 337.3

9.0 V AB

58.6 7.0

V BA

V BC

V CB

SM @ B = 0 9.0 V AB (31.6 x 9.0 x 4.5) + 337.3 V AB = ( 1281.4 - 337.3 ) / 9.0 = V BA = ( 31.6 x 9.0) 104.9 = SM @ C = 0 7.0 V BC (58.6 x 7.0 x 3.5) + V BC = ( 1436.7 +337.3 ) / 7.0 V CB = ( 58.6 x 7.0) 253.4

= 0 104.9 kN 179.9 kN

- 337.3 = 0 = 253.4 kN = 157.1 kN

msy '11

page 6/6 Ref.

Calculations

Output

Shear force and bending moment diagrams 209 275 217 171 105 253 (kN)

3.57 3.71 3.32

4.70 5.39 4.32 318 310 180

135 51 157

491 419 337 (kNm)

374 403 174

156 41 210

Column B 3 Stiffeness, I /L = (bh /12)/L Column K cu = [ 300 x 400 3/12 ]/ 3600 = K cl = [ 300 x 400 3/12 ]/ 4200 = 2

FEM 1 = wL /12

Kcu

58.6 kN/m 0.5KAB

9.0 m

31.6 kN/m

Kcl

4.4 x 105 mm3 3.8 x 105 mm3

0.5KBC

7.0 m

D M = FEM 1 - FEM 2 =

= 58.6 x 9.0 2 /12 = 395.8 kNm 2 FEM 2 = wL /12 = = 396 -

31.6 x 7.0 2 /12 129.2 kNm 129 =

266.6 kNm

Moment in upper column, M = D M x K cu /(K cu + K cl + 0.5K AB + 0.5K BC) = 266.6 x 4.4 / (4.4 + 3.8 + 3.97 + 5.10) = 68.4 kNm Moment in lower column, M = D M x K cl /(K cu + K cl + 0.5K AB + 0.5 K BC) = =

266.6 x 3.8 / (4.4 + 58.6 kNm

3.8 +

68.4 58.6

3.97 + 5.10)

50 msy '11...