CITY UNIVERSITY OF HONG KONG

Course code & title : BST10346

Structural Design

Session : Semester B, 2007-2008

Time allowed : Two hours

This paper has SEVENTEEN pages (including this cover page).

1. This paper consists of four questions.

2. Answer Question 1 and ANY OTHER TWO questions.

3. Start each question on a new page.

4. State all the assumptions clearly if necessary

5. ALL STEPS OF CALCULATION AND SIGN CONVENTION MUST

BE CLEARLY SHOWN.

Materials, aids and instruments permitted to be used during examinations:

Approved Calculator

Special materials (other than standard materials e.g. answer book or

supplementary sheet) to be supplied to students:

Nil

Not to be taken away

1

Question 1 (40 marks)

The bending moment diagram of the continuous beams below from different load cases are

shown below:

a) Calculate the ultimate design moment at mid-span and support. (5 marks)

b) Calculate the As required and As’ required (if any) of the mid-span

moment only for the two concrete section sizes with the following data

provided.

1.) 400 x 600 mm depth

2.) 400 x 500 mm depth

fcu = 45 N/mm²

fy and fyv = 460 N/mm²

Nominal cover = 30 mm

Link diameter = 12 mm

Main bar diameter = 40 mm

Shear link diameter = 12 mm

(10 marks)

c) Suggest three methods in order to avoid provision of compression steel. (3 marks)

d) From the result of b), design the shear reinforcement for the two

sections size if the shear force from DL and LL is 300 kN and 200 kN

respectively.

(5 marks)

Dead Load kNm

180 220 230

Live Load kNm

130 180 200

150 90

2

190 200

e) Calculate the vc of the below section. (5 marks)

Design data:

fcu = 40 N/mm²

fy and fyv = 460 N/mm²

d = 900 mm

Link diameter and spacing

= T12@250c/c 2 legs

+

T16@200c/c 1 leg

f) Calculate the maximum shear capacity in kN of the beam. (10 marks)

g) If only two legs are allowed in this beam, design the other shear

reinforcement for this beam.

(2 marks)

3

Question 2 (30 marks)

a) Prove that 254 x 254 x 73 kg/m Grade 55C UC and 305 x 165 x 54

kg/m Grade 55C UB are adequate against the following ultimate load

and ultimate moments about X-X axis in low shear load case.

Mxx = 100 kNm

Myy = 30 kNm

F = 700 kN

(18 marks)

b) From the question a), discuss why UB has sufficient structural capacity

against the same loadings as UC although UB has lighter self weight.

(2 marks)

c) Design the bolt connection detail according to the diagram below to

fulfill the following ultimate load. The shear plane will be happened in

the threads.

(Remarks: The no. of bolts should not be less than 6)

(10 marks)

4

Question 3 (30 marks)

a) Use force method to determine all reactions for the beam shown below. (15 marks)

b) Sketch the bending moment and shear force diagrams, indicating

maximum and minimum values, and exact location of contra-flexure

point(s), where appropriate.

(12 marks)

c) If the roller support is removed from the beam, sketch the

corresponding bending moment and shear force diagrams.

(Hint: No further calculation shall be required)

(3 marks)

5

Question 4 (30 marks)

A single-bay frame is subject to a vertical UDL of 5 kN/m and a horizontal point load of

20 kN as shown in the figure below. Using an approximate approach, determine, under the

combined vertical and horizontal loads:

a) the reaction forces at the hinge supports D and E (indicate clearly the

direction of reaction force);

(15 marks)

b) the bending moments and shear forces at point B (middle of beam) and

point C (right end of beam). Indicate clearly whether the moment is

sagging or hogging, and whether the shear is clockwise or anti-

clockwise.

(15 marks)

A B C

L = 4m

D E

H = 4m

5 kN/m

20 kN

Hint:

For lateral load analysis, you can assume a point of contra-flexure at mid-span of the

beam.

For vertical load analysis, you can assume a point of contra-flexure located at 0.1L from

each end of the beam of span L.

Moment: Sagging Hogging

Shear: Clockwise Anti-clockwise

- END -6

Fo rmulae K = M / fcu bd²

z = (0.5 + √(0.25 - K/0.9) )d

As req. = M / (0.87fyz)

As′ = (K - 0.156) fcu bd² / ((0.87 fy (d - d′)

As = (0.156 fcu bd²) / (0.87 fy 0.775d) + As′

Mu (conc) = 0.156 fcu bd²

Mu (steel) = 0.87fyAsz

z = d - 0.45x

x = (0.87fyAs) / (0.45fcub0.9)

v = V/bvd

vc = 0.79(100As/bvd)1/3(400/d)1/4/1.25(fcu/25)1/3

Asv/sv = 0.4bv/(0.87fyv)

Asv/sv = bv(v-vc)/(0.87fyv)

Msx = sx nlx²

Msy = sy nly²

Fs/Ps + Ft/Pt 1.4

Ft = Pe y1 / no. of bolt column x y²

F/Aepy + Mx/Mcx + My/Mcy 1

Mc = S py but 1.2pyZ for plastic or compact sections

Mc = pyZ for semi-compact and slender sections

7

AppendicesLoad

Combination

Ultimate Limit State

Dead Imposed Wind Earth &

water

pressure

Serviceability

Limit State

Dead & Imposed

(& earth & water

pressure)

1.4 (or

1.0)

1.6 - 1.4 1.0

Dead & wind (&

earth & water

pressure)

1.4 (or

1.0)

- 1.4 1.4 1.0

Dead & imposed

& wind (& earth

& water

pressure)

1.2 1.2 1.2 1.2 1.0

Conditions:

1. The area of each bay exceeds 30m²,

2. Spans are considered to be approximately equal if they do not differ by more than 15% of

the longest span.

3. Qk/Gk < 1.25,

4. Qk < 5 kN/m² excluding partitions

8

ly / lx 1.0 1.1 1.2 1.3 1.4 1.5 1.75 2.0

sx 0.062 0.074 0.084 0.093 0.099 0.104 0.113 0.118

sy 0.062 0.061 0.059 0.055 0.051 0.046 0.037 0.029

Bending moment coefficient for slabs spanning in two directions at right-angles, simply-

supported on four sides

9

10

Basic span / effective depth ration for rectangular or flanged beams (BS 8110, Table 3.10)

Support conditions Rectangular sections Flanged beams with bw/b 0.3

Cantilever 7 5.6

Simply supported 20 16

Continuous 26 20.8

Modification factor (m1)for tension reinforcement (BS 8110, Table 3.11)

Service stress

N/mm²

M/bd²

0.5 0.75 1.00 1.5 2.0 3.00 4.00 5.00 6.00

100 2 2 2 1.86 1.63 1.36 1.19 1.08 1.01

150 2 2 1.98 1.69 1.49 1.25 1.11 1.01 0.94

(fy = 250) 156 2 2 1.96 1.66 1.47 1.24 1.1 1 0.94

200 2 1.95 1.76 1.51 1.35 1.14 1.02 0.94 0.88

250 1.9 1.7 1.55 1.34 1.2 1.04 0.94 0.87 0.82

(fy = 460) 288 1.68 1.5 1.38 1.21 1.09 0.95 0.87 0.82 0.78

300 1.6 1.44 1.33 1.16 1.06 0.93 0.85 0.8 0.76

Modification factor (m2) for compression reinforcement (BS 8110, Table 3.12)

100 As′ prov./bd Factor 100 As′ prov./bd Factor

0 1 1 1.25

0.15 1.05 1.5 1.33

0.25 1.08 2 1.4

0.35 1.1 2.5 1.45

0.5 1.14 3 1.5

0.75 1.2 - -

Table 6, Design strength py for steel to BS 4360

BS 4360 Grade

Thickness (mm) less than or

equal to (flange thinkness for

rolled section)

Sections, plates and hollow sections

py (N/mm²)

43 A, B and C

16 275

40 265

63 255

100 245

50 B and C

16 355

40 345

63 340

100 325

55 C

16 450

25 430

40 425

63 400

11

12

13

14

15

16

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