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QUESTION 1 (16%)

For each of the concrete members shown in Figure 1 below, sketch a feasible strut and

tie model indicating the position and orientation of the compression struts and tension

ties.

Beam with single ledge support Beam with double ledge support

Irregular deep beam Deep beam

Irregular deep beam Beam-column joint under

closing bending moment

w*

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Irregular deep beam Beam with stepped double ledge support

Figure 1

QUESTION 2 (18%)

A deep reinforced concrete member carries two members with factored loads as

shown in Figure 2. Material properties:fsy= 400 MPa, fc = 50 MPa.

a)Sketch a feasible strut and tie model indicating the position and orientation of thecompression struts and tension ties and calculate the forces in the struts and ties.

b)Check the bearing stresses at where the loads are applied.c)What is the minimum width required for the most heavily loaded compression

strut.

d)Calculate the minimum width required for support bs.Take: u = 0.12 h; w = bs sin + u cos ; fb1 = 0.85 fc , where = either 1.0 or 0.8

1800m

m

1800mm

Figure 2

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QUESTION 3 (16%)

The concrete slab shown in Figure 3 is 7m 5m. The slab is not supported along one

of the 7m long edges (free edge). The other three edges are supported and continuous

over the supports, and therefore can be considered fixed edges. The slab isisotropically reinforced with an ultimate bending moment capacity of 20 kNm/m in

the positive direction. The bending moment, which causes tension at the bottom, is

taken as the positive moment. Along the fixed edges, the slab is reinforced to have a

negative bending moment capacity of 1.33 times the positive bending moment

capacity. Slab is expected to support a uniformly distributed load.

Using Hillerborgs strip method, calculate the ultimate uniformly distributed load that

the slab can support.

7 m

5 m

X

Y

Figure 3

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QUESTION 4 (17%)

The horizontal rectangular reinforced concrete slab of 7m 3m shown in Figure 4 has

fixed supports along the long edges and simple supports along the short edges. Theslab is subjected to a uniform distributed load, q kN/m2. Mx

+ = Mx- = My

+ = My- = 4

kN.m/m. Using yield line analysis, calculate the collapse load q.

Lx = 3 m

Ly= 7m

Y

X

Figure 4

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QUESTION 5 (17%)

A steel frame ABCDE made of 360UB56.7 with plastic moment capacity Msx= 273kNm is subjected to two design loads, 100 kN each, as shown in Figure 5. The

connection at C is a pin joint. An elastoplastic analysis is carried out for the frame

until collapse with three plastic hinges. The results of the analysis for bendingmoments for = 1 are shown in the tables below.

a) Determine the sequence of plastic hinge formation (that is, Case A,..,..,H in thecorrect order) and hence the final bending moments of all members and the load

factor (accurate up to 2 decimal places) at collapse. Show all stages ofcalculations using the standard spreadsheet form given below. (10 marks)

b) What is the lightest possible section from the table given below for the frame atdesign load level (that is, at = 1). (4 marks)

c) If the elastic buckling load factor c of the frame is 5.5, determine the actualcollapse load factor for the frame by taking into account the moment

amplification effect. (3 marks)

Section

(kg/m)

Moment

capacity

Msx (kNm)410UB59.7 324

410UB53.7 304

360UB56.7 273

360UB50.7 242

360UB44.7 222

310UB46.2 197310UB40.4 182

310UB32 134

250UB37.3 140

Figure 5

Bending moments at joints (in kNm)

Case Structure Joint A Joint B Joint D Joint E

A 104.9 26.8 234.2 -209.6

B 345.0 322.5 0 -555.0

A

B

C100 kN

D

4m6m

100 kN

6m

3m

E

A

B

C100 kND

100 kN

E

A

B

C100 kN

D

100 kN

E

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C 193.4 34.7 282.6 0

D 0 -41.0 254.7 -263.3

E 0 150 0 -900

F 0 -120 360 0

G 900 600 0 0

HCollapse

A

B

C100 kN

D

100 kN

E

A

B

C100 kN

D

100 kN

E

A

B

C100 kN

D

100 kN

E

A

B

C100 kN

D

100 kN

E

A

B

C100 kN

D

100 kN

E

A

B

C100 kN

D

100 kN

E

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Analysis Stage No: 1 Critical load factor, cr=Member Joint Moment

Mo

(kNm)

Residual

Plastic

Moment

MP-Mi

(kNm)

Load factor

o

ip

M

MM =

Cumulative

Moment

ocri1i MMM +=+(kNm)

AB AB

BC B

C

CD CD

DE D

E

Analysis Stage No: 2 Critical load factor, cr

=

Member Joint MomentMo

(kNm)

Residual

Plastic

Moment

MP-Mi(kNm)

Load factor

o

ip

M

MM =

Cumulative

Moment

ocrii MMM +=+1(kNm)

AB A

B

BC BC

CD C

DDE D

E

Analysis Stage No: 3 Critical load factor, cr=Member Joint Moment

Mo

(kNm)

Residual

Plastic

Moment

MP-Mi(kNm)

Load factor

o

ip

M

MM =

Cumulative

Moment

ocrii MMM +=+1(kNm)

AB AB

BC BC

CD CD

DE DE

Analysis Stage 4 shows that the structure has collapsed

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QUESTION 6 (16%)

(a) Using the Mechanism (Upper Bound) Method, determine the collapse load factor

from all possible collapse mechanisms for the fixed end beam ABCD shown inFigure 6(a). Plastic moment capacity for the members of the portal frame is Mp = 240

kNm. (6 marks)

(b) A fixed-base portal frame is subjected to a horizontal load 30 kN at B and twovertical loads, one of 10 kN at C and the other 30 kN at D as shown in Figure 6(b).Plastic moment capacity for the members of the portal frame is Mp = 240 kNm.

By using the Mechanism Method, calculate the collapse load factor from allpossible mechanisms. (10 marks)

Figure 6(b)

A

B C

E

2m 1m

10 kN

30 kN

4m

D2m

30 kN

D

2m2m

Figure 6(a)

CBA

30 kN

1m

10 kN

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Formula sheet

L

r fym

y

+( )80 50250

( )2121

22 MM

PPPPMM dd

+=

( )0

2

212

1 =

L

xwwxwSi

265.11

L

Mw

p=

sx

s

sxprx M

N

NMM

=

*

118.1 for bending about the x-x axis

sys

sypry MN

NMM

=

2*

119.1 for bending about the y-y axis.

c

p

1

1

9.0

= .