Solutions to Q1,3,5,13 and 15

8/3/2019 Solutions to Q1,3,5,13 and 15

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CE 2164 Structural design and Materials – Part 1 Reinforced concrete design

Department of Civil Engineering National University of Singapore 1

Take Home Assignments Solutions Questions 1, 3, 6, 13 and 15:

1.1 Under local/tropical climate, it is not uncommon for peak temperature of a mass concrete to

exceed 70oC.

(a) This is due to heat liberated from the hydration reaction between cement and aggregates.(b) This is due to heat liberated from the hydration reaction between cement and water.(c) If such temperature rise is allowed to take place, there is risk of cracking due to early-

age thermal contraction.

(d) If such temperature rise is allowed to take place, there is risk of cracking due to dryingshrinkage.

(e) If such temperature rise is allowed to take place, there is risk of cracking due to delayed

ettringite formation.Ans: (b), (c) & (e)

1.2 The concrete will crack when the tensile stress developed in the concrete exceeds the tensile

strength. Tensile stress will be induced in the concrete:

(a) when there is some form of internal restraint.

(b) only when there is some form of internal and external restraint.(c) when there is no restraint.

(d) when there is some form of external restraint.

Ans: (a) & (d)

1.3 Grade 30 concrete is defined as:

(a) Average standard cube compressive strength of concrete is 30 MPa.

(b) Compressive strength of all the stardand cube specimens exceeds 30 MPa.(c) 95% of the standard cube test results exceeds the compressive strength of 30 MPa

(d) Concrete mix is designed to have 30 MPa compressive strength.(e) Design strength of concrete is 30 MPa.

Ans: (c)

1.4 The contractor misplaced the tension steel bars of a singly reinforced beam at a shorter

effective depth than it should be. It may lead to

(a) a structurally safer structure

(b) improved moment capacity of the beam

(c) loss of ductility(d) a over reinforced beam

Ans: (c) & (d)

1.5. If it is found that the singly reinforced beam is over reinforced and it is desired to makeunder reinforced or balanced, this may NOT be achieved by

(a) Increasing the depth of the section





(b) Making it as a doubly reinforced beam

(c) Decreasing the concrete strength(d) Increasing the breadth of the section

Ans: (c)

1.6 Which of the following statements about moment redistribution is/are not true

(a) Moment capacity of a section is reduced by allowing moment redistribution at thatsection

(b) By allowing the moment redistribution, the overall load taken by fixed- ended beam

decreases(c) A beam is designed for 30% moment redistribution deflects more than that designed for

20% moment redistribution

(d) Higher the neutral axis depth lesser the moment redistribution

Ans: (b)

1.7 Concrete and steel as a composite is a good construction material because(a) the weakness of concrete in tension is overcome by the use of steel(b) reinforced concrete can be cast into any shape and size

(c) both concrete and steel have equally good resistance against fire

(d) Concrete can provide a high alkaline environment to protect steel(e) against corrosion

Answer: (a), (b) and (d)

1.8 In comparing the properties of concrete to steel, it may be stated that:(a) steel is more ductile than concrete

(b) concrete has a lower tensile strength than steel

(c) concrete creeps under normal ambient temperature unlike steel(d) concrete has a higher modulus of elasticity than steel

Answer: (a), (b) and (c)

1.9 The setting and hardening of Portland cement after addition of water is due to

(a) the presence of gypsum

(b) binding action of water

(c) hydration of some of the constituent compounds of cement(d) evaporation of water

Answer: (c)

1.10 The purpose of adding gypsum in the production of Portland cement is

(a) to modify the colour of cement

(b) to significantly increase the strength of cement(c) to prevent the occurrence of flash set

(d) to speed up the hydration of tricalcium aluminate

Answer: (c)





Solution Q3 – ANALYSIS OF OVER-REINFORCED SECTION

Determine the moment of resistance of the section shown.

Calculations Output

Area of steel = 22 25 4π ππ π × × = 981.7 mm2

Assuming that the steel has yielded,

T= .0 87 y s f A = 0.87 × 460 × 981.7 = 392876 N T = 392876 N

Hence the balancing compressive force,

C = . .0 45 0 9 cu f b x× = 0.45 × 25 × 150 × 0.9 x = 392876 N

⇒ x = 259 mm x = 259mm

259300

x d

= = 0.86

.0 64 x d

>

Hence the steel has not yielded.

We shall try to find a value for x, by trial and error, such that

T and C are approximately equal.Try x = 200 mm.

C = . .0 45 0 9 cu f b x× = 0.45 ×××× 25 ×××× 150 ×××× 0.9 ×××× 200 = 303750 N

( ).0 0035 300 200 200 sε εε ε = − = 1.75 ×××× 10-3

Hence, ( ) ( ). 3 31 75 10 200 10 s f −

= × × = 350 Nmm-2

and T = 350 × 981.7 = 343595 N

C =303750 N, T = 343595 N

Since C is slightly less, need to increase the x. For a better

approximation, try x = 205 mm.C = . .0 45 0 9 cu f b x× = 0.45 ×××× 25 ×××× 150 ×××× 0.9 ×××× 205 = 311344 N

( ).0 0035 300 205 205 s

ε εε ε = − = 1.62 ×××× 10-3

Hence, ( ) ( ). 3 31 62 10 200 10 s f −

= × × = 334 Nmm-2

150 mm

300 mm

2T25

25 cu f =

460 y f =

N/mm2

N/mm2





and T = 324 × 981.7 = 318070 N

Since C is again slightly less, need to increase the x. For a

better approximation, try x = 206 mm.C =

. .0 45 0 9 cu f b x× = 0.45 ×××× 25 ×××× 150 ×××× 0.9 ×××× 206

= 312862 N( ).0 0035 300 206 206

sε εε ε = − = 1.59 ×××× 10

-3

Hence, ( ) ( ). 3 31 59 10 200 10 s f −

= × × = 319 Nmm-2

and T = 319 × 981.7 = 313348 N

This approximation is sufficient. x = 206 mm. x = 206 mm

. .0 45 300 0 45 206 z d x= − = − × = 207 mm z = 207 mm

Moment resistance = C.z = 312862 × 207 =64.8 × 106 Nmm

= 64.8 k Nm M=64.8 kNm

Solution Q6 [DESIGN OF RECTANGULAR SECTION]

Design a section of a rectangular reinforced concrete beam to take an ultimate moment of 150

kNm and draw the section showing the reinforcement details,

(1) as a singly reinforced beam of cross section 225 mm × 525 mm as shown in Figure 6A, and

(2) as a beam of cross section 225 mm × 400 mm as shown in Figure 6B.

• You can use either design formulae or design charts provided in your calculations.

• The characteristic strength of concrete is 25 N/mm2 and that of steel is 460 N/mm2 .

• Assume that the moment redistribution is not more than 10%.

(a) singly reinforced beam

5 2 5 m m

225 mm

d’= 50 mm

225 mm

d = 4 7 5 m m

Fig 6A

Fig 6B

4 0 0 m m

d = 3 5 0 m m





2

cu

M K

bd f = ⇒

{ }

6

2

150 100.148 0.156

225 425 25K

×= = <

× ×

⇒ ( )0.5 0.250.9

K z d = + − = ( )0.148425 0.5 0.250.9

+ −

=329.4 mm

⇒ z <0.95d (= 403.7 mm ), Hence OK.

62150 10

1137.860.87 0.87 460 329.4

s

y

M A mm

f z

×= = =

× ×

Hence use 4T20 ( )21257 s A mm=

(b) Overall depth is restricted

If overall depth is restricted to 400 mm, and the b= 225 mm

h = 400 mm , d = 400 – 50 = 350 mm.

Now2

cu

M K

bd f =

{ }

6

2

150 100.218

225 350 25

×= =

× ×> 0.156

Hence, the compression reinforcement is required.

Assume 50 d mm′ =

( )

( )

2

0.87

cu

s

y

K K f bd A

f d d

′−′ =

′−

( )

( )

20.218 0.156 25 225 350

0.87 460 350 50

− × ×=

× × −

= 356 mm2

Use 2T16 ( )2356 s A mm′ =

( )0.5 0.250.9

K z d ′= + − ( )0.156350 0.5 0.250.9

= + −

= 272 mm <0.95 × 350 = 333 mm2 2

20.156 25 225 350356 13440.87 0.87 460 272

cus s

y

K f bd A A mm f z

′ × × ×

′= + = + =× ×

Use 3T25 ( )21473 s A mm=

Hence, Use 3T25 at the bottom and 2T16 at the top.





Solution Q13 [DESIGN OF A SHORT COLUM SUBJECTED TO AXIAL LOAD AND MOMENT]

The details of a column of rectangular cross section are shown in

Fig. Q13. The column is symmetrically reinforced with four

numbers of 32 mm diameter longitudinal bars. The material

properties are given as 40=cu f MPa for concrete and 460= y f

MPa for longitudinal reinforcement as well as shear links. Using

the design charts check whether the load-moment combination N

=2000 kN and M= 225 kNm is acceptable to the column sectionshown. If not, re-design the longitudinal reinforcement for the

section. Sketch the reinforcement details of your re-designed

cross-section.

Solution Q13

For the given rectangular column section( )2100 4 32 4100

2.68300 400

× ×= =

×

s A

bh

π

2000 100016.66

300 400

×= =

×

N

bh

6

2 2

225 104.69

300 400

×= =

×

M

bh

This combination of axial load and moment does not fall under safe region of the interaction

diagram for the given column section. Therefore, the reinforcement has to be re-designed for

this combination.

4 7 5 m m

(a) Singly reinforced beam

4 0 0 m m

225 mm

d = 350 mm

d’= 50 mm

s A′ = 2T16

s = 3T 25

225 mm

d = 425 mm

s A = 4T20

(b) Doubly reinforced beam

Fig. Q13

3 4 0 m m

4 0 0 m m

300 mm

2T32

2T32





For 16.66= N

bh

and2

4.69= M

bh

Hence,100

3=s A

bh 23 300 400

3600100

× ×⇒ = =

s A mm

Hence, provide 8T25. Bars can be placed evenly distributed at top and bottom as shown in the

sketch or 2 in each corner.

( As, provided = 3927 mm2)

3 4 0 m m

4 0 0 m m

300

4T25

4T25





Question 15 [DESIGN OF A SECTION FOR SHEAR]

A simply supported reinforced concrete beam of rectangular cross-section carries a characteristic

live load of 20 kN/m in addition to a characteristic dead load (including self-weight of beam) of

20 kN/m over its clear span of 5.0 m. The breadth of the beam, b is 225 mm and the effective

depth, d is 400 mm. The characteristic strength of concrete is 25 N/mm

2

and that of shear andlongitudinal reinforcements are 250 N/mm2

and 460 N/mm2

respectively. Design the shear

reinforcement required for the beam given that the percentage of tension reinforcement near thesupport is 0.8%. You may use 10 mm diameter bars for links and assume the same spacing of

links throughout the beam.

(Design load, W = 1.4 Gk + 1.6 Qk )

Solution Q15

Uniformly distributed ultimate design load, W = 1.4 Gk + 1.6 Qk

= 1.4×

20 + 1.6×

20= 60 kN/m

Maximum shear force at the support

max

60 5150

2

×= =V kN

Design shear force

Considering the enhanced shear strength near support,

( )150 2500 400126

2500

× −= =designV kN

Check for maximum design shear stress

⇒

32

max

126 101.4225 400

−×

= =×v Nmm <2

0.8 4−

=cu f Nmm

and not more than 5 Nmm-2

Hence OK

Concrete shear resistance

21000.8, 400 , 25= = =

scu

v

Ad mm f Nmm

b d

5.0 m

60 kN/m

V d V max

Shear force diagram

2500 mm

400 mm





From Table 3.9 of BS 8110 20.58 −=

cv Nmm

Shear reinforcement

From Table 3.8 of BS 8110 Since 20.4 ,−> +

cv v Nmm Links have to designed.

Calculation of links spacing

Assuming 10 mm links, Asv= 157.1 mm2

Hence ,

=

Check for links spacing

< 0.75 d = 300 mm; hence OK.Provided shear reinforcement

Provide R10 Links at 175 mm spacing.

( )

0.87

−=

v csv

v yv

b v v A

s f

( )

0.87≤

−

yv sv

v

v c

f As

b v v ( )

0.87 250 157.1185

225 1.4 0.58

× ×=

−mm

Solutions to Q1,3,5,13 and 15

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