The reinforced concrete beam shown in Figure 1 carries an imposed live load of 12kN/m2 and a dead load of 12 kN/m2 as shown in the diagram. Given, fcu = 30 N/mm2, fy = 460 N/mm2, fy, links = 250 N/mm2. The beam is exposed to a mild condition. a) Determine whether the beam is singly or doubly reinforced beam. b) Determine the main reinforcement for the beam using a design chart. c) Determine the shear reinforcement for the beam.

d) Check whether the deflection of the beam is acceptable. e) Sketch the detailed cross section of the beam indicating the reinforcements in (b) and (c).

a) Determine whether the beam is singly or doubly reinforced beam. Solution

i) Characteristic Loads

Dead loads

Superimposed load=12.00kN/m

Before we find the selfweight we need to find the size of beam. The thickness of the concrete cover is determined by using Table 3.2 and Table 3.3 of BS8110. The condition is mild, therefore the thickness of the concrete cover is 25mm. Assumed the reinforcement bar is 25mmØ and link is 10mmØ.

225

400

448

Hence h = d + ½ Ø bar + link + cover

h = 400 + ½ (25) + 10 + 25 = 448mm

Selfweight (24 X 0.225 X 0.448)=2.4kN/m

Total dead load=14.4kN/m

Live loads

Imposed live load=12kN/m

ii) Design Loads

Design load, w = 1.4gk + 1.6qk

= 1.4(14.4) + 1.6(12)

= 39.36kN/m

iii) Ultimate Bending Moment, M

M= wl28

M= 39.36(6)28 = 177.12kN/m

iv) Ultimate Moment of Resistance, Mu

Mu= 0.156 fcu bd2

Mu=0.156 (30) (225) (400)2

=168.48kN/m

M > Mu

Therefore it is doubly reinforced beam.

b) Determine the main reinforcement for the beam using a design chart.

Solution

Compresses Reinforcement, A’s

Assume compressed reinforcement (A’s) is 12mmØ.

d’=cover + ½ Ø bar + link

For mild condition by using Table 3.2 and Table 3.3 of BS8110 the thickness of the concrete cover is 25mm and link size are as assumed in the Solution 1(a)

d’=25 + ½ 12 + 10 = 41mm

z = d0.5+0.25-k'/0.9 and k’ = 0.156

z = 4000.5+0.25-0.156/0.9

= 310.75mm

x = d-z0.45

x = 400-310.750.45

= 198.89mm

d’x = 41198.89 = 0.206<0.37

Therefore compression steel yields.

A’s = M-Mu0.95fy(d-d')

A’s = (177.12-168.48)1060.95(460)(400-41)

= 55.07mm2

By using Table 3.8 the number of compressed reinforcement bars required is 1 (A’s = 133mm2) but need to adjust the minimum number of compressed reinforcement bars required is 2T12 (A’s = 226mm2)

Tension Reinforcement, As

As = Mu0.95fyz +A's

As = 168.48(10)60.95460(310.75) +55.38

= 1296mm2

By using Table 3.8 the number of tension reinforcement bars required is 3, therefore we write 3T25 (As = 1474mm2)

c) Determine the shear reinforcement for the beam.

Solution

Maximum Shear Force, Vmax

Vmax = w(l)2

Vmax = 39.36(6)2

= 118.08kN

Shear Force Diagram

Vmax = 118.08kN

Maximum Applied Shear Stress, v

v = Vbd

v = 118.08(10)3225(400)

= 1.312N/mm2

BS 8110 limits the maximum applied shear stress (v) to 5N/mm2 or 0.8fcu , whichever is the lesser. In this case, with fcu of 30N/mm2, the maximum applied shear stress, v, is less than the limiting values.

Concrete Shear Stress ,vc

Check the concrete shear stress, vc, from Table 3.8 of BS8110. vc = 100Asbd

vc = 1001296225(400)

= 1.44

This value is between 1.0 to 1.5 of Table 3.8 BS 8110, giving the value of concrete shear stress, vc, of between 0.63 to 0.72. From the linear interpolation between 1.0 to 1.5 we get the value of concrete shear stress, vc,:

y-y0vc-x0 = y1-y0x1-x0

1.44-1vc-0.63= 1.5-10.72-0.63

vc = 0.709N/mm2

For concrete other than grade 25

vc= vc(table 3.8) fcu2513

vc= 0.709 302513

= 0.948N/mm2

Therefore, concrete shear stress, vc, is less than the...