QUESTION 1 / SOALAN 1
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
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.
Hence h = d + ½ Ø bar + link + cover
h = 400 + ½ (25) + 10 + 25 = 448mm
Selfweight (24 X 0.225 X 0.448)
Total dead load
Imposed live load
ii) Design Loads
Design load, w
= 1.4gk + 1.6qk
= 1.4(14.4) + 1.6(12)
iii) Ultimate Bending Moment, M
M= 39.36(6)28 = 177.12kN/m
iv) Ultimate Moment of Resistance, Mu
0.156 fcu bd2
0.156 (30) (225) (400)2
M > Mu
Therefore it is doubly reinforced beam.
b) Determine the main reinforcement for the beam using a design chart.
Compresses Reinforcement, A’s
Assume compressed reinforcement (A’s) is 12mmØ.
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)
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
x = d-z0.45
x = 400-310.750.45
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)
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
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.
Maximum Shear Force, Vmax
Vmax = w(l)2
Vmax = 39.36(6)2
Shear Force Diagram
Vmax = 118.08kN
Maximum Applied Shear Stress, v
v = Vbd
v = 118.08(10)3225(400)
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)
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
vc = 0.709N/mm2
For concrete other than grade 25
vc= vc(table 3.8) fcu2513
vc= 0.709 302513
Therefore, concrete shear stress, vc, is less than the...
Please join StudyMode to read the full document