Instructional Objectives: At the end of this lesson, the student should be able to:
5.12.1 Introduction
Lesson 10 illustrates the governing equations of flanged beams and Lesson 11 explains their applications for the solution of analysis type of numerical problems. It is now necessary to apply them for the solution of design type, the second type of the numerical problems. This lesson mentions the different steps of the solution and solves several numerical examples to explain their stepbystep solutions.
5.12.2 Design Type of Problems
We need to assume some preliminary dimensions of width and depth of flanged beams, spacing of the beams and span for performing the structural analysis before the design. Thus, the assumed data known for the design are: D_{f}, b_{w}, D, effective span, effective depth, grades of concrete and steel and imposed loads. There are four equations: (i) expressions of compressive force C, (ii) expression of the tension force T, (iii) C = T and (iv) expression of M_{u} in terms of C or T and the lever arm {M = (C or T ) (lever arm)}. However, the relative dimensions of D_{f}, D and x_{u} and the amount of steel (underreinforced, balanced or overreinforced) influence the expressions. Accordingly, the respective equations are to be employed assuming a particular situation and, if necessary, they need to be changed if the assumed parameters are found to be not satisfactory. The steps of the design problems are as given below.
Step 1: To determine the factored bending moment M_{u}
Step 2: To determine the M_{u,lim} of the given or the assumed section
The beam shall be designed as underreinforced, balanced or doubly reinforced if the value of M_{u } is less than, equal to or more than M_{u,lim.} The design of overreinforced beam is to be avoided as it does not increase the bending moment carrying capacity beyond M_{u,lim }either by increasing the depth or designing a doubly reinforced beam.
Step 3: To determine x_{u}, the distance of the neutral axis, from the expression of M_{u }_{ }
Here, it is necessary to assume first that x_{u} is in the flange. Later on, it may be necessary to calculate x_{u} if the value is found to be more than D_{f} . This is to be done assuming first that D_{f} /x_{u} < 0.43 and then D_{f} /x_{u} > 0.43 separately.
Step 4: To determine the area(s) of steel
For doubly reinforced beams A_{st }= A_{st,lim }+ A_{st2 } and A_{sc} are to be obtained, while only A_{st} is required to be computed for underreinforced and balanced beams. These are calculated employing C = T (for A_{st } and A_{st, lim}) and the expression of M_{u2} to calculate A_{st2 }and A_{sc}.
Step 5: It may be necessary to check the xu and Ast once again after Step 4
It is difficult to prescribe all the relevant steps of design problems. Decisions are to be taken judiciously depending on the type of problem. For the design of a balanced beam, it is necessary to determine the effective depth in Step 3 employing the expression of bending moment M_{u}. For such beams and for underreinforced beams, it may be necessary to estimate the A_{st} approximately immediately after Step 2. This value of A_{st } will facilitate to determine x_{u}.
5.12.3 Numerical Problems
Four numerical examples are solved below explaining the steps involved in the design problems.
Ex.5: Design the simply supported flanged beam of Fig. 5.12.1, given the following: D_{f} = 100 mm, D = 750 mm, b_{w }= 350 mm, spacing of beams = 4000 mm c/c, effective span = 12 m, cover = 90 mm, d = 660 mm and imposed loads = 5 kN/m^{2}. Fe 415 and M 20 are used.
Solution:
Step 1: Computation of factored bending moment
Weight of slab per m2 = (0.1) (1) (1) (25) = 2.5 kN/m2
So, Weight of slab per m = (4) (2.5) = 10.00 kN/m
Dead loads of web part of the beam = (0.35) (0.65) (1) (25) = 5.6875 kN/m
Imposed loads = (4) (5) = 20 kN/m Total loads = 30 + 5.6875 = 35.6875 kN/m
Factored Bending moment = 963.5625 kNm
Step 2: Computation of x_{u,lim }_{ }
Effective width of flange =(l_{o}/6) + b_{w}+ 6 D_{f} = (12000/6) + 350 + 600 = 2,950 mm.
x_{u,max } = 0.48 d = 0.48 (660) = 316.80 mm. This shows that the neutral axis is in the web of this beam.
D_{f }/d = 100/660 = 0.1515 < 0.2, and
D_{f} /x_{u} = 100/316.8 = 0.316 < 0.43
The expression of M_{u,lim } is obtained from Eq. 5.7 of Lesson 10 (case ii a of sec. 5.10.4.2) and is as follows:
M_{u,lim} = 0.36(xu,max /d){1  0.42 (x_{u,max }/d)} f_{ck} b_{w} d^{2} + 0.45 fck (b_{f}  b_{w}) D_{f} (d  D_{f} /2)
= 0.36(0.48) {1  0.42(0.48)} (20) (350) (650) (650)
+ 0.45 (20) (2950  350) (100) (660  50) = 1,835.43 kNm
The design moment M_{u} = 963.5625 kNm is less than M_{u,lim.} Hence, one underreinforced beam can be designed.
Step 3: Determination of x_{u}_{ }
Since the design moment M_{u }is almost 50% of M_{u,lim}, let us assume the neutral axis to be in the flange. The area of steel is to be calculated from the moment equation (Eq. 3.23 of Lesson 5), when steel is ensured to reach the design stress fd = 0.87 (415) = 361.05 N/mm^{2}. It is worth mentioning that the term b of Eq. 3.23 of Lesson 5 is here bf as the T beam is treated as a rectangular beam when the neutral axis is in the flange.
(3.23)
Here, all but Ast are known. However, this will give a quadratic equation of A_{st} and the lower one of the two values will be provided in the beam. The above equation gives:
which gives the lower value of A_{st} as: A_{st }= 4,234.722097 mm^{2}. The reason of selecting the lower value of A_{st} is explained in sec 3.6.4.8 of Lesson 6 in the solution of Design Problem 3.1. Then, employing Eq. 3.16 of Lesson 5, we get
(3.16)
or x_{u} = 71.98 mm. Again, employing Eq. 3.24 of Lesson 5, we can determine x_{u} first and then A_{st} from Eq. 3.16 or 17 of Lesson 5, as explained in the next step. Eq. 3.24 of Lesson 5 gives:
M_{u } = 0.36(x_{u} /d) {1  0.42(x_{u} /d)} f_{ck }b_{f }d^{2}
= 0.36 (x_{u}) {1  0.42 (x_{u} /d)} f_{ck} b_{f }d
963.5625 (10^{6}) = 0.36 (x_{u}) {1  0.42 (x_{u }/660)} (20) (2950) (660)
or x_{u} = 72.03 mm.
The two values of x_{u} are the same. It is thus seen that, the value of x_{u } can be determined either first finding the value of A_{st}, from Eq. 3.23 of Lesson 5 or directly from Eq. 3.24 of Lesson 5 first and then the value of A_{st }can be determined.
Step 4: Determination of A_{st }
Equating C = T , we have from Eq. 3.17 of Lesson 5:
Minimum A_{st } = (0.85/f_{y}) b_{w} d = (0.85/415) (350) (660) = 473.13 mm^{2}
Maximum A_{st} = 0.04 b_{w} D = (0.04) (350) (660) = 9,240 mm^{2}
Hence, A_{st } = 4,237.41 mm^{2} is o.k.
Provide 6  28 T (= 3694 mm^{2}) + 220 T (= 628 mm^{2}) to have total A_{st} = 4,322 mm^{2}
Ex.6: Design a beam in place of the beam of Ex.5 (Fig. 5.12.1) if the imposed loads are increased to 12 kN/m^{2}. Other data are: D_{f} = 100 mm, b_{w} = 350 mm, spacing of beams = 4000 mm c/c, effective span = 12 m simply supported and cover = 90 mm. Use Fe 415 and M 20.
Solution: As in Ex.5, b_{f }= 2,950 mm.
Step 1: Computation of factored bending moment
Weight of slab/m^{2} = 2.5 kN/m^{2} (as in Ex.1)
Imposed loads = 12.0 kN/m^{2} (given)
Total loads = 14.5 kN/m^{2}
Total weight of slab and imposed loads = 14.5 (4) = 58.0 kN/m
Dead loads of the beam = 0.65 (0.35) (25) = 5.6875 kN/m
Total loads = 63.6875 kN/m
Step 2: Determination of M_{u,lim}
M_{u,lim} of the beam of Ex.5 = 1,835.43 kNm. The factored moment of this problem (1,719.5625 kNm) is close to the value of M_{u,lim }of the section.
Step 3: Determination of d
Assuming D_{f} /d < 0.2, we have from Eq. 5.7 of Lesson 10,
M_{u} = 0.36(x_{u,max} /d){1  0.42 (x_{u,max} /d)} f_{ck} b_{w }d^{2} + 0.45 f_{ck }(b_{f}  b_{w}) D_{f }(d  D_{f} /2)
1719.5625 (10^{6}) = 0.36(0.48) {1  0.42(0.48)} (20) (350) d^{2 } + 0.45 (2600) (20) (100) (d  50)
Solving the above equation, we get d = 624.09 mm, giving total depth = 624.09 + 90 = 715 mm (say).
Since the dead load of the beam is reduced due to decreasing the depth of the beam, the revised loads are calculated below:
Loads from the slab = 58.0 kN/m
Dead loads (revised) = 0.615 (0.35) (25) = 5.38125 kN/m
Total loads = 63.38125 kN/m
Approximate value of A_{st: }
Step 4: Determination of Ast (Fig. 5.12.2)
x_{u} = x_{u,max } = 0.48 (625) = 300 mm
Equating T and C (Eq. 5.5 of Lesson 10), we have: 0.87 f_{y} A_{st} = 0.36 x_{u,max }b_{w} f_{ck }+ 0.45 f_{ck} (b_{f}  b_{w}) D_{f}
Maximum A_{st } = 0.04 b D = 0.04 (350) (715) = 10,010.00 mm^{2}
Minimum A_{st} = (0.85/fy) b_{w} d = (0.85/415) (350) (625) = 448.05 mm^{2}
Hence, A_{st } = 8,574.98 mm^{2} is o.k.
So, provide 836 T + 218 T = 8143 + 508 = 8,651 mm^{2}
Step 5: Determination of x_{u}_{ }
Using A_{st} = 8 ,651 mm2 in the expression of T = C (Eq. 5.5 of Lesson 10), we have:
0.87 f_{y} A_{st } = 0.36 x_{u} b_{w} f_{ck }+ 0.45 f_{ck} (b_{f}  b_{w}) D_{f}
So, A_{st} provided is reduced to 836 + 216 = 8143 + 402 = 8,545 mm^{2}. Accordingly,
Step 6: Checking of M_{u}_{ }
D_{f} /d = 100/625 = 0.16 < 0.2 D_{f} /x_{u} = 100/215.7 = 0.33 < 0.43.
Hence, it is a problem of case (iii a) and M_{u} can be obtained from Eq. 5.14 of Lesson 10.
So, M_{u} = 0.36(x_{u} /d) {1  0.42(x_{u} /d)} f_{ck} b_{f }d^{2} + 0.45 f_{ck} (b_{f}  b_{w}) (D_{f}) (d  D_{f} /2)
= 0.36 (295.703/625) {1  0.42 (295.703/625)} (20) (350) (625) (625)
+ 0.45 (20) (2600) (100) (625  50)
= 1,718.68 kNm > (M_{u})design (= 1,711.29 kNm)
Hence, the design is o.k.
Ex.7: Determine the tensile reinforcement A_{st} of the flanged beam of Ex.5 (Fig. 5.12.1) when the imposed loads = 12 kN/m^{2}. All other parameters are the same as those of Ex.5: D_{f }= 100 mm, D = 750 mm, b_{w} = 350 mm, spacing of beams = 4000 mm c/c, effective span = 12 m, simply supported, cover = 90 mm and d = 660 mm. Use Fe 415 and M 20.
Solution:
Step 1: Computation of factored bending moment M_{u}
Dead loads of the slab (see Ex.5) = 2.5 kN/m^{2}
Imposed loads = 12.0 kN/m^{2} Total loads = 14.5 kN/m^{2}
Loads/m = 14.5 (4) = 58.0 kN/m
Dead loads of beam = 0.65 (0.35) (25) = 5.6875 kN/m
Total loads = 63.6875 kN/m
Factored M_{u} = (1.5) (63.6875) (12) (12)/8 = 1,719.5625 kNm.
Step 2: Determination of M_{u,lim}
From Ex.5, the M_{u,lim} of this beam = 1,835.43 kNm. Hence, this beam shall be designed as underreinforced.
Step 3: Determination of x_{u}
Assuming x_{u} to be in the flange, we have from Eq. 3.24 of Lesson 5 and considering b = b_{f},
M_{u} = 0.36x_{u } {1  0.42(x_{u }/d)} f_{ck }b_{f} d 1719.5625 (10^{6})
= 0.36 x_{u} {1  0.42 (xu /660)} (20) (2950) (550)
Solving, we get x_{u} = 134.1 > 100 mm
So, let us assume that the neutral axis is in the web and D_{f} /x_{u} < 0.43, from Eq. 5.14 of Lesson 10 (case iii a of sec. 5.10.4.3), we have:
M_{u } = 0.36(x_{u} /d) {1  0.42(x_{u} /d)} f_{ck} b_{w} d^{2} + 0.45 f_{ck} (b_{f}  b_{w}) (D_{f}) (d  D_{f} /2)
= 0.36 x_{u} {1  0.42 (x_{u} /660)} (20) (350) (660)
+ 0.45 (20) (2600) (100) (660  50)
Substituting the value of
M_{u} = 1,719.5625 kNm
in the above equation and simplifying, x_{u}^{2 }  1571.43 x_{u} + 276042 = 0
Solving, we have
x_{u} = 201.5 mm D_{f} /x_{u} = 100/201.5 = 0.496 > 0.43.
So, we have to use Eq. 5.15 and 5.18 of Lesson 10 for y_{f }and M_{u} (case iii b of sec. 5.10.4.3).
Thus, we have: M_{u} = 0.36 x_{u} {1  0.42( x_{u} /d)} f_{ck }b_{w }d + 0.45 f_{ck} (b_{f}  b_{w}) y_{f} (d  y_{f} /2)
where, y_{f }= (0.15 x_{u} + 0.65 D_{f})
So, M_{u} = 0.36 x_{u} {1  0.42 (xu/660)} (20) (350) (660)
+ 0.45 (20) (2600) (0.15 x_{u }+ 65) (660  0.075 x_{u}  32.5) or 1719.5625 (10^{6} )
= 3.75165 (106) x_{u}  795.15 x_{u}^{2 } + 954.4275 (10^{6})
Solving, we get x_{u }= 213.63 mm. D_{f} /x_{u} = 100/213.63 = 0.468 > 0.43. Hence, o.k.
Step 4: Determination of A_{st }
Equating C = T from Eqs. 5.16 and 5.17 of Lesson 10 (case iii b of sec. 5.10.4.3), we have:
0.87 f_{y} A_{st} = 0.36 f_{ck} b_{w} x_{u }+ 0.45 f_{ck }(b_{f} – b_{w}) y_{f} where, y_{f} = 0.15 x_{u }+ 0.65 D_{f}
Here, using x_{u} = 213.63 mm, D_{f} = 100 mm,
we get y_{f} = 0.15 (213.63) + 0.65 (100) = 97.04 mm
So,
Minimum Ast = (0.85/fy) (bw) (d) = 0.85 (350) (660)/(415) = 473.13 mm2 Maximum Ast = 0.04 bw D = 0.04 (350) (750) = 10,500 mm2 Hence, Ast = 7,780.32 mm2 is o.k.
Provide 636 T + 328 T (6107 + 1847 = 7,954 mm^{2}). Please refer to Fig. 5.12.3.
Step 5: Checking of x_{u } and M_{u} using A_{st} = 7,954 mm^{2}
From T = C (Eqs. 5.16 and 5.17 of Lesson 10), we have
0.87 f_{y} A_{st} = 0.36 f_{ck} b_{w }x_{u} + 0.45 f_{ck} (b_{f }– b_{w}) y_{f }where, y_{f} = 0.15 x_{u} + 0.65 D_{f}
or 0.87 (415) (7954) = 0.36 (20) (350) x_{u} + 0.45 (20) (2600) (0.15 x_{u }+ 0.65 D_{f})
or x_{u} = 224.01 mm D_{f} /x_{u } = 100/224.01 = 0.446 > 0.43.
Accordingly, employing Eq. 5.18 of Lesson 10 (case iii b of sec. 5.10.4.3), we have:
So, M_{u} = 0.36 x_{u} {1  0.42( x_{u} /d)} f_{ck} b_{w} d + 0.45 f_{ck }(b_{f } b_{w}) y_{f} (d  y_{f} /2)
= 0.36 (224.01){1  0.42 (224.01/660)} (20) (350) (660)
+ 0.45 (20) (2600) {(0.15) 224.01 + 65} {(660)  0.15 (112)  32.5}
= 1,779.439 kNm > 1,719.5625 kNm
Hence, o.k.
Ex.8: Design the flanged beam of Fig. 5.12.4, given in following: D_{f} = 100 mm, D = 675 mm, b_{w }= 350 mm, spacing of beams = 4000 mm c/c, effective span = 12 m simply supported, cover = 90 mm, d = 585 mm and imposed loads = 12 kN/m^{2}. Use Fe 415 and M 20.
Step 1: Computation of factored bending moment, M_{u }
Weight of slab/m^{2} = (0.1) (25) = 2.5 kN/m^{2}
Imposed loads = 12.0 kN/m^{2} Total loads = 14.5 kN/m^{2}
Total weight of slab + imposed loads/m = 14.5 (4) = 58 kN/m
Dead loads of beam = 0.575 (0.35) (25) = 5.032 kN/m
Total loads = 63.032 kN/m
Factored M_{u} = (1.5) (63.032) (12) (12)/8 = 1,701.87 kNm
Step 2: Determination of M_{u,lim }
Assuming the neutral axis to be in the web, D_{f} /x_{u} < 0.43 and Df /d = 100 / 585 = 0.17 < 0.2, we consider the case (ii a) of sec. 5.10.4.2 of Lesson 10 to get the following:
M_{u,lim }= 0.36 (x_{u,max} /d) {1 – 0.42 (x_{u,max }/d)} f_{ck} b_{w} d^{2}
+ 0.45 f_{ck} (b_{f }– b_{w}) D_{f} (d – Df /2)
= 0.36(0.48) {1 – 0.42 (0.48)} (20) (350) (585) (585)
+ 0.45(20) (2600) (100) (585 – 50) = 1,582.4 kNm
Since, factored M_{u} > M_{u,lim}, the beam is designed as doubly reinforced.
M_{u}^{2} = M_{u}  M_{u,lim} = 1701.87  1582.4 = 119.47 kNm
Step 3: Determination of area of steel
A_{st,lim } is obtained equating T = C (Eqs. 5.5 and 6 of Lesson 10).
0.87 f_{y} (A_{st}, l_{im}) = 0.36 b_{w} (x_{u,max} /d) d f_{ck} + 0.45 f_{ck} (b_{f}  b_{w}) D_{f }
or
mm^{2}
where f_{sc } = 353 N/mm^{2} for d'/d = 0.1 f_{cc }
= 0.446 f_{ck } = 0.446 (20) = 8.92 N/mm^{2 }
M_{u}^{2} = 119.47 (10^{6}) Nmm
d' = 58.5 mm
d = 585 mm
Using the above values in the expression of A_{sc} (Eq. 4.4 of Lesson 8), we get A_{sc} = 659.63 mm^{2}
Substituting the values of A_{sc}, f_{sc}, f_{cc } and f_{y } we get A_{st}^{2 } = 628.48 mm^{2}
Total A_{st } = A_{st,lim} + A_{st}^{2} = 8,440.98 + 628.48 = 9,069.46 mm^{2}
Maximum A_{st} = 0.04 bw D = 0.04 (350) (675) = 9,450 mm^{2}and minimum A_{st} = (0.85/fy) b_{w} d = (0.85/415) (350) (585) = 419.37 mm^{2}
Hence, A_{st} = 9, 069.46 mm^{2} is o.k.
Provide 836 T + 320 T = 8143 + 942 = 9,085 mm^{2 } for A_{st} and 120 + 216 = 314 + 402 = 716 mm^{2} for A_{sc} (Fig. 5.12.5).
Step 4: To check for x_{u} and M_{u} (Fig. 5.12.5)
Assuming x_{u} in the web and D_{f }/x_{u } < 0.43 and using T = C (case ii a of sec. 5.10.4.2 of Lesson 10 with additional compression force due to compression steel), we have:
0.87 f_{y} A_{st }= 0.36 b_{w} x_{u} f_{ck } + 0.45 (b_{f } b_{w}) f_{ck} D_{f }+ A_{sc} (f_{sc } f_{cc})
or 0.87 (415 ) (908 5) = 0.36 (350) x_{u} (20) + 0.45 (2600) (20) (100)
+ 716 {353  0.45 (20)}
This gives x_{u }= 275.33 mm. x_{u,max } = 0.48 (d) = 0.48 (585) = 280.8 mm.
So, x_{u }< x_{u,max,} D_{f }/x_{u } = 100/275.33 = 0.363 < 0.43 and D_{f }/d = 100/585 = 0.17 < 0.2.
The assumptions, therefore, are correct. So, M_{u} can be obtained from Eq. 5.14 of sec. 5.10.4.3 of Lesson 10 with additional moment due to compression steel, as given below:
So, M_{u} = 0.36 b_{w} x_{u} f_{ck }(d  0.42 x_{u}) + 0.45 (b_{f}  b_{w}) f_{ck} D_{f }(d  D_{f} /2)
+ A_{sc }(f_{sc}  f_{cc}) (d  d')
= 0.36 (350) (275.33) (20) {585  0.42 (275.33)}
+ 0.45 (2600) (20) (100) (585  50) + 716 (344) (585  58.5)
= 325.66 + 1251.9 + 129.67 = 1,707.23 kNm
Factored moment = 1,701.87 kNm < 1,707.23 kNm. Hence, o.k.
5.12.4 Practice Questions and Problems with Answers
Q.1: Determine the steel reinforcement of a simply supported flanged beam (Fig. 5.12.6) of D_{f} = 100 mm, D = 700 mm, cover = 50 mm, d = 650 mm, b_{w} = 300 mm, spacing of the beams = 4,000 mm c/c, effective span = 10 m and imposed loads = 10 kN/m^{2}. Use M 20 and Fe 415.
A.1: Solution:
Step 1: Computation of (M_{u})_{factored }
The design moment M_{u }= 963.5625 kNm is less than M_{u,lim.}
Hence, one underreinforced beam can be designed.
Total loads per m = (12.5) (4) = 50 kN/m
Dead loads of beam = (0.3) (0.6) (25) = 4.50 kN/m
Total loads = 54.50 kN/m
Factored M_{u} = ( 1.5) (54.50) (10) (10)/8 = 1,021.87 kNm
Step 2: Determination of M_{u,lim}
Effective width of the flange b_{f }= l_{o}/6 + b_{w} + 6 D_{f} = (10,000/6) + 300 + 600 = 2,567 mm.
x_{u,max } = 0.48 d = 0.48 (650) = 312 mm
Hence, the balanced neutral axis is in the web of the beam.
D_{f }/d = 100/650 = 0.154 < 0.2 D_{f} /x_{u} = 100/312 = 0.32 < 0.43
So, the full depth of flange is having a stress of 0.446 f_{ck}.
From Eq. 5.7 of Lesson 10 (case ii a of sec. 5.10.4.2), we have,
M_{u,lim } = 0.36 (x_{u,max }/d) {1 – 0.42 (x_{u,max} /d)} f_{ck }b_{w} d^{2 } + 0.45 fck (b_{f }– b_{w}) D_{f} (d – D_{f} /2)
= 0.36(0.48) {1 – 0.42(0.48)} (20) (300) (650) (650)
+ 0.45(20) (2267) (100) (650 – 50)
= 1573.92 kNm > M_{u} (= 1021.87 kNm) So, the beam will be underreinforced one.
Step 3: Determination of x_{u}
Assuming x_{u} is in the flange, we have from Eq. 3.24 of Lesson 5 (rectangular beam when b = b_{f} ).
M_{u} = 0.36 (x_{u }/d) {1 – 0.42 (x_{u} /d)} f_{ck} b_{f} d^{2}
= 0.36 x_{u} {1 – 0.42 (x_{u} /d)} f_{ck }b_{f }d 1021.87 (106)
= 0.36 x_{u} {1 – 0.42(x_{u}/650)} (20) (2567) (650)
x_{u} = 89.55 mm^{2 }< 100 mm (Hence, the neutral axis is in the flange.)
Step 4: Determination of A_{st }
Equating C = T , we have from Eq. 3.17 of Lesson 5:
Minimum
Maximum A_{st }= 0.04 b_{w }D = (0.04) (300) (700) = 8,400 mm^{2}
So, A_{st } = 4,584.12 mm^{2} is o.k.
Provide 628 T + 225 T = 3694 + 981 = 4,675 mm^{2} (Fig. 5.12.6
5.12.6 Test 12 with Solutions
Maximum Marks = 50,
Maximum Time = 30 minutes
Answer all questions.
TQ.1: Determine the steel reinforcement A_{st }of the simply supported flanged beam of
Q.1 (Fig. 5.12.6) having D_{f} = 100 mm, D = 700 mm, cover = 50 mm, d = 650 mm, b_{w} = 300 mm, spacing of the beams = 4,000 mm c/c, effective span = 12 m and imposed loads = 10 kN/m^{2}. Use M 20 and Fe 415.
A.TQ.1:
Solution: Step 1: Computation of (M_{u})_{factored }
Total loads from Q.1 of sec. 5.12.4 = 54.50 kN/m
Factored M_{u } = (1.5) (54.50) (12) (12)/8 = 1,471.5 kNm
Step 2: Determination of M_{u,lim}_{ }
Effective width of flange = l_{o}/6 + b_{w }+ 6
D_{f }= (12000/6) + 300 + 600 = 2,900 mm (Fig. 5.12.7)
x_{u,max } = 0.48 d = 0.48 (650) = 312 mm
Hence, the balanced neutral axis is in the web.
D_{f} /d = 100/650 = 0.154 < 0.2
D_{f} /x_{u} = 100/312 = 0.32 < 0.43
So, the full depth of flange is having constant stress of 0.446 fck. From Eq. 5.7 of Lesson 10 (case ii a of sec. 5.10.4.2), we have
M_{u,lim} = 0.36 (x_{u,max }/d) {1 – 0.42 (x_{u,max }/d)} f_{ck }b_{w }d^{2 } + 0.45 f_{ck} (b_{f }– b_{w}) D_{f} (d – D_{f} /2)
= 0.36(0.48) {1 – 0.42(0.48)} (20) (300) (650) (650) + 0.45(20) (2600) (100) (650 – 50) = 1,753.74 kNm > 1,471.5 kNm
So, the beam will be underreinforced.
Step 3: Determination of x_{u }
Assuming x_{u} to b e in the flange, we have from Eq. 3.24 of Lesson 5 (singly reinforced rectangular beam when b = b_{f} ):
M_{u } = 0.36 x_{u} {1 – 0.42 (x_{u} /d)} f_{ck} b_{f} d
or 1471.5 (106 ) = 0.36 (x_{u}) {1 – 0.42 (x_{u} /650)} (20) (2900) (650)
or x_{u}^{2} – 1547.49 x_{u }+ 167.81 (103) = 0
Solving, we have x_{u} = 117.34 mm > 100 mm
So, neutral axis is in the web. Assuming D_{f} /x_{u} < 0.43, we have from Eq. 5.14 of Lesson 10 (case iii a of sec. 5.10.4.3),
M_{u} = 0.36 x_{u} {1 – 0.42 (x_{u} /d)} f_{ck} b_{w} d + 0.45 f_{ck} (b_{f }– b_{w}) D_{f} (d – D_{f} /2)
= 0.36 xu {1 – 0.42 (xu/650)} (20) (300) (650) + 0.45(20) (2600) (100) (650 – 50)
or x_{u}^{2 }– 1547.62 x_{u }+ 74404.7 = 0
Solving, we have x_{u} = 49.67 < 100 mm
However, in the above when it is assumed that the neutral axis is in the flange xu is found to be 117.34 mm and in the second trial when x_{u} is assumed in the web xu is seen to be 49.67 mm. This indicates that the full depth of the flange will not have the strain of 0.002, neutral axis is in the web and Df /xu is more than 0.43. So, we have to use Eq. 5.18 of Lesson 10, with the introduction of y_{f} from Eq. 5.15 of Lesson 10.
Assuming D_{f} /x_{u} > 0.43, from Eqs. 5.15 and 5.18 of Lesson 10 (case iii b of sec. 5.10.4.3), we have:
M_{u} = 0.36 x_{u} {1 – 0.42 (x_{u} /d)} f_{ck} b_{w }d + 0.45 f_{ck} (b_{f }– b_{w}) y_{f} (d – y_{f} /2)
where, y_{f } = (0.15 x_{u} + 0.65 D_{f}) So,
M_{u} = 0.36 x_{u} {1 – 0.42 (x_{u}/650)} (20) (300) (650) + 0.45(20) (2600) (0.15 x_{u} + 0.65) (650 – 0.075 x_{u} – 0.325 x_{u})
or, 1471.5 (10 6) =  1170.45 x_{u}^{2} + 3.45735 xu + 939.2175 (106)
Solving, we get x_{u} = 162.9454 mm.
This shows that the assumption of D_{f} /x_{u} > 0.43 is correct as D_{f} /x_{u} = 100 / 162.9454 = 0.614.
Step 4: Determination of Ast
Equating C = T from Eqs. 5.16 and 5.17 of Lesson 10 (case iii b of sec. 5.10.4.3), we have
0.87 f_{y} A_{st } = 0.36 f_{ck} b_{w} x_{u }+ 0.45 f_{ck} (b_{f }– b_{w}) y_{f}
= 974.829 + 5,796.81 = 6,771.639 mm^{2}Minimum
A_{st } = (0.85/f_{y}) (b_{w}) (d) = 0.85 (300) (650)/415 = 399.39 mm^{2}
Maximum A_{st }= 0.04 (b_{w}) (D) = 0.04 (300) (700) = 8,400 mm^{2}
So, A_{st }= 6,771.639 is o.k. Provide 236 T + 632 T = 2035 + 4825 = 6,860 mm^{2 } > 6,771.639 mm^{2} (Fig. 5.12.7).
Summary of this Lesson
This lesson explains the steps involved in solving the design type of numerical problems. Further, several examples of design type of numerical problems are illustrated explaining the steps of their solutions. Solutions of practice problems and test problems will give the readers confidence in applying the theory explained in Lesson 10 in solving the numerical problems.
13 videos42 docs34 tests

1. What is a flanged beam and how is it different from a regular beam? 
2. What are the advantages of using flanged beams in civil engineering projects? 
3. How can I calculate the loadbearing capacity of a flanged beam? 
4. Are there any limitations to using flanged beams in civil engineering projects? 
5. Can flanged beams be used in seismicresistant structures? 

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