Designing doubly reinforced beam section: An example

When we provide reinforcement on both sides of a beam section, we call it a doubly reinforced beam.

In most basic beam design, we learn that steel is required mainly to resist tension, since concrete is weak in tension.
So a natural question arises—why do we provide reinforcement in the compression zone?

The answer lies in capacity.

When bending moments are high or beam dimensions are restricted, the compression capacity of concrete alone is not sufficient.
By introducing steel in the compression zone, we increase the beam’s ability to resist compressive stresses.

In other words, compression reinforcement is provided not because concrete cannot take compression, but because we want the beam to carry more moment than what a singly reinforced section can offer.

This is the fundamental idea behind a doubly reinforced beam.

Check this post to know more about a doubly reinforced beam section. 

The design procedure of each beam is different only the underlying principle is the same. In this post I will explain about the step by step procedure of design of a beam, by considering an example problem. 

Problem Statement: Design a doubly reinforced beam and use M20 concrete and Fe415 steel. End condition: two span continuous beam with equal spans. Ends are simply supported and span $L = 5 m$

Trial Section:

Assume cross section of the beam $230 mm \times 380 mm, d = 340 mm, d' = 40 mm,$D_f = 130 mm $.

Compute Maximum load :

$$w_{u \max }=1.5(D L+L L)=1.5(\text { self weight+ slab1 }+\text { slab2})$$ $$\begin{aligned} w_{u,\max} &= 1.5 \left[1.6 + 0.55 \times 6.5 \times 4 + 0.6 \times 6.5 \times 4.5 \times 0.73 \right] \\ &= 43 \, \text{kN/m} \\[6pt] w_{u,\min} &= DL = \left[1.6 + 0.55 \times 5 \times 4 + 0.6 \times 5.0 \times 4.5 \times 0.73 \right] \\ &= 22.4 \, \text{kN/m} \end{aligned}$$

Analysis of beam

Design Moment at support as well as at mid- span $= \pm w_u L^\frac{2}{10}$

$$M_{u.max } =43 \times 5^\frac{2}{10} = 107.5 \mathrm{kN} \cdot \mathrm{m}$$ $$M_{\max } = 0.8 \times 134.4 \text{This corresponds to 20} \% \text{redistribution of moment}$$ $$k_{u .limit} = 0.6- 0.2 = 0.4, x_{u. limit} = 0.4 \times 340 = 136 mm$$

For this,

$$\begin{aligned} M_{\text { ur. limit }} & =0.36 \times 20 \times 0.4 \times(1-0.42 \times 0.4) \times 230 \times 340^2 \times 10^{-6} \\ & =63.7 \mathrm{kN} . \mathrm{m}<107.5 \mathrm{kN} . \mathrm{m} \end{aligned}$$ $$M_{ u2} = 107.5 -63.7 =43.8 kNm$$

Main Steel

$$\begin{aligned} & A_{st1}=\frac{63.7 \times 10^6}{ 0.87 \times 415 \times(340-0.42 \times 136)}= 624 \mathrm{~mm}^2 \\ & A_{st2}=\frac{43.8 \times 10^6}{ 0.87 \times 415 \times(340-40)}=404 \mathrm{~mm}^2 \\ & \therefore A_{st} =A_{st1 }+A_{ st2 } = 624+ 404 = 1028 \mathrm{~mm}^2 \end{aligned}$$

$\frac{d_c}{d} = \frac{40}{340} = 0.117, A_{sc} = 1.063 \times 404 = 430 mm^2$

At mid span

$$M_{u \cdot \max }=107.5 \mathrm{kN} . \mathrm{m}>M_{u r \cdot \max }=73.38 \mathrm{kN} . \mathrm{m}\left(=2.76 \times 230 \times 0.34^2\right)$$

Therefore, assistance will have to be taken of the flange.

$$L_o \quad=0.7 L=0.7 \times 5000= 3500 \mathrm{~mm}$$ $$b_f \quad=(\frac{3500}{6}+6 \times 130)+ 230= 1593 \mathrm{~mm}$$

When $x_u = D_f$,

$$\begin{aligned} M_{ur} = 0.36 \times 20 \times 1593 \times 130 \times (340 - 0.42 \times 150) \times 10^{-6} \\ = 425 kN.m > M_{u.max}( = 107.5 kN.m) \therefore x_u < D_f \end{aligned}$$ $$A_{st} = \frac{0.5 \times 20}{415} \left[ 1 - \sqrt{ 1 - \frac{4.6 \times 107.5 \times 10^6} {20 \times 1593 \times 340^2} } \right] \times 1593 \times 340 = 908 \,\text{mm}^2$$

Detailing:

Required A_{st} in mm^2	Simple support	At mid-span	At Continuous end
At top At bottom	------ ------	------ 908	1028 430
No. dia at top No. dia at bottom	2 # 10 3 # 20	2 #10 3 #20	2 #10 + 3 #20 2 #20
Provided $A_{st}$ at top $A_{st}$ at bottom	157 942	157 942	1099 628

Conclusions

In this blog post, we have delved into a practical problem that provides an explanation of a DRB. As per IS:456:2000, Indian Standards, we have designed a R.C.C beam section.

So, let us discuss the key points from this problem as follows:

• Significance of Design and Analysis: It gives the need to consider various factors, including load distribution, to ensure the function of a beam.

• Consideration of Critical Factors: Throughout our design procedure, we have discussed significant factors such as bending moments, reinforcement, and deflection limits.

• Integration of Theory and Practice: All in one, by following the reinforcement detailing guidelines as per Indian codes, engineers can develop structurally sound and efficient solutions that meet safety and performance criteria.