The design approach to designing reinforced concrete beam to BS 8110:1:1997 is to design for ultimate limit state and check whether serviceability limit state criteria are met. The foremost of the ultimate limit state to design for is bending, afterwards ultimate limit state of shear is designed for and then the beam is checked for serviceability limit state of deflection.
2.0 Flexural Design
The beam is designed to successfully resist the internal bending stress that develops in the beam as a result of the effect of the external load and self-weight. This bending phenomenon, in a typical simply supported beam for example, causes the beam section above the neutral axis to be in compression while those portion below the neutral axis are subjected to tension. The tension and compression are greatest at bottommost fibers and topmost fibers of the beam respectively.
Do you remember this buzzword: Steel is good in tension while concrete is good in compression?
Can you connect the dots?
Since it is established that the beam section above the neutral axis is subjected to compression while those below the neutral axis are subjected to tension; concrete which is ordinarily good in compression resists the compression at the top of the beam while reinforcement steel is provided at the bottom to resist tension at the bottom of the beam below the neutral axis. This is the basic philosophy of reinforced concrete design!
Now, how consistent is this philosophy? Have you not seen concrete beams that have reinforcement at the top?
To rationalize this, we need to understand the concept of singly and doubly reinforced beam.
2.1 Singly and Doubly Reinforced Beam
Simply put: A singly reinforced beam is a beam that the concrete section in compression is adequate to resist the compressive stress developed in the beam due to loading without being complemented by reinforcement bars. For the beam to successfully resist compressive stress, the moment of resistance of the concrete must be larger than the moment induced in the beam by loading.
i.e.: Mu > M
The moment of resistance (Mu) is given by:
Mu = 0.156bd2 fcu (where 0.156 = K’)
Once it is ascertained that the beam can be designed as a singly reinforced section, then we can now proceed to determine the area of reinforcement needed to resist the tensile stress.
The area of reinforcement needed is given by:
$A_{s\,\,=\,\,\frac{M}{0.87f_yZ}}
$
Where Z, which is the lever arm, is given as:
$
Z\,\,=\,\,d\left( \text{0.5}+\sqrt{\text{0.25}-\,\,\frac{K}{0.9}} \right) \,\,\,\,\leqslant \,\,0.95d
$
Click here to study a worked example on the design of singly reinforced beam to BS 8110
As for a doubly reinforced beam, the concrete portion in compression is not sufficient to resist the compressive stress induced in the beam so additional reinforcement is required to complement the concrete section in compression. These complementary reinforcements are called compression reinforcement and their area is denoted by As’.
For a doubly reinforced section, the moment of resistance will be less than the design moment
I.e.: Mu < M
The compression reinforcement is given by:
$
Asc\,\,=\,\,\frac{M\,\,-\,\,0.156fcubd^2}{0.87f_y\left( d\,\,-\,\,d’ \right)}
$
The tensile reinforcement is given by:
$
As\,\,=\,\,\frac{K’fcubd^2}{0.87f_yZ}\,\,+\,\,Asc
$
Where Z for doubly reinforcement is given by:
$
Z\,\,=\,\,d\left( \text{0.5}+\sqrt{\text{0.25}-\,\,\frac{K’}{0.9}} \right) \,\,\,\,\leqslant \,\,0.95d
$
The above deductions imply that for a simply supported beam which typically has its bottom fiber and top fiber subjected to tension and compression respectively, the reinforcement at the bottom below the neutral axis are always the tensile reinforcement while that above the neutral axis are the compression reinforcement.
Does this mean that in all cases, all simply supported beams with reinforcement at the top are designed as doubly reinforced?
The answer is – No!
Click here to study a worked example on the design of doubly reinforced beam to BS 8110
When a beam is singly reinforced, in spite that we do not need reinforcement at the top, we still nonetheless have to provide minimum area of reinforcement at the top as this reinforcement prevent undue cracking due to shrinkage and thermal stress. We also need to provide at least minimum area of reinforcement so that a reinforcement cage can be formed. These reinforcements serve as hanger bars on which shear links are anchored.
So what’s the function of links?
To properly put this into perspective, we will have to understand how to design beams against shear force.
3.0 Shear design
Shear forces are internal force that tend to make the part of a structural member slide against another. It is often significant close to supports and under point load. According to BS 8110:1:1997, beams are designed against shear by checking the shear stress (v) against the shear strength of the concrete (Vc). The shear stress (v) is calculated by dividing the shear force (V) in the section by the effective cross-section area of the section.
ie: v = V/bd
The shear strength (Vc) of the concrete section is affected by a few numbers of variables. The strength of the concrete is a chief factor, likewise, do the interlocking action and dowel action of the coarse aggregate and longitudinal reinforcement bars respectively contribute also considerably. This can be observed in table 3.8 of the code where the percentage longitudinal reinforcement is to be checked against the effective depth of the concrete section to obtain the shear strength of the section.
According to table 3.7 which is established to guide the form and area of required shear reinforcement, when the shear stress (v) is lesser than the shear strength (Vc) plus 0.4N/mm2 (ie: v < Vc + 0.4) minimum links is to be used.
i.e.: if v < Vc + 0.4 – Asvmin
where:
$
Asv_{\min}=\,\,\frac{\text{0.4}x\,\,b_v\,\,x\,\,s_v}{0.9fy_v}\,\,
$
However, if the shear stress (v) is greater than Vc + 0.4 then the shear reinforcement should be designed for using:
$
Asv_{}=\,\,\frac{b_v\,\,x\,\,s_v\left( v-v_c \right)}{0.95fy_v}\,\,
$
It is nonetheless pertinent to note that on no account should the shear stress be greater than the lesser of 5N and 0.8xfcu0.5 should this happen then the section must be resized and redesigned.
When the shear stress is extremely low such that it is lesser than half of the concrete shear strength as it happens in lightly loaded element such as lintel, the shear links should not be provided theoretically. However, for all practical purpose, it is pragmatic to provide minimum links so that a reinforcement cage can be formed and large cracks can be prevented.
Under no circumstances should the spacing of the links exceed 0.75d. And at right-angles to the span, the spacings of the main tension bar should be within 150 mm from a vertical leg. This spacing between the longitudinal bars should never be more than the effective depth of the beam for a shallow beam.
4.0 Deflection Check
Deflection is an important serviceability limit state that must be checked whether its limit is exceeded. Excessive deflection can compromise the aesthetic of the structure, give impression of unsafe structure, cause damage to brittle finishes, cause problem to fixtures etc. So ensuring deflection limit is not exceeded is one of the important criteria of beam design.
The factors affecting deflection of beams are numerous likewise is calculating deflection for heterogenic material like reinforced concrete very tedious. In order to simplify the problem, the code uses the concept of span-depth ratio to estimate deflection. Hence, table 3.9 gives limiting span to depth ratio for rectangular and flange beams. Allowance for factors such as amount of tension and compression reinforcement are also catered for in tables 3.10 and 3.11 respectively. Consequently table 3.10 and 3.11 provide modification factors which modifies the limiting span-depth ratio base on amount of reinforcement bars provided.
The table is based on limiting the deflection of the beam to L/250. This value is expected to decrease to about L/500 after the construction of cladding and other finishes which would consequently stiffen the member.
The span of the member is expected not to be more than 10m, if the span exceeds 10m, then it is expected that the value in table 3.9 should be multiplied by 10/actual span.
In determining whether the beam satisfy deflection criteria, the actual span to effective depth is to be calculated and checked against the limiting value in table 3.9 after it has been duly modified with appropriate factors from table 3.10 and 3.11 respectively. If the actual value is lesser, then the beam has passed the deflection criteria but would have failed if otherwise.
Reference(s)
BS 8110-1-1997: Structural Use of Concrete: Code of practice for design and Construction
