Design of Beams
In general beams are designed basically for carrying Bending Moment and Shear Forces. In some cases beams are designed also for Torsion in conjunction with bending moment and shear forces. Beams are made either of same material or different materials, such as Timber and steel beams are of same material. These beams are called homogeneous beams. Whereas, reinforced concrete beams are made of reinforcing steel and concrete. In this article I will explain briefly how homogeneous beams are designed. I will provide answer to questions like, What are the steps to design a beam?, What are the types of beams, etc. Explanation will be provided in simple language so that the people with limited knowledge can also get benefit from this article. Engineering students and those who want to learn structural engineering will get benefit from this article.
Types of beam: From its arrangement point of view, beams can be categorised basically as Simply supported, Continuous, Cantilever and Fixed beams. Other types can also be created by using combination of these types.
From material point of view the beams can be categorised mostly as Steel, Timber and Reinforced Concrete beams. Design method for reinforced concrete beams is different as compared to the method for homogenous beams. I will describe this method in the upcoming articles.
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| Basic arrangement of beams |
Steps for designing a beam: Basic steps for designing a beam is a) Analysis of beam to determine internal forces, such as Bending Moment, Shear Force, Torsion, etc. b) After determining the internal forces we determine required minimum cross section of a beam that is adequate to carry the developed internal stresses safely.
Unlike to rods and columns, beams have both tensile and compressive stresses in their cross section. The line that separates the two types of stresses is called a Neutral Axis (N.A.).
When a beam is loaded it deflects as shown below. In this case the top surface of the beam undergoes compression and bottom surface undergoes tension causing compressive and tensile stresses at top and bottom surfaces respectively. These stresses that develop due to bending moment are called flexural stresses.
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| Deflected shape of a beam |
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| Flexural stresses in a simply supported beam |
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| Example Floor System |
Solution:
The dimensions are in mm. When we divide these dimension by 1000 we will get all these dimension in meter.
Weight of floor panels, Gp = 1200/1000*1.0*5/1000*4.0=0.024 kN/m (per meter run on beam)
Let's assume size of the beam to be 50 mm wide by 150 mm deep,then
Weight of beam, Gb = 50/1000*150/1000*1.0*4= 0.03 kN/m (per meter run on beam)
Live load on beam, LL= 2.4*1200/1000*1.0= 2.88 kN/m
We will design the beam according to Load Factor method. Let's use load factors in the load combination as shown below:
1.5*(LL)+1.2*(DL)
Then the uniformly distributed load on the beam per m will be
w*= 1.5*(2.88)+1.2*(0.024+0.03) =4.38 kN/m
The beam is assumed to be simply supported on a wall or on a girder. For the simply supported beam,
Maximum Bending moment, M* = (4.38*(4500/1000)^2 )/8= 11.10 kNm
Maximum Shear Force, V*= 4.38*(4500/1000)/2= 9.86 kN
Moment of Inertia of the beam section, I = (50*150^3) /12= 14062500 mm4
Since stress, σ=(M/I)*(D/2), M=σ*I/(D/2)
Therefore, Bending capacity of the beam, øM = 0.8* σb*I/(D/2) =0.8*80*14062500*(150/2)
= 12000000 Nmm = 12000000/1000000=12.00 kNm. Here, 0.8 is bending capacity reduction factor
Similarly, Shear capacity of the beam, øV =0.6*τ *B*D =0.6*24*50*150= 108000 N =108/1000
= 108.0 kN. Here, 0.6 is shear capacity reduction factor
Load/Capacity ratio for bending = 11.10/12.0=0.93 < 1.0 Good
Load/Capacity ratio for shearing = 9.86/108.0= 0.09 <1.0 Good
Strength wise the 50x150 beam is adequate.
Design for serviceability: From the Serviceability point of view, we need to check the actual deflection of the beam. In order to calculate actual deflection, we will use load factor=1.0 for all type of loads.
w= 1.0*(2.88)+1.0*(0.024+0.03) =2.93 kN/m= 2.93*1000N/m=2.93*1000/1000=2.93 N/mm
Deflection, δ = 5/384* (w.L^4)/E.I)= 5/384*2.93*4500^4/(19200*14062500)=57.9 mm >L/250=
=4500/250=18 mm. No good. We need to select a deeper beam.
So let's select 50x250 mm beam then, after calculating the parameters similarly, we will get
M*=11.16 kNm V*=9.92 kN
I = (50*250^3) /12= 65104167 mm4 øM = 33.33 kNm øV = 180 kN
Load / Capacity ratio will be lower which is on conservative side, okay.
Deflection, δ = 5/384*2.93*4500^4/(19200*65104167)=12.72 mm < L/250 = 18 mm Good.
Since the two end beams receive lesser load, for the end beams we use the same beam size as the middle beam.
Now the 50x250 mm beam is good from both strength and serviceability point of view. A smaller section, such as 50x220 can also be used if available. So use the beam with following dimension.
Some other coefficients may be used to further reduce capacity of the beam depending on type of material it is made of, for example, in case of timber beams. Since this article only introduces the basic concept of designing a beam, I have not discussed about these coefficients here which is applicable for design of timber beams.
Lateral Torsional Buckling: The bending capacity calculated above does not take into account the lateral supports of the beams that are used in between two supports. Even if the bending capacity is adequate the long span slender beams can fail in buckling before it reaches its bending capacity. The bending capacity calculated without taking into account the lateral supports can be termed as Section Capacity of a beam. The lateral supports of a beam is called Lateral Restraints. The bending capacity of a beam calculated taking into account the lateral restraints or the section capacity of a beam that is not greater than Critical Moment is called Member Capacity of a beam.
Explanation about critical moment is available below at the end of this article. Buckling of a beam can be separated in two effects, i.e., Lateral Buckling and Torsional Buckling.
Lateral Buckling: The situation when a beam buckles laterally or displaces laterally is called Lateral Buckling.
Torsional Buckling: The situation when a beam buckles by twisting itself is called Torsional Buckling.
In general, beams fail by the combined buckling effect which is called Lateral Torsional Buckling.
If critical moment is inadequate, the overall moment capacity of a beam can be increased by providing lateral restraints that restrict the beam form rotating about its longitudinal axis and moving away from its longitudinal axis.
Following figures show how lateral restraints can be provided in the timber beam system.
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| Methods of restraining timber beams |
Other methods, such as cross bracings can be used in the case of beams made of other materials.
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| Cross bracings for steel beams/ bridge |
Critical Moment: In the similar way, a compression member capacity is limited by Critical Load, also the beams have Critical Moment which limits its bending capacity. The maximum amount of bending moment beyond which a beam starts buckling is called Critical Moment. Since a long slender beams can fail in buckling prior to it reaching its bending capacity, before we finalize the overall bending capacity of a beam, we need to check whether the beam will fail earlier by buckling or not. For the rough estimate, following relationship may be used to determine the elastic critical moment.
Mcr = π/L*(E.Iy.G.J)^0.5
Where,
L: Distance between points which have lateral restraints, E: Elasticity modulus, Iy: Minor axis moment of inertia, J: Polar moment of inertia and G: Shear modulus
Now imagine a thin, deep and long span beam which can have an adequate bending capacity but the beam fails by buckling. Let's consider the above mentioned example, let's use the following parameters:
B=10 mm D=390 mm σb = 250 MPa, Shear strength, τ = 150 MPa G= 80000 MPa, E=200000 MPa
Then, Minor moment of inertia, Iy= 390*10^3/12=32500 mm4
Say, Polar moment of inertia, J= 0.32*390*10^3=124800 mm4
For simplicity, let's assume the same BM and SF,
M* = 11.10 kNm V*= 9.86 kN
Capacities calculated in the same way,
øM = 50.7 kNm øV = 351 kN, Load/Capacity ratios,
M*/øM =11.10/50.7=0.219<1 Good, V*/øV = 9.86/351=0.028<1 Good
Deflection, δ= 17.01 mm<18 mm okay
All beam capacities are good without checking the critical moment.
Now Critical Moment, Mcr=3.14/4500*(32500*200000*80000*124800)^0.5=5621162 Nmm
Mcr=5.62 kNm <11.10 kNm No Good.
The beam with 10x390 section will buckle before it reaches its calculated moment capacity. Here, the member capacity (5.62 kNm) < Section capacity (11.10 kNm). In order to make this section safe against torsional failure a sufficient number of lateral restraints shall be provided such a way that the Critical moment (Mcr) is equal or less than Bending capacity (øM).
Because of thin section the beam fails by buckling before it reaches its bending capacity. Worth noting, lesser the unsupported length greater is the beam's overall bending capacity. Laterally unsupported slender beams like, "I" beams are more prone to failure by buckling.
Similarly, in the case of compression capacity, different national standards, also called national codes, provide methods for determination of actual bending capacity of a beam. These methods are modified version of the above mentioned basic relationship which take into account the lateral restraints of a beam.
With this, I have explained briefly the basic concept of designing a beam. Related national standards are recommended to be referenced for precise calculation of the bending capacity of a beam that take into account the lateral restraints.
In the next article I will explain briefly, how a member with combined actions (Compression and Bending, Tension and Shear, etc.) are designed.
Previous article: Design of compression members.
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