Challenge

To model beams and bracings for industrial steel structures using TSD

CC Mallika15 Feb 2023

Aim -To determine the loading on the beam indicated and analyze the beams and draw bending moment and shear force diagram.

Given Data -

i) The floor system is 120 mm thick normal weight concrete floor

ii) The floor finish and mech loads on the floor is 3 kPa = 3kN/m^2

iii) The live load on the floor is 5 kPa = 5kN/m^2

iv) All beams have cross section 500 mmx750 mm.

So, I =bd^3 /12 = 0.5*0.75 ^3 /12 =0.0176 m^4

v) Modulus of elasticity of concrete is 10 GPa. = 1*10^6 KPa

Procedure -

i) Self weight of slab = 0.12*25 = 3kn/m

ii) Self weight of beam = 0.5*0.75*25 = 9.375Kn/m

iii) Floor finish = 3kn/m

iv) Live load = 5kn/m

Total weight = self wt + live load + floor finish

= 3 + 5 + 3 = 11kN/m

Therefore , Load on beam ,

i) Load on beam = Total weight * Tributary width

So , Let us give the names to all the beams in order as B1,B2,B3,B4 starting from left side

i) Load on B1 = W * tributary width+ self weight of beam

= 11* 0.5(3+3)+9.375

= 42.375 Kn/m

ii) Load on B2 = W * tributary width + self wt of beam

= 11* 0.5(3+3)+9.375

= 42.375Kn/m

iii) Load on B3 = W * tributary width + self wt of beam

= 11* 0.5(3+3)+9.375

= 42.375Kn/m

iv) Load on B4 = W * tributary width + self wt of beam

= 11 * 0.5(3+3)+9.375

= 42.75 kN/m

The figure is as shown below for 1st Case - Considered a Continuous beam with simply supported

Step 1 ) Fixed End Moments

For Span AC -

Mac = -wL^2/12 = -42.375*9*9/12 = -286.03Knm

Mca = wL^2/12 = 42.375*9*9/12 = +286.03Knm

For Span CD -

Mcd = -wL^2/12 = -42.375*9*9/12 = -286.03Knm

Mdc = wL^2/12 = 42.375*81/12 = +286.03Knm

For Span DE -

Mde = -wL^2/12 = -42.375*81/12 = -286.03Knm

Med = wL^2/12 = 42.375*81/12 = +286.03Knm

For Span EB -

Meb = -wL^2/12 = -42.375*81/12 = -286.03Knm

Mbe = wL^2/12 = 42.375*81/12 = +286.03Knm

Step 2) End Moment Equation -

For Span AC -

MA = MFab + 4EI/L * Ta + 2EI/L * Tc

= -286.03 + 78125 Ta + 39062.5 Tc

MC = MFca + 4EI*Tc/L + 2EI*Ta/L

= 286.03 + 78125Tc + 39062.5Ta

For Span CD -

MC = MFcd + 4EI*Tc/L + 2EI*Td/L

= -286.03 + 78125Tc + 39062.5Td

MD = MFdc + 4EI*Td/L + 2EI*Tc/L

= 286.03 + 78125*Td + 39062.5*Tc

For Span DE -

MD = MFde + 4EI*Td/L + 2EI*Te/L

= -286.03 + 78125Td + 39062.5Te

ME = MFed + 4EI*Te/L + 2EI*Td/L

= 286.03 + 78125Te + 39062.5Td

FOR Span EB -

ME = MFeb + 4EI*Te/L + 2EI*Tb/L

= -286.03 + 78125Te + 39062.5Tb

MB = MFbe + 4EI*Tb/L + 2EI*Te/L

= 286.03 + 78125Tb + 39062.5Te

Step 3 ) Using Moment Equilibrium eqn

i) Ma = 0

-286.03 + 78125Ta + 39062.5Tc = 0 ......(1)

Considering the equilibrium at C, Mc = 0

ii) Mc = Mcd + Mca

78125Tc + 78125Tc + 39062.5Ta + 39062.5Ta + 39062.5Td = 0

156250Tc + 78125Ta + 39062.5Td = 0.........(2)

Considering the equilibrium at D, Md = 0

iii) Md = Mdc + Mde

286.03 + 78125*Td + 39062.5*Tc - 286.03 + 78125Td + 39062.5Te = 0

156250Td + 39062.5Tc + 39062.5Te = 0 .............(3)

Considering equilibrium at E, Me = 0

iii) Me = Med + Meb

= 286.03 + 78125Te + 39062.5Td - 286.03 + 78125Te + 39062.5Tb = 0

= 156250Te + 39062.5Td + 39062.5Tb = 0 ........(4)

Considering the equilibrium at B, Mbe = 0

iv) Mb = 0

286.03 + 78125Tb + 39062.5Te = 0 .....(5)

Solving above equation

we get ,

Tc = 1.046 x 10^-3 rad

Tb =4.18*10^-3 rad

Te = 1.046*10^-3 rad

Td = -1.84*10^-6 rad

Ta = 4.0*10^-3 rad

Step 4) Substitute the above unknown values in moment equations and get the moment values.

i) MA = MFac + 4EI*Ta/L + 2EI*Tc/L

= -286.03 + 78125Ta + 39062.5Tc

MA = -0.00

ii) MC = MFca + 4EI*Tc/L + 2EI*Ta/L

= 286.03 + 78125Tc + 39062.5Ta

= 367.7Knm

iii) MD = MFdc + 4EI*Td/L + 2EI*Tc/L

= 286.03 + 78125Td + 39062.5Tc

MD = 245.17Knm

iv) ME = MFed + 4EI*Te/L + 2EI*Td/L

= 286.03 + 78125Te + 39062.5Td

ME = 367.7Knm

v) MB = MFbe + 4EI*Tb/L + 2EI*Te/L

= 286.03 + 78125Tc + 39062.5Ta

MB = 0 KNm

Step 5) To determine the shear forces , consider each span individually,

i) Consider span AC,taking moment about A,

Hc * 9 + 367.59 - 42.375*9*4.5 = 0

Hc = -149.84 KN

Ha + Hc = 42.375*9

Ha + 149.84 = 381.375

Ha = 231.53KN

ii) Consider span CD,taking moment about c,

-367.59 + 245.78 + Hd * 9 - 42.375*9*4.5 = 0

Hd = -204.28Kn

Hc+Hd = 381.375

Hc = 177.09Kn

iii) Consider span DE,taking moment about D,

-225.65 + 367.7 + He*9 - 42.375*9*4.5 = 0

He = -174.9Kn

Hd= 206.45Kn

iv) Consider span EB,taking moment about E,

9Hb - 367.7 - 42.375*9*4.5 = 0

Hb = - 231.535Kn

He = 149.84Kn

SFD

BMD

analysed using check model and analysis is run.

after that we get the different values for area of steel at each beam as shown below.

bars calculation is done along grid A and 3.

calculated number of bars is represented in autocad interface as shown in figure.

There are a number of properties of a beam that an Engineer should be aware of as they dictate beam behaviour when subject to a load and ultimately represent possible areas or mechanisms for failure. The main ones being:

Second moment of area (also referred to as the second moment of inertia): this depends on the cross section profile of the beam and is a measure of the resistance of the shape of the beam to bending.
Bending moment: usually illustrated on a bending moment diagram, and often related the deflection of the beam, can be used to calculate regions subject to maximum bending forces and consequently most likely to yield. It also illustrates which sections of the beam are in compression or tension.
Beam deflection: beam deflection tends to be undesirable and correlates to the bending moment.
Shear diagrams: these are used to illustrate stress concentrations along the beam and provide a means to identify areas of maximum shear forces where the beam is more likely to fail by shear.

>Second moment of area The second moment of area (I) is a property of the shape used to predict the resistance of the beam to bending and deflection. It is calculated from the physical cross sectional area of the beam and relates the profile mass to the neutral axis (this being a region where the beam is subject to neither compression or tension, as labelled in figure 5. It is dependant on the direction of loading; for most beams except both hollow and solid box and circular sections, the second moment of area will be different when loaded from a horizontal or vertical direction.

imply supported beam of length l with no force; b) simply supported beam subject to point load (force) F at centre creating bending. The second moment of area can be calculated from first principles for any cross section profile using the equation:

However, for common beam profiles standard formula are used: I – beam / Universal beam

I-beam cross-section profile with loading parallel to web. The I – beam or Universal beam has the most efficient cross sectional profile as most of its material is located away from the neutral axis providing a high second moment of area, which in turn increases the stiffness, hence resistance to bending and deflection. It can be calculated using the formula:

As shown in figure 6, this is only suitable for loading parallel to the web, as loading perpendicular to the web would be less efficient. Box section