The document analyzes the structural design of an I-beam roof on a shuttle kiln. It calculates the load on the beam, draws the shear force and bending moment diagrams, and determines the maximum bending stress, deflection, and linear expansion of the beam. The results show the beam design is safe with the maximum bending stress less than the allowable stress at 150 degrees C, deflection of 1.5mm is negligible, and a 2.25mm expansion gap is needed on both sides of the beam.

1. CASE STUDY
AIM:
To check the structural design of roof of shuttle kiln using shear force and bending moment diagram.
TO FIND:
1. Maximum bending stress on ‘I’ beam
2. Maximum deflection of ‘I’ beam
3. Linear expansion of the I beam @ 150 Deg.C
SOLUTION:
As per the structural design I beams carry major loads where as other L angles are used for stiffening
purpose. Let us consider single I beam at the middle of the roof for design. This I beam carries the load of
structural members and ceramic fiber modules in the distance of 1500mm.
Let us say, ‘I’ beam acts as a simply supported beam with uniform distributed load. Hence, let us determine
the load acting on the ‘I’ Beam.
Load Acting on the ‘I’ beam:
a) Plate - 1.5m x 3.75m x .005 x 7850 = 220 Kg
b) Double L angle - 7.7 Kg x 5 x 1.5m x 2 = 115 Kg
c) Single L angle - 7.7 Kg x 3.75m = 30 Kg
d) Pyrobloc/ Blanket - for the area of 1.5 x 3.75m = 620 Kg
----------------------------------
Total Mass acting on the beam = 985 Kg
------------------------------------
Weight acting on the beam = 985 x 9.81 N
= 9662.85 N
Weight acting per meter of the Beam = 9662.85/3.75 N/m
Weight acting per meter of the Beam = 2576 N/m
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2. Shear force and bending moment diagram:
From the above diagram, a simply supported beam with load has two reactions RA & RB at both ends.
To find out the reactions RA & RB, following equation has to be resolved.
From the above diagram,
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3. Summation of reactions is zero.
∑R = 0
RA +RB = 2576 x 3.75
RA +RB = 9660 -------------- (1)
From the above diagram,
Summation of bending moment about point A is zero.
∑MA = 0
RB x 3.75 – [2576 x 3.75 x 3.75/2] = 0 ---------------- (2)
RB = 18112.5/ 3.75
RB=4830 N
Substitute this value in Eqn.1
Therefore,
RA = 4830 N
From these result we can draw the shear force diagram. Shear force diagram clearly depicts that max shear
force will be at the both end of I beam and zero at the mid of I beam.
The formula for calculation max. bending moment of simply supported beam having uniform distributed
load is,
M max = wl2
/8
Where,
M max - Max. Bending moment (Nm)
w – Load per unit length (N/m)
l - Length of the I beam (m)
M max = 2576 x 3.75 x3.75/ 8
M max =4528 Nm
As we known, bending moment will be max, at the point where shear force is zero. Hence, the max.
bending will happen only at the middle of the beam.
Calculation of bending stress:
Max. bending stress acting at the middle of I beam can be calculated by means of bending equation.
M/I = σb/y = E/R
Here,
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4. M - Max. bending moment (Nm)
I - Moment of inertia (m4
) (Ixx = 2235.4 x 10-8
m4
)
σb - Bending stress (N/m2
)
y - Distance between neutral axis to extreme fiber (m) [ for ISMB 200 x 100, Y = 100mm)
Hence,
σb = M x y/I (N/m2)
= 4528 x 0.1/ (2235.4 x 10-8)
= 20.25 MN/ m2
σb = 20.25 N/mm2
Calculation of Allowable stress:
Let us consider, factor of safety as 4. Yield strength of structural steel is 250 N/mm2
Allowable stress = Yield strength/ FOS
= 250/4
Allowable stress = 62.5 N/mm2
From the above results, it is very clear bending stress is three times less than allowable stress. Hence our
design is in safe side. i.e., σb < Allowable stress.
Here we need to consider temperature as a factor as it is affecting the performance of the beam. Let us
consider the max. cold face temperature is 150Deg.C.
Hence we need to calculate the yield strength of the beam at 150 Deg.C.
We know,
E (young’s modulus) = yield stress / strain and formula for coefficient of thermal expansion is,
ά = [Δl/ LxΔT]
Here, Δl – change in length (m)
L – Length of the beam (m)
ΔT – change in temperature (Deg.C)
The above eqn. can be written like this,
ά x ΔT = [Δl/ L] = Strain
Incorporate this formula in young’s modulus formula,
Yield Stress = ά x ΔT x E
= 11 x 10-6
x (150 – 40) x 200 x 109
= 242 106
N/m2
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5. Yield stress = 242 N/mm2
Now, let us calculate the allowable stress again for this value.
Allowable stress = Yield strength @ 150 Deg.C/ FOS
= 242/4
Allowable stress = 60.5 N/mm2
From the above result we can say, σb < Allowable stress.
Calculation of deflection of beams:
Formula to compute the Max. deflection of simply supported beam having uniform distributed load is,
Max. deflection = (5/384) x (wl4
/ EI) (m)
Here,
w – Load per unit length (N/m)
l - Length of the I beam (m)
E - Young’s modulus = 206842 x 106
N/m2
I - Moment of inertia (m4) (Ixx = 2235.4 x 10-8
m4
)
Max. deflection = (5/384) x (2576 x 3.754
)/(206842 x106
x 2235 x10-8
)
Max. deflection = 1.4348 x 10-3
m
Max. deflection =1.4348 mm
Consideration of linear thermal expansion:
Δl = ά x LxΔT
= 11x10-6
x 3750 x (150-40)
Δl = 4.5 mm
Hence, as per the above result we need to maintain 2.25mm expansion gap on the both side of the beam.
Result:
1. Maximum bending stress is less than allowable stress @ 150 Deg.C; hence design is in safer side.
2. Beam deflection is apprx. 1.5mm and it is negligible.
3. Linear Expansion of 2.25mm has to be considered on both the side of the I beam.
Reference:
• Structural drawing 10224801 R1
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6. • Hand book for structural engineers – Bureau of Indian standards SP: 6(1)-1964 revised 1998
• Strength of Materials - Dr.R.K. Bansal
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Company Registration Number: 18-9622, Registered Office : Dare House Complex, No.2, N.S.C. Bose Road, Chennai – 600 001. INDIA

7. • Hand book for structural engineers – Bureau of Indian standards SP: 6(1)-1964 revised 1998
• Strength of Materials - Dr.R.K. Bansal
www.morganthermalceramics.com
Company Registration Number: 18-9622, Registered Office : Dare House Complex, No.2, N.S.C. Bose Road, Chennai – 600 001. INDIA