Week 4 Challenge

Design of Beams

Question No: 1. Section classification problem for listed beam shapes

Calculate section classification of following beam shapes

• ISMB 500
• ISMC 200
• PG 1300 x 450 x 20 x 12

• Double angle (Back to back) – 2 x ISA 75x75x6
• RHS - 150 x 100 x10
• Builtup box – 1000 x 450 x 20 x 12

Aim:

To be determine the section classification for the given beam shapes.

Given Data:

For Beams & Channel sections

• ISMB 500
• ISMC 200

• PG 1300 x 450 x 20 x 12

For Built up Box section

• Double angle (Back to back) – 2 x ISA 75x75x6
• RHS - 150 x 100 x10
• Builtup box – 1000 x 450 x 20 x 12

Procedure:

1. ISMB 500

= b / t_f

= 180 / 17.2

= 10.46 < 10.50

= d / t_w

= 500 / 10.2

= 49.2 < 84

Its limits are in compact section. So it is classified as compact section.

1. ISMB 200

= b / t_f

= 75 / 11.4

= 6.58 < 9.4

= d / t_w

= 200 / 6.1

= 32.80 < 84

Its limits are in plastic section. So it is classified as plastic section.

• PG 1300 x 450 x 20 x 12

= b / t_f

= 450 / 20

= 22.5 < 29.30

Its limit is in plastic section. So it is classified as plastic section.

1. Double angle (Back to back) – 2 x ISA 75x75x6

= b / t_f

= 75 / 6

= 12.5 < 15.7

Its limit is in semi compact section. So it is classified as semi compact section.

1. RHS – 150 x 100 x 10

= d / t_w

= 150 / 10

= 15 < 42

Its limits are in plastic section. So it is classified as plastic section.

1. Built- up box section – 1000 x 450 x 20 x 12

= d / t_w

= 1000 / 12

= 83.33 < 84

Its limits are in plastic section. So it is classified as plastic section.

Result:

Thus the section classification for the given shapes of beams are classified.

Question No: 2. Effective length problems

Calculate effective length of rafter made up of Built-up I section about major (x-x) and minor axis (y-y) having purlin spaced at 1.2 m.

Aim:

To be determine the effective length of rafter for made up of built up I section.

Given Data:

Data are taken from the given diagram.

Procedure:

Calculation of the effective Length of rafter for made up of built up I section.

Effective Length of the Rafter (E_L) = 0.85 x L

Length of the Rafter (L) = Top chord Length / Sin 15 ̊

(L) = 5.36 / Sin 15 ̊

(L) = 20.71 m

Effective Length of the Rafter (E_L)

(E_L) = 0.85 x L

(E_L) = 0.85 x 20.71

Effective Length of the Rafter (E_L) = 17.60 m

Result:

Thus the Effective Length of the Rafter (E_L) = 17.60 m for the made up of Built-up I section.

Question No: 3. Deflection of a beam problem

• Calculate deflection at the centre of the beam shown below. EI = 20 x 10⁶ Nm²

• Calculate deflection at the centre of the beam shown below. EI = 25 x 10⁶ Nm²

• A simply supported beam is 10 m long with udl of 10 kN/m Calculate the Flexural stiffness (EI) of the beam which limits the deflection to 5 mm at the middle.

Aim:

To be determine the deflection and Flexural Stiffness at the center of the beam for the given problem

Given Data:

Data are taken from the given problem diagram.

Procedure:

1. a) Deflection at the center of the beam (Y)

(Y) = (wl³/ 48 EI) ((3a₁ / L) – 4 x (a₁ /L)³) + (wl³/ 48 EI) ((3a₂ / L) – 4 x (a₂ / L)³) + (5wl³/ 384 EI)

(Y) = (100 x 7³/ 48 x 20 x 10⁶) ((3 x 2 / 7) – 4 x (2 / 7)³) + (100 x 7³/ 48 x 20 x 10⁶) ((3 x 5 / 7) – 4 x (5 / 7)³) + (5 x 6 x 7³/ 384 x 20 x10⁶)

(Y) = 0.0536 m.

1. b) Deflection at the center of the beam (Y)

(Y) = (wl³/ 48 E_I) ((3a₁ / L) – 4 x (a₁ /L)³) + (wl³/ 48 E_I) ((3a₂ / L) – 4 x (a₂ / L)³) + (5wl³/ 384 E_I)

(Y) = (2000 x 5³/ 48 x 20 x 10⁶) ((3 x 1.5 / 5) – 4 x (1.5 / 5)³) + (200 x 5³/ 48 x 20 x 10⁶) ((3 x 2.5 / 5) – 4 x (2.5 / 5)³) + (5 x 1500 x 5³/ 384 x 20 x10⁶)

(Y) = 0.783 m.

1. c) Flexural stiffness (E_I) of the beam

(Y) = 5 wl³ / 384 E_I

From that, (E_I) = 5 wl³ / 384 (Y)

Value of the (Y) = 5 mm

(E_I) = 5 x 10 x 1000 x l0³ / 384 x (5 x 10³)

(E_I) = 26.04 x 10⁶ N/m²

Result:

1. a) Deflection at the center of the beam (Y) = 0.0536 m.
2. b) Deflection at the center of the beam (Y) = 0.783 m.
3. c) Flexural stiffness (E_I) of the beam = 26.04 x 10⁶ N/m²

Question No: 4. Moment capacity determination problem

Select a standard hot rolled section for the beam which is simply supported of length 6 m spaced at 3 m c/c, carrying a floor Dead load of 5 kN/m² (including Self weight) & Live load of 6 kN/m² and a concentrated load of 100 kN at mid span of beam. The beam is laterally restrained and assumes bearing length of 50 mm

Tip steps:

• Calculate maximum bending moment & required plastic modulus
• Select suitable trial section
• Calculate design shear force and design shear strength
• Check design capacity of the section
• Check for deflection, web buckling, bearing

Aim:

To be determine the moment capacity for the given problems by using the given data.

Given Data:

• Supported Length = 6 m
• Spaced Length = 3 m/cc
• Dead Load (including Self weight) = 5 kN/m²
• Live load = 6 kN/m²
• Concentrated load = 100 kN
• Bearing length = 50 mm

Procedure:

Load Calculation:

DL = 5 x 3 = 15 kN/m

LL = 6 x 3 = 18 kN/m

Total UDL = 33 kN/m

Point load at the center of span = 100 kN

Shear Force:

R_A + R_B = 33 x 6 + 100 = 298

R_A = R_B

Taking moment about A

R_B x 6 = 100 x 3 + 33 x 6 x 3

R_B = 149 kN

R_A = 149 kN

SF at A = 149 kN

SF at C (Centre) = 50 kN

SF at B = -149 kN

Hence Maximum Shear force = 149 kN

Bending Moment:

M_A = 149 x 6 – 100 x 3 – 33 x 6 x 3 = 0

M_C = 149 x 3 - 33 x 3 x 1.5 = 298.5 kN m

M_B = 0

Factored Moment = 1.5 x 298.5 = 447.75 kN m

Section Modulus:

Z_P (required) = M x γ_mo / f_y = 447.75 x 10⁶ x 1.1 / 250 = 1970.1 cm³

Choosing Section ISWB 600 @ 133.7 kg/m, Section properties from IS 808 : 1989 Table 2.2.

Shape Factor = Z_p / Z_e = 1.14

So, Z_e (required) = 1970.1 / 1.14 = 1728.16 cm³

Z_P (provided) = 4035.6 cm³ _> Z_P (required) = 1970.1 cm³

Z_e (provided) = 3540 cm³

r₁ = 17

h = 600 mm

b_f = 250 mm

t_f = 21.3 mm

t_w = 11.2 mm

I_xx = 106198 cm⁴

I_yy­ = 4700 cm⁴

r_xx = 250 mm

r_yy = 52.5 mm

d = h – 2(r₁ + t_f) = 600 – 2 x (17 + 21.3) = 523.4 mm

Classification of section:

As per IS 800 : 2007 Table 2

(b_f / 2) / t_f = (250 / 2) / 21.3 = 5.86 < 9.4

d / t_w = 523.4 / 11.2 = 46.73 < 84

Hence the section is Plastic.

Calculation of Shear Capacity of section:

V_d = (A_v x f_y) / (√3 x γ_mo) = (600 x 11.2 x 250) / (√3 x 1.1) = 881797 N

V_d = 881.797 kN > 1.5 x 149 = 223.5 kN

Hence Safe in Shear.

0.6 V_d =0.6 x 881.797 = 529.08 kN > V = 223.5 kN

Hence Low shear.

Calculation of Design Capacity of Moment – By Interpolation method

Design Strength for laterally unsupported beam

kL / r­_y = 6000 / 52.5 = 114.28

h_f / t_f = (600 – 21.3) / 21.3 = 27.16

f_cr,_b IS 800 : 2007 from table 14

f_cr,_b = 240 N / mm²

λ_LT = √(f_y / f_cr)

= √(250 / 240)

λ_LT = 1.02

ɸ_LT = 0.5 [1 + α_LT (λ_LT - 0.2) + λ²_LT]

= 0.5 [1 + 0.21 (1.02 - 0.2) + 1.02²]

ɸ_LT = 1.1063

X_LT = 1 / [ɸ_LT + (ɸ²_LT - λ²_LT)^0.5 ≤ 1.0

= 1 / [1.1063 + (1.1063² – 1.02²)

X_LT = 0.652 < 1

Hence OK.

f_bd = X_LT x f_y / γ_mo

= 0.652 x 250 / 1.1

f_bd = 148.18 N / mm²

M_d = β_b x Z_p x f_bd

= 1.0 x 4035.6 x 1000 x 148.18

M_d = 598 kN m > M (447.75 kN m)

Hence Section is safe in bending stress.

Check for deflection:

δ = [5wl⁴ / 384EI] + [Wl³/48EI]

= [ 5 x 18 x 6000⁴ / 384] + [100 x 6000³ / 48] x 1 / (20 x 10⁴ x 106198 x 10⁴)

δ = 1.43 + 21.18 = 22.61 mm

Permissible maximum deflection as per IS 800 Table 6 for Live load with Elastic cladding

= L / 240 = 6000 / 240 = 25 mm > 22.61 mm

Hence safe in deflection.

Check for Web Buckling:

d / t_w = 523.4 / 11.2 = 46.73 < 67

Hence OK

So no need to check for web buckling.

Check for Web Crippling

n₁ = 2.5 x (t_f + r₁) = 2.5(21.3 + 17) = 95.75 mm

As per IS 800 : 2007 Clause 8.7.4

F_w = [(b₁ + n₂) x t_w x f_y)] / γ_mo

Where b₁ = Length of stiff bearing = 50 mm

n₂ = Length of load dispersion

= [(50 + 95.75) x 11.2 x 250] / 1.1

= 371000 N

= 371 kN > V = 223.5 kN

Hence safe in Bearing

Result:

Thus, it is concluded that the beam ISWB 600 at 133.7 kg/m, is capable to resist bending moment of 548.70 kN m safely.

Question No: 5. Purlin design problem

Design channel purlin, considering sag rod at mid span for the roof truss having following data

• Span of truss : 20 m
• Spacing of Truss : 4 m c/c
• Roof slope : 20 degrees
• Spacing of purlin : 2 m
• Wind suction : 1 kN/m²
• Live Load : 0.75 kN/m² ,Dead Load 1.4 kN/m²

Given Data:

• Span of truss = 20 m
• Spacing of Truss = 4 m c/c
• Roof slope = 20 degrees
• Spacing of purlin = 2 m
• Wind suction = 1 kN/m²
• Live Load = 0.75 kN/m²
• Dead Load = 1.4 kN/m²

Procedure:

Calculation of Load:

Dead Load perpendicular to the rafter = 1.4 Cos 20°

= 1.4 x 0.939 = 1.3146 kN/m²

Dead Load per meter (Load/m² x spacing of purlin) = 1.3146 x 2 = 2.6292 kN/m

Live load perpendicular to the rafter = 0.75 Cos 20°

= 0.75 x 0.939 = 0.7042 kN/m²

Live Load per meter (Load/m² x spacing of purlin) = 0.7042 x 2 = 1.4084 kN/m

Wind Load perpendicular to the rafter = 1.0 Cos 0°

= 1.0 kN/m²

Wind Load per meter (Load/m² x spacing of purlin) = 1.0 x 2 = 2.0 kN/m

Load Combination (1) = 1.5 x (DL + LL)

= 1.5 x (2.6292 + 1.408)

= 6.0558 kN/m

Load Combination (2) = 1.5 x (DL + LL – WL)

= 1.5 x (2.6292 + 1.408 – 2.0)

Worst Load combination = 6.0558 kN/m

Dead load parallel to the rafter = 1.4 Sin 20°

= 1.4 x 0.342

= 0.4788 kN/m²

Dead Load per meter (Load/m² x spacing of purlin)

= 0.4788 x 2

= 0.9576 KN/m

Live load parallel to the rafter = 0.75 Sin 20°

= 0.75 x 0.342 = 0.2565 kN/m²

Live Load per meter (Load/m² x spacing of purlin) = 0.2565 x 2 = 0.513 kN/m

Wind load parallel to rafter = 0.00

1.5 x (DL+LL ) = 1.5 x (0.9576 +0.513) = 2.2059 kN/m

Bending moment calculation:

BM about Z-Z M_z = wl² / 10 = 6.0558 x 4² / 10 = 9.689 kN m

(Purlin is continuous over PR of Truss)

BM about Y-Y M_y = wl² / 10 = 2.2059 x 4² / 10 = 3.529 kN m

Calculation of Shear Force:

F_z = wl/2 = 6.0558 x 4 / 2 = 12.117 kN

F_y = wl/2 = 2.2059 x 4 / 2 = 4.419 kN

Selection of Trial Channel section for purlin:

ISMC 200 selected for purlin

Properties taken from Steel Tables

h = 200 mm

b_f = 75 mm

t_f = 11.4 mm

t_w = 6.1 mm

h₁ = 150.2 mm

I_zz = 1819.3 mm⁴

Z_pz = M_z x γ_mo / f_y + (2.5 D / b_f ) M_y x γ_mo / f_y

= 9.689 x 10⁶ x 1.1 / 250 + (2.5 x 200 / 75) x 3.529 x 10⁶ x 1.1 / 250

= 42631.6 + 6.6666 x 15527.6

= 146147 mm³

X – Coordinate C.G of the channel section

200 x 6.1 + 2 x 11.4 X = 2 x (68.9 – X) x 11.4

X = 7.69 mm

Z_py = (600 x 6.1) (6.1/2 + 7.69) + 2[(68.9 x 11.4) (68.9/2 – 7.69)]

= 39308.4 + 42037.819

= 81346.2 mm³

Section Classification:

d / t_w = 150.2 / 6.1 = 24.62 < 42 ε

ε = 1

b / t_f = 75 / 11.4 = 6.57 < 9.4 ε

Hence the section is Plastic.

Calculation of Shear Capacity of section:

About Z axis

V_d = (A_v x f_y) / (√3 x γ_mo) = (200 x 6.1 x 250) / (√3 x 1.1) = 160.083 kN

0.6 V_d =0.6 x 160.08 = 96.05 kN > F_z = 12.117 kN

Hence Safe.

About Y axis

V_d = (A_v x f_y) / (√3 x γ_mo) = (2 x 75 x 11.4 x 250) / (√3 x 1.1) = 224.38 kN

0.6 V_d =0.6 x 224.38 = 134.628 kN > F_y = 4.419 kN

Hence it is Safe.

Design Moment in Z direction

M_dz = β_b x Z_pz x f_y / γ_mo

= 1.0 x 146147 x 250 / 1.1 = 33215227.3 N mm

M_dz = 33.215 kN m > M_z (9.689 kN m)

Hence Section is safe in moment in Z direction.

Design Moment in Y direction

M_dy = β_b x Z_py x f_y / γ_mo

= 1.0 x 81346.2 x 250 / 1.1 = 18487772 N mm

M_dy = 18.48 kN m > M_y (3.529 kN m)

Hence Section is safe in moment in Y direction.

Overall Strength condition:

M_z / M_dz + M_y / M_dy ≤ 1

9.689 / 33.215 + 3.529 / 18.48 = 0.292 + 0.191 = 0.483 < 1

Hence safe.

Check for deflection:

Actual deflection = 5wl⁴ / 384EI

= (5 x 1.4084 x 4000⁴) / (384 x 20 x 10⁴ x 1819.3 x 10³)

= 12.89 mm.

Permissible maximum deflection as per IS 800:2007 Table 6 for Live load with Elastic cladding = L / 240 = 4000 / 240 = 16.66 mm.

Actual deflection is lesser than Permissible deflection.

Hence it is safe in deflection.

Result: -

• It is concluded that for purlin of truss ISMC 200 section is safe in shear, moment, strength and deflection.