Week 4 Challenge

Question 1: -

Section classification problem for listed beam shapes

Calculate section classification of following beam shapes

Answer: -

Assume all Sections are F_y 250.

ε = (250/f_y)^1/2 = (250 / 250)^1/2 = 1

ISMB 500

From Table,

d = 500 mm

b = 180 mm

t_f = 17.2 mm

t_w = 10.2 mm

As per IS 800 : 2007 Table 2

b / t_f = 180 / 17.2 = 10.47 < 10.5 ε

d/t_w = 500 / 10.2 = 49.02 < 84 ε

Hence it is classified as a Compact section (Class 2).

ISMC 200

From Table,

h = 200 mm

b = 75 mm

t_f = 11.4 mm

t_w = 6.1 mm

As per IS 800 : 2007 Table 2

b/t_f = 75 / 11.4 = 6.58 < 9.4 ε

d/t_w = 200 / 6.1 = 32.79 < 84 ε

Hence it is classified as a Plastic section (Class 1).

PG 1300 x 450 x 20 x 12

As per IS 800 : 2007 Table 2

b/t_f = 450 / 20 = 22.5 < 29.3 ε

Hence it is classified as a Plastic section (Class 1).

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

From Table,

b = 75 mm

t_f = 6 mm

As per IS 800 : 2007 Table 2

b/t_f = 75 / 6 = 12.5 < 15.7 ε

Hence it is classified as a Semi Compact section (Class 3).

RHS - 150 x 100 x10

d/t_w = 150 / 10 = 15 < 42 ε²

Hence it is classified as a Plastic section (Class 1).

Built-up box – 1000 x 450 x 20 x 12

d/t_w = 1000 / 12 = 83.33 < 84 ε

Hence it is classified as a Plastic section (Class 1).

Question 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.

Answer: -

Rafters are in compression which is a Continuously supported.

Length of Rafter = 5.36/Sin 15° = 20.71 m

As per IS 800 : 2007 Table 15,

Effective Length = 0.85 L

Therefore, Effective length of Rafter = 0.85 x 20.71 = 17.60 m

Question 3: -

Deflection of a beam problem

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

Given data: -

a = 2 m

b = 3 m

L = 7 m

W = 100 N

w = 6 N/m

EI = 20 x 10⁶ Nm²

Answer: -

Deflection at Centre = WL³ /24EI x [ 3a/L – 4(a/L)³] + 5wL³ / 384EI

                              = 100 x 7³ / (24 x 20 x 10⁶) x [(3 x 2 / 7) – 4 x (2 / 7)³]

                                  + 5 x 6 x 7³ / (384 x 20 x 10⁶)

      ∆_max(at centre) = 0.056 mm

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

Answer: -

Given data: -

a = 1.5 m

b = 1 m

c = 2.5 m

L = 5 m

W₁ = 2 kN

W₂ = 5 kN

w = 1.5 kN/m

EI = 25 x 10⁶ Nm²

Answer: -

Deflection at Centre = W₁L³ /48EI x [ 3a/L – 4(a/L)³] + W₂L³ /48EI x [ 3c/L – 4(c/L)³] + 5wL³ / 384EI

                              = 2000 x 5³ / (48 x 25 x 10⁶) x [(3 x 1.5 / 5) – 4 x (1.5 / 5)³]

                                 + 5000 x 5³ / (48 x 25 x 10⁶) x [(3 x 2.5 / 5) – 4 x (2.5 / 5)³]

                                 + 5 x 1500 x 5³ / (384 x 25 x 10⁶)

      ∆_max(at centre) = 0.783 mm 

• 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.

Answer: -

Given data: -

L = 10 m

w = 10 kN/m

∆_max(at centre) = 5 mm

Answer: -

∆_max(at centre) = 5wL³ / 384EI

EI = 5wL³ / (384 x ∆_max)

    = 5 x 10000 x 10³ / (384 x 5 x 10^-3)

EI = 26.04 x 10⁶ N / m²

Question 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

Given data: -

Length = 6 m

Dead Load = 5 kN/m² 

Live Load = 6 kN/m² 

A Point load at midpoint = 100 kN

Bearing Length = 50 mm

Answer: -

Step 1: -

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

Step 2: -

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

Step 3: -

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

Step 4: -

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

Step 5: -

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.

Step 6: -

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.

Step 7: -

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) = 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]

      = 1.1063

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

      = 1 / [1.1063 + (1.1063² – 1.02²)

      = 0.652 < 1

Hence OK.

f_bd = X_LT x f_y / γ_mo

     = 0.652 x 250 / 1.1 = 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.

Step 8: -

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.

Step 9: -

Check for Web Buckling:

d / t_w = 523.4 / 11.2 = 46.73 < 67ε

Hence OK

So no need to check for web buckiling.

Step 10: -

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 @ 133.7 kg/m, is capable to resist bending moment of 548.70 kN m safely.

Question 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

Answer: -

Step 1: -

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

Step 2: -

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

Step 3: -

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

Step 4: -

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

Step 5: -

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.

Step 6: -

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 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.

Step 7: -

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.

Step 8: -

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 safe in deflection.

Result: -

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