Week 4 Challenge

Aim:

To understand the concepts in analysis and design of flexural members like beam, girders, etc.

Introduction:

Flexural members like Beams are structural members that support transverse loads and are therefore subjected primarily to flexure, or bending. If a substantial amount of axial load is also present, the member is referred to as a beam–column.

Design of Beams

1. Section classification problem for listed beam shapes

Calculate section classification of following beam shapes

1) ISMB 500

Section classification:

Taking f_y = 250MPa

For ISMB500

b = 180mm

d = h – 2(t_f +R) = 500 – 2(17.2 + 17) = 431.6mm

ɛ = (250/f_y)^1/2 = (250/250)^1/2 = 1

b/t_f = 180/2x17.2 = 5.23 < 9.4ɛ

d/t_w = 431.6/10.2 = 42.3 < 84ɛ

Hence the section is classified as plastic section.

2.ISMC 200

Section classification:

Taking f_y = 250MPa

For ISMC200

b = 75mm

d = h – 2(t_f +R) = 200 – (11.4 + 11) = 177.6mm

ɛ = (250/f_y)^1/2 = (250/250)^1/2 = 1

b/t_f = 75/11.4 = 6.58 < 9.4ɛ

d/t_w = 177.6/6.2 = 28.65 < 42ɛ

Hence the section is classified as plastic section.

3)PG 1300 x 450 x 20 x 12

Taking f_y = 250MPa

For PG 1300 x 450 x 20 x 12

b = 450/2 = 225mm

d = h – 2(t_f ) = 1300 – 2(20) = 1260mm

ɛ = (250/f_y)^1/2 = (250/350)^1/2 = 0.845

b/t_f = 225/20 = 11.25 < 13.6ɛ=13.6x0.845=11.492

d/t_w = 1260/12 = 105 < 126ɛ=126x0.845=106.47

Hence the section is classified as Semi Compact section.

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

Taking f_y = 250MPa

For ISA75x75x6

b = 75mm

d = h – t_f = 75 – 6 = 69mm

ɛ = (250/f_y)^1/2 = (250/250)^1/2 = 1

d/t_f = 69/6 = 11.5 < 9.4ɛ

Hence the section is classified as plastic section.

5)RHS - 150 x 100 x10

Taking f_y = 250MPa

For RHS 150 x 100 x 10

b = 100-(10x2) = 80mm

d = 150-(10x2) = 130mm

ɛ = (250/f_y)^1/2 = (250/250)^1/2 = 1

b/t_f = 80/10 = 8 < 9.4ɛ

d/t_w = 130/10 = 13 < 84ɛ

Hence the section is classified as plastic section.

6)Builtup box – 1000 x 450 x 20 x 12

Taking f_y = 250MPa

For BOX 1000 x 450 x 20 x 12

b = 450-24 =426mm

d = h – 2(t_f ) = 1000 – 2(20) = 960mm

ɛ = (250/f_y)^1/2 = (250/250)^1/2 = 1

b/t_f = 426/20= 21.3 < 15.7ɛ =>> Slender

d/t_w = 960/12 = 80 < 105ɛ =>> Compact

Hence the section is classified as Slender section.

2. Effective length problems

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

Length of Rafter = (20²+5.36²)^1/2 = 20.71m

Ly = Minor axis effective length = 1.2m

Since, Purlin spaced @ 1.2m

Lz = Major axis effective length = 0.8L = 0.8(2x20.71) = 33.13m

3. Deflection of a beam problem

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

Deflection due to UDL = (5/384)wL⁴/EI = (5/384)*6x7⁴/20x10⁶ = 9.38 x10^-6 mm

Deflection due to Point Load = (Fa/24EI)x(3L² – 4a²) = (100x2/24x20x10⁶)x(3x7² – 4x2²) = 5.46 x10^-5 mm

Where L = Total Length

F = Concentrated load

a = Distance from start of beam & point of application of load

Total Deflection = 9.38x10^-6 mm + 5.46x10^-5 mm = 6.398x10^-5 mm

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

Deflection due to UDL = (5/384)wL³/EI = (5/384)*1.5x10³x5³/25x10⁶ = 9.77 x 10^-6 mm

Deflection due to Point Load of 2kN = (wL³/48EI)x(3a/L – 4(a/L)³)

= (2x10³x5³ /48x25x10⁶ )x(3x1.5/5 – 4(1.5/5)³) = 0.000165 mm

Deflection due to Point Load of 5kN = (wL³/48EI)x(3a/L – 4(a/L)³)

= 5x10³x5³ /48x25x10⁶ (3x2.5/5 – 4(2.5/5)³) = 0.00052 mm

Total Deflection = 9.77x10^-6 mm + 0.000165 mm + 0.00052= 6.95x10^-4 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.

Deflection due to UDL = (5/384)wL⁴/EI

(5/384)*10x10³x10⁴ /EI = 5 x10^-3 m

EI = 26.0417 x10⁷ Nm²

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

STEP 1:

Dead load = 5 x 3 = 15kN/m

Live load = 6 x 3 = 18kN/m

Assumed bearing length = 50 mm

Calculation of factored loads:

Dead load = 1.5 x 15 = 22.5 kN/m

Live load = 1.5 x 18 = 27 kN/m

Concentrated load on beam = 1.5 x 100 = 150 kN

Total UDL on beam = 22.5+27 = 49.5 kN/m

STEP 2:

Maximum bending moment = WL²/8 + WL/4

= (49.5 x 6² / 8) + (150 x 6 / 4) = 447.75 kNm

STEP 3:

Section modulus required (Zp)

Zp required = (M x Ymo) / fy = (447.75 x 10⁶ x 1.1 / 250) = 197 x 10⁴ mm³

STEP 4:

Choose a trial section of ISMB 500

Max. Deflection allowed = (Span/360) = 6000/360 = 16.67mm

To satisfy this deflection, Iz = ?

Deflection = ((5/384)*18x6000³/2x10⁵ x I) + (100 x 6000³ )/(48x2x10⁵ x I)

I = 150157 mm⁴

Section is OK

Section classification:

ɛ = SQRT(250/fy) = SQRT(250/250) = 1

b/tf = (190/2) / 19.3 = 4.922 < 9.4ɛ

d/tw = [550 - 2(19.3+18)] / 11.2 = 42.45 < 84ɛ

Hence the section is plastic section

STEP 5:

Adequacy of the section including self-weight of the beam

Factored self-weight of the beam = 1.5 x 1.037 = 1.5555 kN/m

Maximum bending moment = 299.25 + 1.56 x 6² /8 = 306.27 kNm

Zp required = (M x Ymo) / fy = (306.27 x 10⁶ x 1.1 / 250)

= 1347.6 x 10³ mm³ < 2711.98 x 10³ mm³

Hence the section is adequate.

STEP 6:

Shear force V = (49.5x6/2) + (150/2) = 223.5 kN

STEP 7:

Design shear strength of the section

Design shear strength Vd = (fy/Ymo x √3) x h x tw = (250/1.1x√3)x500x10.2 = 669.2 kN

Also 0.6Vd = 401.52kN > 223.5 kN, Hence Safe.

The design shear force V<Vd

Step 8 :

Check for design capacity of the section

d/tw = 42.45 which is less than 67ɛ

Md = BbZp x fy / Ymo < (1.2 x Ze x fy) / 1.10

Bb = 1.0 since the section is plastic

Md = (1.0 x 2711.98 x 10³ x 250)/1.1 = 616.36 kNm

< (1.2 x 2359.8 x 10³ x 250) / 1.1 = 643.58

Hence the design capacity of the member is more than the maximum bending moment Md.

Check for deflection :

d = (5/384)wL³/EI + (WL³/48EI)

= (5/384)*5x10³x6000⁴/2x10⁵x45200x10⁵ + (100 x10³ x 6000³/48x2x10⁵x45200x10⁵)

= 0.0156 mm + 0.4977 mm = 0.5133 mm

Allowable deflection = L/300 = 6000/300 = 20 mm

Actual deflection is less than the allowable deflection.

Step 9:

Check for web buckling at support

Assume stiff bearing length b1=75mm

Depth of web d=465.6mm

Ieff web = (b1 x tw³)/12 = (75 x 10.2³)/12 = 6632.55 mm⁴

Aeff web = b1 x tw = 75 x 10.2 = 765 mm²

r = √(Ieff web / Aeff web =√(6632.55/765) = 2.94mm

Effective length of web deff = 0.7 times the depth of the web (d)

𝜆 = (deff/r) = (0.7x465.6)/2.94 = 110.9

Therefore fcd = 103.4 N/mm² (From table 9c of the code)

n1 =550/2 = 275mm

b1 + n1 = 75+275 = 350 mm

A_b = 350 x 11.2 = 3920 mm²

Buckling resistance = fcd x A_b

= 103.4 x 3920 / 1000 = 405.3 kN > 266kN

Hence the web is safe

Step 10:

Check for web bearing

Fw = (b1 + n2) x tw x fy / Ymo

b1 = 75mm

n2 = 2.5 x (root radius + flange thickness)

= 2.5 x (17+17.2) = 85.5mm

tw = 10.2mm & fy = 250 N/mm2

Fw = (75+85.5) x 10.2 x 250 / 1.10 = 372.06kN > 266Kn

Hence the web is Safe.

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

Load Calculation:

Dead Load = 0.6 x 2 = 1.2 kN/m

Live load = 0.75 x 2 = 1.5 kN/m

Wind load = 2 x 2 = 4 kN/m (suction)

W_dz = 1.2 x cos 20^o = 1.128 kN/m

W_iz = 1.5 x cos 20^o = 1.41 kN/m

W_wz = -4 kN/m

W_dy = 1.2 x sin 20^o = 0.41 kN/m

W_iy = 1.5 x sin 20^o = 0.51 kN/m

Note that W_wy is zero as wind pressure is perpendicular to the surface on which it acts, i.e., normal to the rafter.

Factored load combination

Z-direction

1.2WL+1.2DL+1.2LL = -(1.2x4) + (1.2x1.128) + (1.2x1.41) = -1.75 kN/m

1.5DL + 1.5LL = (1.5x1.128) + (1.5x1.41) = 3.807 kN/m

Y-direction

1.5DL + 1.5LL = (1.5x0.41) + (1.5x0.51) = 1.38 kN/m

Bending moment and shear force calculation:

Mz = 3.8 x 4²/8 = 7.6 kNm

My = 1.38 x 4²/8 = 2.769 kNm

Fz = 3.8 x 4/2 = 7.6 kNm

Fy = 1.38 x 4/2 = 2.769 kNm

Since the channel section purlin is provided with a sag rod at mid-span , the bending moment in the Y-direction will be reduced considerably.

My = 1.38 x 4² /32 = 0.69 kNm

Mz = 7.6 kNm

Required section modulus = (Mz x Ymo / fy) + 2.5(d/b)(My Ymo/fy)

Assuming ISMC 150 with d = 150mm and b= 75mm

Required Z = (7.6 x 10⁶ x 1.1 /250) + 2.5 x (150/75) x (0.69 x 10⁶ x 1.1/250)

= 48.62 x 10³ mm³

Provide ISMC 150

Section classification:

ɛ = SQRT(250/fy) = SQRT(250/250) = 1

b/tf = (150/2) / 9.0 = 8.33 < 9.4ɛ

d/tw = [150 - 2(9.0+9.5)] / 5.4 = 20.93 < 42ɛ

Hence the section is plastic section.

Shear capacity is not being checked since the shear force will be adequate.

Design capacity of the section

M_dz = (Zpz x fy/Ymo) = (119.82 x 10³ x 250/1.1 x 10⁶) = 27.23 kNm

<= (1.2 x Ze fy /Ymo) = [(1.2 x 105 x 10³ x 250/(1.1 x 10⁶) = 28.63

Zpy = 2t_f b_f²/4 + (h-2t_f )t_w²/4 = 2 x 9 x 75²/4 + (150 – 2 x 9) x 5.7²/4 = 26384.6 mm³

M_dy = (Zpy x fy/Ymo) = (26384.6 x 250/1.1 x 10⁶) = 6.0 kNm

<= (1.5 x Ze fy /Ymo) = [(1.5 x 19.5 x 10³ x 250/(1.1 x 10⁶) = 6.6 kNm

Hence the section is safe.

Overall member strength

For overall member strength, the following interaction equation must be satisfied.

(Mz / Mdz) + (My / Mdy) <= 1.0

(7.6 / 27.23) + (0.69 / 6.0) = 0.39 < 1.0

Hence the member strength is satisfactory.

Check for deflection :

Allowable deflection = L/180 = 4000/180 =22.22mm

d = (5/384)wL⁴/EI

= (5/384)*1.41x10³x4000⁴/2x10⁵x788x10⁴

= 2.98mm < 22.22 mm

Hence the section is safe.