(a) Three bodies of weights 200 N, 800 N and 400 N are connected with inextensible strings, passing over the frictionless pulleys…

(a) Virtual work: force F for impending motion (10 marks)

Give the three blocks (weights 200 N, 800 N and 400 N) a small virtual displacement x in the direction of impending motion. The cord is inextensible and the pulleys are frictionless, so every block moves through the same distance x: the 200 N and 400 N blocks move to the left horizontally, and the 800 N block moves up the 45° incline. With μ = 0.30 the limiting friction forces are:

• 200 N block: N = 200 N, so F₁ = 0.30 × 200 = 60 N.
• 800 N block: N = 800 cos 45° = 565.7 N, so F₂ = 0.30 × 565.7 = 169.7 N, and the component of its weight along the incline is 800 sin 45° = 565.7 N.
• 400 N block: N = 400 N, so F₃ = 0.30 × 400 = 120 N.

Friction opposes the motion and the weight component along the incline resists the upward movement. The virtual work of all forces is zero at impending motion:

F·x − 60x − 169.7x − 565.7x − 120x = 0

F = 60 + 169.7 + 565.7 + 120 = 915.4 N

The unknown cord tension does no net work because it acts as an internal force pair, which is the advantage of the virtual-work method.

(b)(i) Design bending strength and UDL of ISMB 400 (10 marks)

For a laterally supported plastic section, IS 800:2007 gives M_d = β_b Z_p f_y/γ_m0, with β_b = 1 for plastic and compact sections. For ISMB 400, Z_p = 1176 × 10³ mm³ and Z_e ≈ 1023 × 10³ mm³, with f_y = 250 MPa (E250) and γ_m0 = 1.10.

M_d = 1 × 1176 × 10³ × 250/1.10 = 267.3 × 10⁶ N·mm = 267.3 kN·m

The upper limit 1.2 Z_e f_y/γ_m0 = 1.2 × 1023 × 10³ × 250/1.10 ≈ 279 kN·m is larger, so it does not govern. For simple support over L = 5 m, the factored UDL is

w_u = 8M_d/L² = 8 × 267.3/5² = 85.5 kN/m

The design load is the factored load, so the service (unfactored) intensity, with the partial safety factor 1.5 for gravity loads, is 85.5/1.5 = 57.0 kN/m, including the self-weight of about 0.604 kN/m. The superimposed service UDL is therefore about 56.4 kN/m. The shear check is satisfied: V_d = A_v f_y/(√3 γ_m0) ≈ 467 kN against V_u = 85.5 × 5/2 = 213.8 kN, which is below 0.6 V_d, so there is no moment-shear interaction.

(b)(ii) Stiffness matrix of cantilever element AB (10 marks)

Coordinate 1 is the downward deflection at B and coordinate 2 is the clockwise rotation at B, with A fixed and l = 2 m. K is the force system needed to hold unit displacements.

Flexibility route. Applying unit forces at B gives the flexibility coefficients:

f₁₁ = l³/3EI, f₁₂ = f₂₁ = l²/2EI, f₂₂ = l/EI.

Inverting F, with |F| = f₁₁f₂₂ − f₁₂² = l⁴/12E²I², gives K = F⁻¹:

k₁₁ = f₂₂/|F| = 12EI/l³, k₁₂ = k₂₁ = −f₁₂/|F| = −6EI/l², k₂₂ = f₁₁/|F| = 4EI/l.

Unit-displacement route. Hold B at Δ₁ = 1 with Δ₂ = 0: the cantilever then needs a downward force k₁₁ = 12EI/l³ and a moment of magnitude 6EI/l² that acts against the positive (clockwise) rotation, so k₂₁ = −6EI/l². Hold B at Δ₂ = 1 with Δ₁ = 0: the beam needs a moment k₂₂ = 4EI/l and a force of magnitude 6EI/l² acting against the positive (downward) direction, so k₁₂ = −6EI/l². Both routes agree.

For l = 2 m, so l² = 4 and l³ = 8:

K = EI [[1.5, −1.5], [−1.5, 2.0]]

The matrix is symmetric (Maxwell-Betti reciprocity) and its determinant, EI² × (3 − 2.25) = 0.75 EI², is positive.

(c) Moment capacity of the gantry girder section (20 marks)

The section is an ISMB 600 with an ISMC 300 laid flat, web on top, over the top flange. Take the ISMB 600 properties as A = 15600 mm², I_zz = 91800 × 10⁴ mm⁴, I_yy = 2650 × 10⁴ mm⁴, b_f = 210 mm, t_f = 20.8 mm and t_w = 12 mm. The ISMC 300 has A = 4630 mm², I_zz = 6420 × 10⁴ mm⁴, I_yy = 313 × 10⁴ mm⁴, flange width 90 mm, t_f = 13.6 mm, t_w = 7.8 mm and C_yy = 23.5 mm. Laid flat, its I_zz acts about the vertical axis.

Section properties. The ISMB centroid is 300 mm above the base and the channel centroid is 607.8 − 23.5 = 584.3 mm above the base. Then

A = 15600 + 4630 = 20230 mm²

ȳ = (15600 × 300 + 4630 × 584.3)/20230 = 365.1 mm above the base, so the top fibre is 242.7 mm above the centroid.

I_zz = [91800 × 10⁴ + 15600 × 65.1²] + [313 × 10⁴ + 4630 × 219.2²] = 1.210 × 10⁹ mm⁴

Z_top = 1.210 × 10⁹/242.7 = 4.98 × 10⁶ mm³ and Z_bot = 1.210 × 10⁹/365.1 = 3.31 × 10⁶ mm³.

I_yy = 2650 × 10⁴ + 6420 × 10⁴ = 9.07 × 10⁷ mm⁴, so Z_ey = 9.07 × 10⁷/150 = 0.605 × 10⁶ mm³ for the whole section, while the top flange alone (ISMB flange plus channel) has I_yy = 16.05 × 10⁶ + 64.2 × 10⁶, so Z_ey,top = 0.535 × 10⁶ mm³.

The plastic neutral axis divides the area equally, 10115 mm² on each side. It lies 490.7 mm above the base, 117 mm below the channel back, so the plastic modulus is Z_p = 4.32 × 10⁶ mm³.

Classification (ε = 1 for f_y = 250). ISMB flange: b/t_f = 5.05 ≤ 9.4. ISMB web: d/t_w = 43.2 ≤ 84. Channel legs: (90 − 7.8)/13.6 = 6.0 ≤ 9.4. The channel web is supported on the flange, and its unsupported strip is about 4 t_w wide, so it is not slender. The section is plastic and β_b = 1.

Bending about z-z, with no lateral restraint. Torsion constant I_t ≈ 1.77 × 10⁶ mm⁴ and warping constant I_w ≈ 4.445 × 10¹² mm⁶ (an approximation for the built-up shape), G = 76923 MPa, span L = 7.5 m, and c₁ = 1.132 for a UDL on a simply supported span:

M_cr = c₁ (π²EI_y/L²) √(I_w/I_y + GI_tL²/π²EI_y)

The square root is √(49009 + 42780) = 303 mm, so M_cr = 1.132 × 3.183 × 10⁶ × 303 ≈ 1092 kN·m. Then

λ_LT = √(Z_p f_y/M_cr) = √(4.32 × 10⁶ × 250/1092 × 10⁶) = 0.995

For a rolled section (α_LT = 0.21), φ_LT = 0.5[1 + 0.21(0.995 − 0.2) + 0.995²] = 1.079, and χ_LT = 1/(φ_LT + √(φ_LT² − λ_LT²)) = 0.669, so f_bd = 0.669 × 250/1.10 = 152.1 MPa and

M_dz = Z_p f_bd = 4.32 × 10⁶ × 152.1 = 657.1 kN·m

Bending about y-y. The surge (lateral) moment is resisted by the top flange, so M_dy = Z_ey,top f_y/γ_m0 = 0.535 × 10⁶ × 250/1.10 = 121.6 kN·m.

Interaction. With M_z = 685.53 and M_y = 19.84 kN·m:

M_z/M_dz + M_y/M_dy = 685.53/657.1 + 19.84/121.6 = 1.043 + 0.163 = 1.21 > 1

The section is therefore not adequate for the given moments on these assumptions (c₁ = 1.132, α_LT = 0.21, approximate I_w). A deeper section, a heavier channel or lateral restraint of the compression flange along the span is required.