(a) A circular log of weight 1000 N and radius 18 cm is supported by a pair of brackets, one of which is shown in the figure below. Bar PN is hinged at 'P' and held by a string MN 67 cm long. To induce minimum tension at MN, determine the value of 2θ

Question

(a) A circular log of weight 1000 N and radius 18 cm is supported by a pair of brackets, one of which is shown in the figure below. Bar PN is hinged at 'P' and held by a string MN 67 cm long. To induce minimum tension at MN, determine the value of 2θ, as shown for equilibrium. Consider all contact surfaces smooth. Also find the value of minimum tension. (10 marks)

(b) Determine the moment of resistance of the T-beam as shown in the figure. Use limit state method. Take M 20 grade of concrete and Fe 415 HYSD bars. (10 marks)

(c) The fixed-end bar ABCD consists of three prismatic segments as shown in the figure. The end segments AB and CD have cross-sectional area 800 mm² and length 300 mm. The middle segment has cross-sectional area 1200 mm² and length 500 mm. Two axial loads P_B = 28·5 kN and P_C = 20 kN are acting in the bar as shown in the figure. Young modulus E is same for all three segments.
(i) Determine the reaction forces at fixed supports.
(ii) Determine the compressive axial force in the middle segment. (10 marks)

(d) A flow meter when tested in a laboratory gives a pressure drop of 100 kN/m² for a discharge of 0·10 m³/s in a 150 mm diameter pipe. If a geometrically similar model is tested in 600 mm diameter pipe at identical conditions of fluid, determine the corresponding discharge and pressure drop in the model. (10 marks)

(e) A sample of dry cohesionless soil whose angle of internal friction is 35°, is subjected to a triaxial test. If the minor principal stress (σ₃) is 105 kPa, at what values of deviator stress (Δσ) and major principal stress (σ₁) will the test specimen fail? (10 marks)

UPSC Answer Check · Accepted Answer

(a) Since the log is supported by a pair of identical brackets, the vertical load on the bracket shown is W = 1000/2 = 500 N. The bar weight is neglected because it is not given. Let r = 0.18 m and L = 0.67 m. The centre G is at equal normal distance r from the smooth vertical wall and the smooth inclined bar, so G lies on the bisector PG of the angle 2θ at P. Hence PG = r/sin θ. The foot of the perpendicular from G to PN is J, so PJ = PG cos θ = r cot θ. The reaction of the bar on the log is normal to PN. If its magnitude is R_J, vertical equilibrium of the log on this bracket gives R_J sin 2θ = W, so R_J = W/sin 2θ. The equal and opposite force on the bar acts normal to PN at J; its moment arm about P is PJ. The string MN is horizontal, so the tension T at N is horizontal. Since N is on PN and is horizontally L from the wall, its height above P is L cot 2θ, which is the moment arm of T about P. The hinge reaction at P has no moment about P. Taking moments about P for bar PN: T L cot 2θ = R_J PJ = (W/sin 2θ)(r cot θ). Therefore T = W r cot θ/(L cos 2θ). To minimise T, minimise f(θ) = cot θ/cos 2θ. Differentiating ln f gives -tan θ - cot θ + 2 tan 2θ = 0. Using tan θ + cot θ = 2/sin 2θ, this becomes tan 2θ = 1/sin 2θ, so sin² 2θ = cos 2θ. Let c = cos 2θ. Then 1 - c² = c, i.e. c² + c - 1 = 0, giving c = (√5 - 1)/2. The admissible range 0 < 2θ < 90° selects the positive root, and T tends to infinity at both ends of the range, so this is the minimum. Thus 2θ = arccos((√5 - 1)/2) = 51.83°. At this angle, cot² θ = (1 + c)/(1 - c) = 2 + √5. Hence T_min = (500 × 0.18/0.67) × √(2 + √5)/((√5 - 1)/2) = 447.3 N. 2θ = 51.83°, minimum tension = 447 N.

(b) The flange width is b_f = 1000 mm, flange thickness D_f = 100 mm, total depth = 500 mm, and the centroid of the four 25 mm bars is 50 mm from the bottom, so d = 500 - 50 = 450 mm. The steel area is A_s = 4 × (π/4) × 25² = 625π = 1963.5 mm². For M20 concrete, f_ck = 20 N/mm²; for Fe415 steel, f_y = 415 N/mm². In the IS 456 limit-state stress block, the concrete compression is C = 0.36 f_ck b x_u, the steel tension is T = 0.87 f_y A_s, and the resultant concrete force is taken at 0.42 x_u from the top. First assume the neutral axis lies in the flange and use b = b_f. Equating C and T gives x_u = (0.87 f_y A_s)/(0.36 f_ck b_f) = (0.87 × 415 × 1963.5)/(0.36 × 20 × 1000) = 98.46 mm. Since x_u < D_f = 100 mm, the neutral axis is indeed in the flange, so the rectangular flange assumption is valid. Also x_u < 0.48d = 216 mm, so the section is under-reinforced. The web width is not used because the compression zone is wholly in the flange. The internal force is C = T = 0.87 × 415 × 1963.5 = 708.9 kN. The lever arm is d - 0.42 x_u = 450 - 0.42 × 98.46 = 408.65 mm. Therefore M_R = 708.9 × 10³ N × 408.65 mm = 2.897 × 10⁸ N mm = 289.7 kN m. Moment of resistance = 289.7 kN m.

(c) Take rightward forces as positive and axial force positive in tension. Let R_D be the reaction at D on the bar, positive to the right. The applied loads are P_B = +28.5 kN and P_C = -20 kN. Considering the portion to the right of a section, the internal forces are N_CD = R_D, N_BC = R_D + P_C = R_D - 20 kN, and N_AB = R_D + P_C + P_B = R_D + 8.5 kN.

(i) The bar is fixed at both ends, so the total change in length is zero. Since E is the same for all segments, compatibility gives (300/800)(R_D + 8.5) + (500/1200)(R_D - 20) + (300/800)R_D = 0. In exact fractions, (3/8)(R_D + 17/2) + (5/12)(R_D - 20) + (3/8)R_D = 0. Collecting terms gives (7/6)R_D - 247/48 = 0, so R_D = 247/56 = 4.411 kN, acting to the right. Global equilibrium gives R_A + R_D + 28.5 - 20 = 0, so R_A = -R_D - 8.5 = -723/56 = -12.911 kN. The negative sign means the support at A pulls the bar to the left. Substitution of N_AB = 12.91 kN, N_BC = -15.59 kN, and N_CD = 4.41 kN into the compatibility sum gives zero, confirming the result. Reactions: A = 12.91 kN left, D = 4.41 kN right.

(ii) The middle segment force is N_BC = R_D - 20 = 4.411 - 20 = -15.589 kN. Negative tension is compression. Compressive axial force in the middle segment = 15.59 kN.

(d) Let p denote the 150 mm laboratory pipe and m the 600 mm model pipe. The diameter ratio is D_m/D_p = 600/150 = 4. For a geometrically similar flow meter with identical fluid, the corresponding dynamically similar condition requires the same Reynolds number, Re = ρVD/μ. Hence V_m/V_p = D_p/D_m = 1/4. The area ratio is (D_m/D_p)² = 16, so the discharge ratio is Q_m/Q_p = (V_m/V_p)(D_m/D_p)² = (1/4) × 16 = 4. Therefore Q_m = 4 × 0.10 = 0.40 m³/s. The pressure drop of a flow meter can be written Δp = K(ρV²/2), where K depends on Re and geometry. With the same Re and the same fluid, K_m = K_p, so Δp_m/Δp_p = (V_m/V_p)² = (1/4)² = 1/16. Hence Δp_m = 100/16 = 6.25 kN/m². Model discharge = 0.40 m³/s, model pressure drop = 6.25 kN/m².

(e) For dry cohesionless soil, cohesion c = 0 and total stresses equal effective stresses. The Mohr-Coulomb failure criterion in principal stresses is σ₁/σ₃ = (1 + sin φ)/(1 - sin φ) = tan²(45° + φ/2). Here φ = 35° and σ₃ = 105 kPa. Using sin 35° = 0.5736, σ₁/σ₃ = (1 + 0.5736)/(1 - 0.5736) = 3.690. Therefore σ₁ = 105 × 3.690 = 387.5 kPa. The deviator stress in a triaxial test is Δσ = σ₁ - σ₃, so Δσ = 387.5 - 105 = 282.5 kPa. Failure deviator stress = 282.5 kPa, major principal stress = 387.5 kPa.

(a) A circular log of weight 1000 N and radius 18 cm is supported by a pair of brackets, one of which is shown in the figure…

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