SOLUTIONS A-03 APPLIED MECHANICS    (June 2003)

Q.1      a.         A         The resultant of any two forces, would be in the plane of these two                                            forces and must be equal, opposite and collinear with the third one.

            b.         D         For a perfect plane truss, the relation between the number                                                         of members n and the number of joints k is n = 2k -3.

            c.         A         The total reaction must be vertically upward to balance the weight                                              of the body acting vertically downward.

            d.         D         The magnitude of the total acceleration a = (a_c² + a_t²)^1/2, where                                                 centripetal acceleration a_c = ω²r, tangential acceleration a_t = .                                              

            e.         B          For the beam span l, the support reactions are R₁ = - R₂ = M/l. The                                           B.M. at a distance x from the support is R₁x – M<x – l/2>.                                                       Maximum B.M. is at the centre of the beam x = l/2, i.e. M/2.

            f.          D         The stiffness of a close-coiled spring k = P/δ is proportional to d⁴.                                             If the diameter d is doubled the stiffness would be 2⁴ = 16 times.

            g.         C         The vacuum pressure is the pressure below the atmospheric                                           pressure.

            h.         B          The runner vanes of a reaction turbine are made adjustable for                                       optimizing the efficiency at part loads.

Q.2.

The F.B.D. of the ladder AB with the man at point D, a distance d up along the ladder is shown in Fig.2. The normal reaction of the floor N_A and the friction force f act on the end A of the ladder. The normal reaction of the wall N_B is at the end B of the ladder. The 800 N weight of the man acts at D. The coefficient of friction μ = tan15.

The equilibrium equations for the ladder give

Σ F_x = 0 → N_B – f = 0                                     (1)

Σ F_y = 0 → N_A – 800 = 0                                 (2)

Σ M_A = 0 → 800dsinα - N_B× 6cosα = 0            (3)

f ≤ μN_A = N_A tan15                                          (4)

Solving equations (1) to (4), d ≤ 6 tan15/tanα.

For α = 30, maximum d = 6 tan15/ tan30 = 2.78 m.     

For d = 6 m, tanα ≤ tan15, i.e.  α ≤ 15⁰.

Q.3.

The F.B.Ds. of the whole frame and members CD and ABC are shown in Fig.3.

 As the end A of the frame is fixed, the reactions at A are the horizontal force H_A, the vertical force R_A and a couple C_A. From the equilibrium equations of the frame

H_A = 0 , R_A = 1000 N and C_A = 1000×0.8 = 800 Nm.

The member CD is a two force member and hence the forces T at the ends C and D must be collinear with CD.

Considering the equilibrium of ABC and taking moment about B to eliminate the unknown reactions H_B, R_B at B from the equation,

ΣM_B = 0 → C_A - T×0.9sin45 – 1000×0.1 = 0 → T = 1100 N.

Q.4a.

A circular area A of radius R in the xy plane is shown in Fig.4a. Consider an infinitesimal element of area dA = rdθdr. The second moment of the area I of the circular area A about the z axis, normal to the area and passing through the centre O, would be

Q.4b.

Let the superscripts 1 and 2 refer to the uniform thin disc of radius R and the hole of radius R/2, respectively. Then, the coordinates of the centroid C of the disc with the hole would be

From symmetry about the axis x, y_c = 0.

Its second moment of area I^C_zz about an axis through C and parallel to the z axis would be

Q.5.

The F.B.Ds. of the pulleys 1, 2 and masses A, B, C are shown in Fig.5. As the pulleys are light and frictionless, the tension in a string on both sides of a pulley would be the same. Also from the F.B.D. of the pulley 2,

T₁ = 2T₂                       (1)

Let a_A, a_B, a_C be the accelerations of the masses A, B, C, respectively and a₂ the acceleration of the pulley 2. Then

a₂ = - a_A                       (2)

a_B – a₂ = - (a_C – a₂)     (3)

The equations of motion for the masses A, B, C are

60 – T₁ = 6a_A               (4)

30 – T₂ = 3a_B               (5)

20 – T₂ = 2a_C               (6)

Solving equations (1) t0 (6),

a_A = 1.11 m/s², a_B = - 1.11 m/s² and  a_C = - 1.11 m/s².

Q.6.

Let d be the diameter of the rod. From strength consideration,

σ =P/A = 1000g/(πd²/4) ≤ σ_allowable = 150×10⁶ → d ≥ 0.0092 m = 9.2 mm.

From stiffness consideration,

δ = PL/AE = 1000g×5/[(πd²/4)×210×10⁹] ≤ δ_allowable = 3×10^-3 → d ≥ 0.01 m = 10 mm.

Hence d = 10mm.

Spring constant of the rod k = P/δ = 1000g/(3×10^-3) = 10⁷/3 N/m.

The frequency f = (1/2π)√(k/m) = (1/2π)√[(10⁷/3)/1000] = 9.19 Hz.

Q.7.

The loading on the cantilever beam and the support reactions at the built in end are as shown in Fig. 7.  

Considering the equilibrium of the cantilever, the reactions at the built in end A are

R_A = wb and C_A = wb(L-b/2).

Using singularity functions, the shear force V and the bending moment M at any section x are    

V = - wb + w<x - (L - b)>,    

M= -wb(L–b/2)+wbx –w<x - (L - b)>²/2.  

The S.F. and B.M. diagrams are also shown in Fig.7.

Their maximum values are at A, x = 0,      

V_max = -wb, M_max = -wb(L-b/2).

Let v be the deflection of the elastic line at x, EId²v/dx² = M. Then,

EId²v/dx² = - wb(L–b/2) + wbx – w<x - (L - b)>²/2

Integrating,

EIdv/dx = - wb(L–b/2)x + wbx²/2 – w<x - (L - b)>³/6 + C₁

EIv = - wb(L–b/2)x²/2 + wbx³/6 – w<x - (L - b)>⁴/24 + C₁x +C₂.

Using the boundary conditions v = 0 and dv/dx = 0 at x = 0 → C₁= 0 and C₂ = 0.

The maximum deflection occurs at the free end B i.e. x = L,

v_max = [- wb(L–b/2)L²/2 + wbL³/6 – w<L - (L - b)>⁴/24]/EI  = - wb(L³/3 – bL²/4 + b³/24).

Q.8a.

The spring is under an axial pull P. Let R be the radius of the coil and d be the wire diameter. The F.B.D. of one part of the spring cut by a section with normal along the spring wire is shown in Fig.8a. Any coil section is subjected to a direct shear force P and a moment T = PR. For a close coiled spring the moment T is a twisting moment. Using the torsion formula τ = Tr/I_p, the maximum shear stress due to torsion would be

τ_max = PR(d/2)/(πd⁴/32) = 16PR/(πd³).

Let n be the number of turns, G the shear modulus of the wire material and δ the deflection. Then the strain energy U would be

U = Pδ/2 = T²L/(2GI_p)  = (PR)²(2πRn)/[2G(πd⁴/32)]  → δ = 64PR³n/Gd⁴.

Q.8b.

Let d be the diameter of the solid shaft, and d_o, d_i the outer and internal diameters, respectively of the hollow shaft.  From the torsion formula, the torque transmitted T for the same maximum shear stress τ_max in the shafts would be T = τ_maxI_p/r_max.

For the solid shaft T_solid = τ_max (πd⁴/32)/(d/2) = τ_max πd³/16.

For the hollow shaft T_hollow = τ_max [π(d_o⁴ – d_i⁴) /32]/(d_o/2) = τ_maxπ(d_o⁴ – d_i⁴)/(16d_o).  

As the shafts are of the same material length and weight, d_o² – d_i² = d².

Hence, the ratio T_hollow/T_solid = (d_o⁴ – d_i⁴)/d³d_o = (d_o² + d_i²)/dd_o = d_o/d + d_i²/dd_o > 1.

Q.9a.

A cube floating in water, with its sides vertical, is shown in Fig.9a. Let M be the metacentre, G the centre of gravity and B the centre of buoyancy. If h is the height of immersion in water, the weight of the water displaced equals the weight of the cube, i.e.

1000hb² = 1000γb³ → h = bγ.

BG = b/2 – h/2 = b/2 – bγ/2 = b(1 - γ)/2

BM = I/V = b(b³/12)/ b²h = b/12γ

MG = BM – BG = b/12γ - b(1 - γ)/2 = 0

→ γ² – γ +1/6 = 0 → γ = (1±√3)/2 = 0.789, 0.211.

Q.9b.

The velocity components u = 2x – x²y + y³/3 and v = xy² -2y +x³/3.

The continuity condition for an incompressible 2D flow is ∂u/∂x + ∂v/∂y = 0.

∂u/∂x + ∂v/∂y = (2 – 2xy) + (2xy - 2) = 0. → It is a possible 2D flow.

The irrotational flow condition for a 2D flow is ∂v/∂x - ∂u/∂y = 0.

∂v/∂x - ∂u/∂y = (y² + x²) - (– x² + y²) = 2 x² ≠ 0. → The flow is not irrotational.

Q.10a.

Consider a 2D inviscid steady flow in the xz plane. The gravity acts in the - z direction. A differential control volume with the forces acting on it is shown in Fig.10a.

The mass in the control volume m = ρdxdydz.

The sum of the forces in the x direction,

∑F_x = - [p+(∂p/∂x)dx]dydz + pdydz.

The total acceleration in the x direction,

Du/Dt = u∂u/∂x+ w∂u/∂z.

The equation of motion mDu/Dt = ∑F_x yields the Euler’s equation in the x direction,

→ u∂u/∂x+w∂u/∂z = - (1/ρ) ∂p/∂x.

Similarly, mDw/Dt = ∑F_z yields the Euler’s equation in the z direction,

→ u∂w/∂x+w∂w/∂z = - (1/ρ) ∂p/∂z - g.

Q.10b.

Let subscripts 1 and 2 refer to the inlet and outlet, respectively of the draft tube. The continuity equation yields the velocity at the outlet V₂ as

V₂ = V₁A₁/A₂ = 5(π×3²/4)/(π×5²/4) = 1.8 m/s.

The Bernoulli’s equation between the inlet and outlet sections is

p₁/γ + V₁²/2g + z₁ = p₂/γ + V₂²/2g + z₂ + losses.

Hence the pressure head p₁/γ at the inlet would be

(p₁/γ - p₂/γ) = (z₂ - z₁) + (V₂²– V₁² )/2g + losses = -5 + (1.8² – 5²)/(2×9.81) + 0.1= - 6.01 m.

Q.11.

A bucket of a Pelton wheel with its inlet and outlet velocity diagrams is shown in Fig. 11. The bucket speed is v and the turning angle is θ. Let subscripts 1 and 2 refer to the inlet and outlet, respectively. Let u₁, u₂ be the absolute jet velocities and w₁, w₂ the relative velocities. As there is no friction,

w₂ = w₁ = u₁ – v.

The peripheral jet velocity at the outlet is,

v + w₂cosθ = v + (u₁ – v)cosθ.

Force R on the jet would be

R = ρQ[v + ( u₁ – v)cosθ – u₁]

   = - ρQ(u₁ – v)(1- cosθ).

The force F on the bucket, F = - R = ρQ(u₁ – v)(1- cosθ).

The power developed P = F×v = ρQv(u₁ – v)(1- cosθ).

The input energy E = ρQu₁²/2. The efficiency η = P/E = 2((v/u₁)(1- v/u₁)(1- cosθ).

For maximum efficiency,

dη/dv = 0. → v = u₁/2, i.e. the bucket speed must be half the absolute jet speed at inlet.

SOLUTIONS A-03 APPLIED MECHANICS    (December 2003)

Q.1.     a.         A         Any horizontal section of the block is subjected to a shear force.

            b.         B          The specific speed N_s = N√P/H^5/4 with speed N in rpm, power P                                              in kW and head H in m of a Francis turbine is from 60 to 300.

            c.         C         T =τ_maxI_p/r_max→T_hollow/T_solid=I_phollow/I_psolid =[d_o⁴ –(d_o/2)⁴]/d_o⁴ =15/16.

            d.         A         The slope and deflection under the load are Wa²/2EI and Wa³/3EI.                                           Free end deflection = Wa³/3EI + (l- a)( Wa²/2EI) = (3l-a)Wa²/6EI.

            e.         B          The first moment of area of a semicircle about its diameter D is

.

            f.          B          A rigid body is in translation if all its points have the same velocity                                  V(t) (which may change with time t). Hence, it can move along a                                          straight or curved path.

            g.         D         A point of the rigid body or its hypothetical extension, having zero                                  velocity always exists for plane motion.

            h.         C         Due to the phenomenon of surface tension, a quantity of liquid                                       tries to minimize its free surface area.

Q.2a.

As the resultant of the three forces acting on the lever passes through O (refer Fig. 1 of Q.2a), the sum of their moments about O must be zero.

∑M_o = P×250cos20 – 120×200- 80×400 = 0 → P = 238.4 N.

The expression for the moment ∑M_o does not depend on the angle θ and consequently, the force P does not depend on the angle θ.

Q.2b.

Let R_x, R_y be the x, y components, respectively of the resultant R of the three forces acting on the eye bolt (refer Fig. 2 of Q.2b.).

R_x = ∑F_x = 6 + 8cos45 -15cos30 = - 1.33 kN,

R_y = ∑F_y = 8sin45 + 15sin30 = 13.16 kN.

Hence R = (R_x² + R_y²)^1/2 = [(-1.33)² + (13.16)²]^1/2 = 13.23 kN.

The angle θ which R makes with + x axis is

θ = cos^-1(R_x/R) = cos^-1(-1.33/13.23) = 95.8⁰.

Q.3

Let H_A, R_A be the support reactions at A and R_D the support reaction at D as shown in Fig.3(i).

Considering the equilibrium of the whole truss,

∑F_x = 0 → H_A + 400 = 0 → H_A = - 400 N.

∑M_A = 0 →12R_D -9600 -1200 = 0 → R_D = 900 N.

∑F_y = 0 → R_A + R_D -1200 = 0 → R_A= 300 N.

The sides AG = GC = ED = √(4²+3²) = 5 m.

Imagine the truss to be cut by a section 1-1 and consider the equilibrium of the portion to the left of the section 1-1 as shown in Fig.3(ii). The forces shown in the members are tensile.

∑F_y= 0 → F_AG(3/5) + R_A = 0.

→ F_AG = - 500 N = 500 N (C).

∑F_x = 0 → F_AG(4/5) + F_AB + H_A = 0.

→ F_AB = 800 N (T).

Imagine the truss to be cut by a section 2-2 as shown in Fig.3(iii). Consider the equilibrium of the portion to the left of the section 2-2.

∑F_x = 0 → F_AG(4/5) + F_BC + H_A = 0.

→ F_BC = 800 N (T).

∑F_y= 0 → F_AG(3/5) + F_BG + R_A = 0.

→ F_BG = 0.

Imagine the truss to be cut by a section 3-3 and consider the equilibrium of the portion to the right of the section 3-3 as shown in Fig.3(iv).

∑F_y= 0 → - F_CE + R_D = 0

→ F_CE = 900 N (T).

∑M_E = 0 → - F_DC ×3 + R_D×4 = 0.

→ F_DC = 1200 N (T).

∑F_x = 0 → - F_EG - F_DC + 400 = 0.

→ F_EG = - 800 N = 800 N (C).

Finally, imagine it to be cut by a section 4-4 and consider the equilibrium of the portion above the section 4-4 as shown in Fig.3(v).

∑M_G = 0 → - [F_DE (3/5) + F_CE]×4 = 0.

 → F_DE = -1500 N = 1500 N (C).

∑M_E = 0 → [F_AG (3/5) +F_BG +F_CG (3/5)]×4 = 0.

→ F_CG  =  500 N (T).

Q.4.

The F.B.Ds. of the bodies A, B and the weight W for impending motion of the bodies A and B down the planes are shown in Fig.4. This would correspond to the least magnitude of W = W_min.

From the equilibrium of body A,

N_A = 1000 cos20 = 939.7 N.

T_A = 1000 sin20 – 0.2 N_A = 154.1 N.

From the equilibrium of body B,

N_B = 800 cos30 = 692.8 N.

T_B = 800 sin30 – 0.25 N_B = 226.8 N.

From the equilibrium of weight W in the vertical direction

W_min = T_Asin45 + T_Bsin60 = 305.4 N.

For horizontal equilibrium, additional horizontal force is required.

The impending motion of the bodies A and B up the planes correspond to the maximum magnitude of W = W_max. In this case, the direction of frictional forces on both the blocks would be reversed and must act down the planes. Considering the equilibrium of the bodies A and B, the normal reactions remain the same. Then,

T_A' = 1000 sin20 + 0.2 N_A = 530.0 N.

T_B' = 800 sin30 + 0.25 N_B = 573.2 N.

W_max = T_A'sin45 + T_B'sin60 = 871.2 N.

Q.5.

Let subscripts 1 refer to the rectangular area ABGD, 2 to the triangular area DGC and 3 to the semicircular area EFB as shown in Fig.5. Then the given area A would be

A = A₁ + A₂ – A₃.

The moment of inertia of area A₁, (I_BC)₁ about BC and (I_AB)₁ about AB, would be

(I_BC)₁ = 8×16³/12 + (8×16)(16/2)² = 10922.7 cm⁴.

(I_AB)₁ = 16×8³/12 + (8×16)(8/2)²  = 2730.7 cm⁴.

The moment of inertia of area A₂, (I_BC)₂ about BC and (I_AB)₂ about AB, would be

(I_BC)₂ = 4×16³/36+ (4×16/2)(16/3)²  = 1365.3 cm⁴.

(I_AB)₂ = 16×4³/36 + (4×16/2)(8+4/3)²  = 2816 cm⁴.

The moment of inertia of area A₃, (I_BC)₃ about BC and (I_AB)₃ about AB, would be

(I_BC)₃ = (π×4⁴/4)/2 + (π×4²/2)×4² = 502.7 cm⁴. (I_AB)₃ = (π×4⁴/4)/2 = 100.5 cm⁴.

The moments of inertia for the area A, I_BC about BC and I_AB about AB, would be

I_BC = (I_BC)₁ + (I_BC)₂ - (I_BC)₃ = 10922.7  + 1365.3 - 502.7 = 11785.3 cm⁴.

I_AB = (I_AB)₁ + (I_AB)₂ - (I_AB)₃  = 2730.7 + 2816 - 100.5 = 5446.2 cm⁴.

Q.6.

Let the common velocity after impact be V. The conservation of momentum yields,

(800+500)V = 800×12 + 500×9 → V = 10.9 m/s.

The loss of kinetic energy (K.E.) due to impact would be

Initial K.E. – Final K. E. = 800×12²/2 + 500×9²/2 - (800+500)(10.9)²/2 = 623.5 J.

Q.7.

The F.B.D. of the beam is shown in Fig 7. Considering the equilibrium of the beam,

∑M_A = 7R_D – (4×2)×7 -5×3 –5 = 0. → R_D = 76/7 = 10.9 kN.

∑F_y= 0 → R_A + R_D -5 - 4×2.

→ R_A = 15/7 = 2.1 kN.

The S.F. V and B.M. M at various sections is:

0 ≤ x ≤ 3m

V = - 2.1, M = 2.1x.

3m ≤ x ≤ 5m

V = - 2.1 + 5 = 2.9,

M = 2.1x + 5 - 5(x - 3).

5 m ≤ x ≤ 7 m

V = - 2(x - 9) + 10.9,

M = - 2(9 - x)²/2 + 10.9(7 - x).

7 m ≤ x ≤ 9 m

V = - 2(x - 9),

M = - 2(9 - x)²/2.

9 m ≤ x ≤ 11 m

V = 0, M = 0.

The S.F. and B.M. diagrams are also shown in Fig.7.

The maximum S.F.

V_max = 6.9 kN at D, i.e. x = 7 m.

The maximum B.M. M_max = 11.3 kNm at B, i.e. x = 3 m.

From, M = - 2(9 - x)²/2 + 10.9(7 - x) = 0, →  x = 6.36 m, is the point of contraflexure.

Q.8.

Let d_o be the outside diameter and d_i = 0.6 d_o the inside diameter of the shaft.

The polar moment of inertia I_p= π (d₀⁴- d_i⁴)/32 = πd_o⁴(1- 0.6⁴)/32 = 0.0272πd_o⁴.

Using the torsion formula, from stiffness consideration,

θ = TL/GI_p = 25000×3/[85×10⁹×0.0272πd_o⁴] ≤ 2.5π/180.

→ d_o⁴ ≥ 25000×3×180/[85×10⁹×0.0272π×2.5π] → d_o ≥ = 0.124 m = 12.4 cm.

Using the torsion formula, from strength consideration,

τ_max = Tr_max/I_p = 25000(d_o/2)/[ 0.0272πd_o⁴] ≤ 90×10⁶

→ d_o³ ≥ 25000/[2×90×10⁶×0.0272π] → d_o ≥ 0.118 m = 11.8 cm.

Hence, d_o = 12.5 cm should be selected. Then, d_i = 0.6 d_o = 7.5 cm.  

Q.9a.

Consider a vertical surface BD in the xz plane, submerged in a liquid with free surface at atmospheric pressure p_o as shown in Fig.9a.

The relation between the pressure p at a depth z in a static incompressible fluid of density ρ is

p = p_o + ρgz.

The force dF on an elemental area dA would be dF = pdA.

The resultant force F_R = ∫_A pdA = = p_oA + ρg∫_AzdA.

If C is the centroid of the area A, ∫_AzdA = z_CA.

The pressure  at the centroid C, p_C = p_o + ρgz_C . Then,

F_R = p_oA + ρgz_CA = (p_o + ρgz_C) A = p_CA.

The resultant force F_R acts at the centre of pressure P(x_P, z_P) such that the moment of the resultant F_R about the x and z axes must be the same as the moment of the distributed pressure loading on the surface.

z_PF_R = z_P p_CA =∫_AzdF = ∫_AzpdA = ∫_Az(p_o + ρgz)dA = p_oz_CA + ρg∫_Az²dA

As ∫_Az²dA = I_xx, the moment of inertia of the area A about the x axis,

z_P p_CA = p_oz_CA + ρgI_xx. → z_P = (p_oz_CA + ρgI_xx)/p_CA.

x_PF_R = x_P p_CA =∫_AxdF = ∫_AxpdA = ∫_Ax(p_o + ρgz)dA = p_ox_CA + ρg∫_A x zdA

As ∫_AxzdA = I_xz, the product of inertia about the x,z axes,

x_P p_CA = p_ox_CA + ρgI_xz. → x_P = (p_ox_CA + ρgI_xz)/p_CA.

Q.9b.

Consider an inclined surface BD in the xz plane at an angle θ to the horizontal, submerged in a liquid with free surface at atmospheric pressure p_o as in Fig.9b. The relation between the pressure p at a depth z in a static incompressible fluid of density ρ is

p = p_o + ρgh = ρgzsinθ.

The force dF on an element dA would be dF = pdA.

The resultant F_R = ∫_A pdA =  p_oA + ρgsinθ∫_AzdA.

If C is the centroid of the area A, ∫_AzdA = z_CA.

The pressure  at the centroid C, p_C = p_o + ρgz_Csinθ. Then, F_R = p_oA + ρgz_CsinθA = (p_o + ρgh_C) A = p_CA.

The resultant force F_R acts at the centre of pressure P(x_P, z_P) such that the moment of the resultant F_R about the x and z axes must be the same as the moment of the distributed pressure loading on the surface.

z_PF_R = z_P p_CA =∫_AzdF = ∫_AzpdA = ∫_Az(p_o + ρgzsinθ)dA = p_oz_CA + ρgsinθ∫_Az²dA

As ∫_Az²dA = I_xx, the moment of inertia of the area A about the x axis,

z_P p_CA = p_oz_CA + ρgsinθI_xx. → z_P = (p_oz_CA + ρgzsinθI_xx)/p_CA.

x_PF_R = x_P p_CA =∫_AxdF = ∫_AxpdA = ∫_Ax(p_o + ρgzsinθ)dA = p_ox_CA + ρgsinθ∫_A x zdA

As ∫_AxzdA = I_xz, the product of inertia about the x,z axes,

x_P p_CA = p_ox_CA + ρgsinθI_xz. → x_P = (p_ox_CA + ρgsinθI_xz)/p_CA.

Q.10.

The stream function ψ = 3x² – y³.

The velocity component u in the x direction, u = ∂ψ/∂y = - 3y².

The velocity component v in the y direction, v = - ∂ψ/∂x = - 6x.

The velocity components at the point P(3,1) are u_P = -3 and v_P = - 18.

Hence at the point (3,1), the velocity vector v = -3i - 18j.

Magnitude v = √(3² +18²) = 18.25, inclination with x axis θ = tan^-1(18/3) -180 = -99.5⁰.

The flow is derived from a stream function and hence is a possible 2D flow. The stream function ψ = 3x² – y3 does not satisfy the Laplace equation,

∂²ψ/∂x² + ∂²ψ/∂y² = 6 – 6y ≠ 0. Therefore, the flow is not irrotational and the potential function would not exist for this flow.

Q.11.

The continuity equation between the inlet section 1 and the outlet section 2 is,

Q = A₂V₂ = A₁V₁ = (π×6²/4 )×15 = 424.115 m³/s.

→ V₂ = Q/A₂ = 424.115/ (π×4.8²/4) = 23.4375 m/s.

The Bernoulli’s equation between the inlet sections 1 and the outlet section 2 would be

P₂/ρg + V₂²/2g + z₂ = P₁/ρg + V₁²/2g + z₁.

→ P₂ = P₁ + ρ(V₁²  - V₂²)/2 + (z₁ – z₂)

= 282×10³+ 0.9×10³(15² – 23.4375²)/2 = 136.1×10³ Pa = 136.1 kPa.

The gage pressure at the inlet and outlet are,

P_g1 = 282 – 101.325 = 180.675 kPa and P_g2 = 136.1 – 101.325 = 34.775 kPa.

The momentum equation in the x direction yields:

- F_x + P_g1A₁ – P_g2A₂cos60 = ρQ(V₂cos60 - V₁).

→ F_x = P_g1A₁ – P_g2A₂cos60 - ρQ(V₂cos60 - V₁)

= 180.675×10³(π×6²/4) - 34.775×10³(π×4.8²/4)cos60

– 0.9×10³×424.115(23.4375cos60 -15) = 6046.3×10³ N = 6046.3 kN.

The momentum equation in the y direction yields:

F_y – P_g2A₂sin60 = ρQV₂sin60.

→ F_y = P_g2A₂cos60 + ρQV₂sin60

 = 34.775×10³(π×4.8²/4)sin60 + 0.9×10³×424.115× 23.4375sin60

= 8292.6×10³ N = 8292.6 kN.

SOLUTIONS A-03 APPLIED MECHANICS    (June 2004)

Q.1.     a.         C         The resultant force magnitude R = (P² + P² + 2PP cos120)^1/2 = P.                                            Hence, the acceleration magnitude = R/m = P/m.

            b.         C         The simplest resultant of a system of parallel forces is either a                                         force or a couple.

            c.         B          The block is in equilibrium, i.e. ∑F_h=0. The       frictional force must                                           be equal and opposite to the applied force P/2.

            d.         D         The second moment of area of a square area about any centroidal                                             axis in the plane of the area is the same, i.e. b⁴/12.

            e.         A         The total distance traveled d = 20 + 20 = 40 km. the time to travel t                              = 20/20 + 20/60 = 4/3 h. average speed = d/t = 40/(4/3) = 30 km/h.

            f.          B          The nominal stress = load/original area of cross-section is                                                          maximum at the ultimate load.

            g.         D         The B.M. is constant. The curvature d²v/dx² = M/EI = constant.                                                Hence, the deflection v would have a quadratic variation.

            h.         A         A manometer connected to a pipeline is used to measure the static                                             pressure.

Q.2.

The F.B.Ds. of the sphere B and the cylindrical tube C are as shown in Fig.2. The forces on the sphere B are its weight W, the radial reaction P from the tube C and the reaction Q from the sphere A along the common normal. From the geometry of the spheres inside the tube, 2R = 2r + 2rcosθ → cosθ = (R - r)/R.

Considering the equilibrium of sphere B,

P = Qcosθ and W = Qsinθ → P = W/tanθ.

The tube C would be subjected to its weight W_C, the radial reactions P and P' from the spheres B and A, respectively and the vertical reactions N₁, N₂ from the horizontal table. From the force equilibrium equation in the horizontal direction,

P' = P = W/tanθ.

At impending clockwise tipping of the tube, the vertical reaction N₁ vanishes, i.e. N₁ = 0.

Considering the moment equilibrium about the point of application of N₂,

W_C×R - P×2rsinθ < 0 → r/R < (1- W_C/2W).

Q.3.

The F.B.D. of the truss is shown in Fig.3(i). As the support A is hinged, the reaction at A has both a horizontal component H_A and a vertical component, R_A. At the roller support C, the reaction R_C is vertical. The equilibrium equations of the truss,

∑F_x = H_A + 80 = 0 → H_A = - 80 kN.

∑M_A = R_C×8 - 80×3 - 40×4 = 0 → R_C = 50 kN.

∑F_y = R_A + R_C - 40 = 0 → R_A = - 10 kN.

Also tanθ = 3/4. → sinθ = 3/5, cosθ = 4/5.

The tensile force (T) in a member would be given a positive sign. Consider the equilibrium of the joints whose F.B.Ds are shown in Figs.3(ii) to (vi).

Consider Joint A:

∑F_x = F_AB + H_A = 0 → F_AB = - H_A = 80 kN (T).

∑F_y = F_AF + R_A = 0 → F_AF = - R_A = 10 kN (T).

Consider Joint F:

∑F_x = F_EF cosθ + F_BF cosθ = 0 → F_EF = - F_BF

∑F_y = -F_AF  + F_EF sinθ - F_BF sinθ = 0

→F_EF = F_AF/2sinθ = 8.3 kN(T),

F_BF = - 8.3 kN, i.e. 8.3 kN(C).

Consider Joint E:

∑F_x = F_DE cosθ – F_EF cosθ = 0. → F_DE = F_EF = 8.3 kN(T).

∑F_y = -F_BE  - F_DE sinθ – F_EF sinθ – 40 = 0. →F_BE = -50 kN, i.e. 50 kN(C).

Consider Joint D:

∑F_x = -F_DE cosθ – F_BDcosθ + 80 = 0. → F_BD = 275/3 = 91.7 kN(T)

∑F_y = -F_CD  + F_DE sinθ – F_BD sinθ = 0 → F_CD = -50 kN, i.e 50 kN(C).

Consider Joint C:

Considering the equilibrium equation of the joint C in the x direction, ∑F_x = - F_BC = 0. The member BC is a zero force member.

Q.4.

The unequal Z section is divided into three parts 1, 2, 3 as shown in Fig.4. The area of the Z section is A and x_c, y_c are the coordinates of its centroid. Let A_i refer to the area and x_i, y_i the coordinates of the centroid of its i^th part.

x_c = ∑A_ix_i/∑A_i = [20×5 + 24×1 + 12×(-1)]/ (20 + 24 + 12) = 112/56 = 2 cm.

y_c = ∑A_iy_i/∑A_i = [20×1 + 24×8 + 12×15)]/ (20 + 24 + 12) = 392/56 = 57/8 = 7 cm.

Let I^c_xx and I^c_yy be the second moment of area of the Z section about centriodal axes through C parallel to the x,y axes.

I^c_xx = ∑(I^c_xx)_i = ∑[b_ih_i³/12 + A_i(y_i – y_c)²]

= 10×2³/12 + 20(1 - 7)²

+ 2×12³/12 + 24(8 - 7)²

 + 6×2³/12 + 12(15 - 7)²

= 1810.67 cm⁴.

I^c_yy = ∑(I^c_yy)_i = ∑[h_ib_i³/12 + b_ih_i(x_i – x_c)²]

= 2×10³/12 + 20(5 - 2)²

+ 12×2³/12 + 24(1 - 2)²

 + 2×6³/12 + 12(-1 - 2)²

= 522.67 cm⁴.

The polar moment of the area I^c_zz about an axis through C, would be

I^c_zz = I^c_xx + I^c_yy = 1810.67 + 522.67 = 2333.3 cm⁴.

Q.5a.

The train starts from rest, i.e. initial speed u = 0. It moves with uniform tangential acceleration a_t and reaches a speed v₁ = 36 km/h in a distance s₁ = 0.6 km. Therefore, using the relation v² = u² + 2a_ts,

a_t = v₁²/2s₁ = 1080 km/h².

The speed v₂ at the middle of the distance s₂ = 0.3 km, would be

v₂ = √(2a_ts₂) = √648 = 25.456 km/h.

The centripetal acceleration a_n2 at the mid-distance s₂ is a_n2 = v₂²/R = 810 km/h².

The total acceleration a = √(a_n² + a_t²) = √(810² + 1080²) = 1350 km/h².

Q.5b.

Let v₁' and v₂' be the velocities of spheres of m₁ and m₂, respectively, just after impact.

The momentum is conserved,

m₁v₁' + m₂v₂' = m₁v₁ + m₂v₂ → m₂(v₂' – v₂) = m₁(v₁ – v₁')       (1)

As the impact is perfectly elastic, the velocity of separation = the velocity of approach,

v₂' - v₁' = v₁ - v₂ → v₂' + v₂ = v₁ + v₁' (2)

Multiplying equations (1) and (2),

m₂(v₂'² – v₂²) = m₁(v₁² – v₁'²) → m₁v₁'² + m₂v₂'² = m₁v₁² + m₂v₂². → (K.E.)_final = (K.E.)_initial.

Thus, the kinetic energy is conserved.

Q.6.

The F.B.D. of the cylinder is shown in Fig. Q.6.The forces on the cylinder are the weight mg, normal reaction N and the frictional force f.

Let a_c be the acceleration of the centre C parallel to the plane and

α the angular acceleration of the cylinder.

As there is no slip,

a_c = αR.                        (1)

The equations of motion parallel to the plane and for rotation are

mgsinθ – f = ma_c.         (2)

fR = Iα = (mR²/2)α.      (3)

From equations (1) to (3), α = 2gsinθ/3R, a_c = 2gsinθ/3, and f = mgsinθ/3.

As the centre of mass C has no acceleration normal to the plane, N = mgcosθ and the frictional force f ≤ μN,                           

mgsinθ/3 ≤ μmgcosθ → tanθ ≤ 3μ.

Q.7a.

As the pin is in double shear, for determining the diameter d of the pin,

τ ≤ P_max/(2πd²/4) → d ≥ (2P_max/πτ)^1/2 = [2×78.5×10³/(π×80×10⁶)]^1/2 = 0.025 m = 25 mm.

For the tension member,

σ ≤ P_max/[(b – d )t] = P_max/(d t), as b = 2d.

 → t ≥ P_max/(σ d) = 78.5×10³/(157×10⁶ ×0.025) =  0.020 m = 20mm.

Q.7b.

Consider a V notch with an angle θ as shown in Fig. 7b. The liquid is at a level H above the base point. The discharge dQ through an elementary strip of depth dh at a depth h below the free liquid level would be

dQ = VdA = √(2gh)bdh.

The discharge Q through the whole notch would be .

For a V notch, b = 2(H – h)tan(θ/2). Hence,

Q.8.

Let d_i and d_o be the internal and external diameters, respectively of the shaft. The polar moment of the cross-sectional area would be I_p = π(d_o⁴ – d_i⁴)/32.               (1)

Using the torsion formula, from stiffness consideration, θ = TL/GI_p.                    (2)

Using the torsion formula, from strength consideration, τ_max = T(d_o/2)/I_p.             (3)       

Eliminating I_p From equations (2) and (3),

d_o = 2τ_maxL/(Gθ) = 2×82×10⁶×2.5/(82×10⁹×2π/180) = 0.144 m = 14.4 cm.                (4)

Using equations, (1), (2) and (4),

d_i⁴ ≤ d_o⁴ - 32I_p/π = (32TL/Gθ/π) =32×25000×2.5/(82×10⁹×π/90×π)

→ d_i = 0.118 m = 11.8 cm.

Let d be the diameter of the solid shaft. Then, I_p = π d⁴/32.

From stiffness consideration, θ ≤ TL/GI_p = 32TL/(Gπd⁴)

→ π/90 ≤ 32×25000×2.5/(82×10⁹× π d⁴) → d ≥ .123m =12.3cm.

From strength consideration, τ_max ≤ T(d_o/2)/I_p = 16T/(πd³).

82×10⁶ ≤ 16×25000/(πd³) → d ≥ .116 m = 11.6 cm.

Hence d = 12.3 cm.

The % increase in weight = 100[d² – (d_o² – d_i²)]/(d_o² – d_i²)

= 100[12.3² – (14.4² – 11.8²)]/(14.4² – 11.8²) = 122.1

Q.9.

The beam with the loading and support reactions is shown in Fig.9. From the equilibrium equations of the beam,

∑M_B = R_A×L – (wL/2)L/4 = 0 → R_A = wL/8.

∑F_y= R_A + R_B  + wL/2 = 0 → R_B = 3wL/8.

The S.F. V at any section x of the beam, using singularity functions would be,

V = - wL/8 + w<x - L/2>.

The S.F. diagram is also shown in Fig. 9. The maximum S.F.

V_max= 3wL/8 at the right support, x = L.

V = - wL/2 + w<x - L/2> = 0 at x = 5L/8.

The B.M. M at any section x is

M = (wL/8)x + w<x - L/2>²/2 .

The B.M.diagram is also shown in Fig.9. The maximum B.M.

M_max = 9wL²/128 at x = 5L/8.

The maximum bending stress σ_max in the beam would be at x = 5L/8 at the top and bottom fibers, y = ± h/2.

| σ_max| = M_max(h/2)/I

= (9wL²/128) (± h/2)/(bh³/12)

= 27wL²/(64bh²).

Q.10a.

The F.B.D. of the wooden block is shown in Fig. 10a. Assume the length of the block normal to the plane of paper to be unity. At the pivot A, it is subjected to the reactions R and H. The weight W acts at the centre of gravity G. It is also subjected to a linear pressure distribution on the left from 0 at D to p_B at B and a constant pressure distribution p_B at the bottom from B to A. Let γ be the specific gravity of the wood. Take the density of water ρ = 1000 kg/m³. Then,

W = γρgL² = 1000(1.2)²γg = 1440γg.

p_B = ρgh = 1000(0.6)g = 600g.

Considering the moment equilibrium about the pivot A, ∑M_A = 0.

→ W×0.6 – (p×0.6/2)×0.6/3 - (p×1.2)×0.6 = 0.

→1440γg×0.6 - (600g×0.6/2)×0.6/3 - (600g ×1.2)×0.6 = 0. → γ = 0.542.

Q.10b.

Let the subscripts i and o refer to the nozzle inlet and outlet, respectively. Applying the continuity equation for incompressible flow,

Q = A_iV_i = A_oV_o = 50×0.02 = 1→ V_i = A_oV_o/A_i = 0.02×50/0.1 = 10 m/s.

Now applying the Bernoulli’s equation between the nozzle inlet and outlet,

p_i/ρg + V_i²/2g + z_i = p_o/ρg + V_o²/2g + z_o,

the gauge pressure (p_i - p_o) at the inlet would be,

(p_i - p_o) = ρ(V_o² - V_i²)/2 + ρg(z_o - z_i) = 1.23×(50² – 10²)/2 + 0 = 1476 Pa = 1.476 kPa.

If R is the axial force required to hold the nozzle in place,

R + (p_i - p_o) A_i = ρQ(V_o - V_i)

→ R = ρQ(V_o - V_i) - (p_i - p_o) A_i = 1.23×(50 – 10) - 1476×0.1 = - 98.4 N.

Q.11.

The inlet and outlet velocity triangles are as shown in Fig.11. Let subscripts 1 and 2 refer to the inlet and outlet diagrams, respectively. As water enters the runner blades in the radial direction and leaves the runner blades axially,

V_f1 = V_r1 and V_f2= V₂.

From the inlet velocity triangle,

u₁ = V_f1/tanα = 8/tan15 = 29.856 m/s = V_w1.

Let D₁ and D₂ be the inlet and outlet diameters of the runner. As u₁ = πD₁N/60 → D₁ = 60×29.856/(π×350) = 1.629 m.

D₂ = 0.6D₁ = 0.977 m.

The head applied

H = V_w1u₁/g + V₂²/2g =(29.856)²/9.81 + 8²/(2×9.81) = 48.69 m.

From the outlet velocity diagram, tanβ = V_f2/u₂.

The flow velocity is constant, V_f2 = 8 m/s, and the blade velocity at the outlet u₂ = 0.6u₁.

Hence, the blade angle at outlet β = tan^-1[8/(0.6×29.856)] = 24.06⁰.

The discharge Q = K(πD₁b₁)V_f1 = 0.95(π×1.629×0.1×1.629)×8 = 6.34 m³/s.

The power output P = ρQV_w1u₁ =1000×6.34×29.856×29.856 = 5651000 W = 5.651 MW.

SOLUTIONS A-03 APPLIED MECHANICS    (December 2004)

Q.1.

            a.         D         Force, velocity and Linear momentum all follow the parallelogram                                              law of addition.                                                                                          

            b.         A         At the top of the trajectory, the speed is vcosθ and centripetal                                       acceleration g. Hence radius of curvature R = (vcosθ)²/g.

            c.         B          As the impact is perfectly elastic the kinetic energy is conserved.                                                The impulse from the fixed plane changes the momentum.                                                                                                    

            d.         C         Force = md²x/dt² = md²(Asinωt)/dt² = - mAω²sinωt. Hence, the                                              maximum force = mAω².

            e.         A         Yield stress is a material property.

            f.          D         As the bending moment is maximum under the load, the curvature                                              is also maximum there.    

            g.         C         Froude number is (inertia force/gravity force)^1/2.                                               

            h.         B          The energy gradient represents the total head     and the hydraulic                                              gradient line the pressure and datum head only.                         

Q.2a.

F_R = (ΣF_x) i + (ΣF_y)j =100 i - 75 j N.

Equating the moment of the resultant and the given force system about O,

x_Ri × F_R = 50k + 2.5i × (-75)j + 0.4j × 100i                                                               

→ -75x_Rk = 50k – 187.5k – 40k = -177.5k → x_R = 2.37 m.                                                

Q.2b.

The F.B.D. of the unit length of the dam is shown in Fig.2b. It is subjected to its own weight υ₁ah, the linearly increasing pressure on the left from 0 at the top to υh at the bottom, the shear force F and the normal reaction N from the foundation.

Considering the equilibrium of the dam,

ΣF_x = (υh)h/2 -F  = 0  → F = υh²/2.                                       

ΣF_y = N – υ₁ah = 0 → N = υ₁ah.                                            

ΣM_A = Nx_N – (υ₁ah)a/2 – (υh²/2)h/3 = 0. 

→ x_N = a/2 + υh²/(6υ₁a).

Q.3a.

Consider the equilibrium of the portion of the truss to the right of the section XX as shown in Fig.3a. The forces acting on this portion are the 500 N loads at D, C and the tensile forces of the sectioned members F_GF, F_BF and F_BC.                  

Taking moment of all the forces about D

ΣM_D = 10×F_BF +5×500 = 0

→ F_BF = -250 N, i.e. 250 N (C).                    

Taking moment about F

 ΣM_F = -10sin30⁰×F_BC -5×500 - 10×500 = 0

→ F_BC = - 1500 N, i.e. 1500 N(C).

Q.3b.

There is inevitable play between the column and the collar and hence the collar will be in contact with the column at A and B. The F.B.D. of the collar is as shown in Fig.3b with load P, normal reactions N_A, N_B and frictional forces f_A, f_B.

At impending slip  f_A = μN_A, f_B = μN_B.

Considering the equilibrium of the collar,

ΣF_H = - N_A + N_B = 0 → N_A = N_B = N. Hence, f_A = f_B = f = μN.

ΣM_C = N_Ba – f_A(x + b/2) - f_B(x - b/2) = 0. → Na – 2fx =0.

Hence x = a/2μ.                                                                                                          

Q.4.

The channel section is divided in parts 1, 2, 3 as shown in Fig.4. Let a_i be the area and x_i, y_i the coordinates of the centroid C_i of the i^th part. C is the centroid of the Channel section. Then,

x_C = (a₁x₁ + a₂x₂ + a₃x₃)/(a₁ + a₂ + a₃)           

    = (8×5 + 48×2 + 4×5)/(8 + 48 + 4) = 2.6 cm.

y_C = (a₁y₁ + a₂y₂ + a₃y₃)/(a₁ + a₂ + a₃)

    = (8×2 + 48×6 + 4×11)/( 8 + 48 + 4) = 5.8 cm.     

Let I^C_xx and I^C_yy be the second moments of area about the x, y axes through C. Then,

I^C_xx = ∑[b_ih_i³/12 + a_i(y_i – y_C)²]

= (2×4³/12) + 8(2 – 5.8)² + (4×12³/12) + 48(6 – 5.8)² + (2×2³/12) + 4(11– 5.8)²

= 813.6 cm⁴.

I^C_yy = ∑[h_ib_i³/12 + a_i(x_i – x_C)²]

= (4×2³/12) + 8(5 – 2.6)² + (12×4³/12) + 48(2 – 2.6)² + (2×2³/12) + 4(5 – 2.6)²

= 154.4 cm⁴.                                                   

Polar moment of area about the axis through centroid I^C_zz = I^C_xx + I^C_yy= 968 cm⁴

Q.5a.

Tangential acceleration in the positive x direction is a_t = 3 m/s².                         

Centripetal acceleration in the positive y direction is a_n = V²/R = 4²/4 = 4 m/s².  

The total acceleration vector a = 3i + 4j m/s².                                                  

Magnitude a = √(3² + 4²) = 5 m/s² at angle θ = tan^-1(4/3) = 53.1⁰ with the x axis.        

Q.5b.

The initial velocity of the car is V_i1 = 8 km/h = 8×1000/3600 = 20/9 m/s.

As the impact with the rigid wall is perfectly plastic, the final velocity V_f1 = 0.     

Energy absorbed by the bumper during impact E_b= mV_i1²/2 = 1100(20/9)²/2 = 2716 J.     

Let U be the maximum initial speed of the moving car at which it can hit a similar stationary car without causing any damage. As the impact is perfectly plastic, the common velocity after impact would be V for both the cars.

            From linear momentum conservation: 1100U = 1100V + 1100V → V = U/2.              

Initial kinetic energy KE₁ = 1100U²/2.                                                 

Kinetic energy after impact KE₂ = (1100 +1100)V²/2 = 1100U²/4.                     

Energy to be absorbed by the bumpers during impact = KE₁ - KE₂ = 1100U²/4.          

The energy which can be absorbed by the two bumpers without damage is: 2E_b = 5432 J.

Therefore, 1100U²/4 = 5432 → U = 4.444 m/s = 16 km/h.                             

Q.6a.

The reference xyz is fixed to the bent rod and at the instant of interest have the same orientation as the ground reference XYZ.

Unit vectors along x, y, z are i, j, k and along the X, Y, Z are I, J, K, respectively.

Angular velocity of the disc C,

ω_C = ω₁j + ω₂K = 10 j + 5 K rad/s = 10 J + 5 K rad/s at this instant.              

Angular acceleration of the disc C

α_C = (dω_C/dt)_XYZ = (dω₁/dt)j + ω₁dj/dt + (dω₂/dt)K + ω₂dK/dt = ω₁dj/dt = ω₁(ω₂K ×j)

   = ω₁ω₂i  = 50i rad/s^{2 .}= 50I rad/s² at this instant.                                                       

Q.6b.

As the string breaks, the F.B.D. of the rod is shown in Fig.6b.                          

The rod AB would start rotating about the pinned end A.

At this instant, its angular velocity ω = 0,

and angular acceleration is α.

The equation of motion for rotation is

ΣM_A = I_Aα → -mgL/2 = (mL²/3)α

→ α = -3g/2L.                                    

Acceleration of the centre of mass C is a_Cx = 0 and a_Cy = αL/2 = -3g/4.                                   

The equations of motion for the centre of mass give the reactions at the hinge A

H = ma_Cx = 0.

 R – mg = m a_Cy = m(-3g/4) = -3mg/4 → R = mg/4.                          

Ans.7(a)

The bar is imagined to be cut by a plane at 45⁰ to the cross-section and the F.B.D. of the portion to the left is shown in Fig.7a.                         

The area of the inclined section A' = A/cos45⁰ = A√2.

The axial force P can be resolved into components normal to the area A' and in the plane of the area A'.

Normal Force P_n =Pcos45⁰ =P/√2

→Normal stress =P_n/A' =(P/√2)/A√2 = P/2A.

Shear force P_t = Psin45⁰ =P/√2

→ Shear stress = P_t/A' = (P/√2)/A√2 = P/2A.             

Q.7b.

Hoop stress σ_θθ = pd/2t = 0.8×10⁶×2000/2×10 = 80×10⁶ Pa.

Axial stress σ_zz = pd/4t = .8×10⁶×2000/4×10 = 40×10⁶ Pa.                                        

Hoop strain ε_θθ = (σ_θθ – νσ_zz)/E = (80×10⁶ – 0.25×40×10⁶)/200×10⁹ = 35×10^-5.         

Change in diameter Δd = ε_θθd = 35×10^-5×2000 = 0.7 mm.                                           

Q.8.

The F.B.D. of the beam is shown in Fig.8. From equilibrium of the beam,

ΣF_x = 0 → H = 0.

ΣM_B=0→8R_A = -40 +10×4 + 40×2 =80

→ R_A = 10 kN

ΣF_y = 0 → R_B = 50 – R_A = 40 kN.

The S.F. at a section x is

V = R_A - 10<x-4>

V_max = 40 kN at the right support B.

The B.M. at a section x is

 M = R_Ax + 40<x - 2>⁰ - 10<x - 2>

        -10<x - 2>²/2.

M_max = 80 kNm at the centre C.

The S.F. and B.M. diagrams are also shown in Fig.8.               

Q.9a.

Let D be the diameter of the solid shaft in mm.

The polar moment of the cross-section I_p = πD⁴/32.                                                     

If τ_shaft  is the maximum shear stress in the shaft, the torque transmitted

T = τ_shaftI_p/(D/2) = τ_shaft πD³/16.                                               (1)                                           

Number of bolts n = 8, diameter of bolts d = 12.5 mm, pitch circle radius R =115 mm.

If τ_bolt is the average shear stress in a bolt, the torque transmitted

T = n×τ_bolt(πd²/4)×R = 8×τ_bolt(π×12.5²/4)×115                        (2)

As the torque transmitted T is the same and τ_shaft = τ_bolt, from equations (1) and (2)

πD³/16 = 8×(π×12.5²/4)×115 → D = 83.2 mm.

Q.9b.

The inlet and outlet velocity diagrams are shown in Fig.9b. The subscripts 1 and 2 refer to the inlet and outlet conditions. Bucket speed U₂ = U₁ = 15 m/s. Inlet jet velocity

V₁ = C_v√(2gH) = 0.985√(2×9.81×42) = 28.27 m/s.         

From inlet velocity triangle

V_r1 = V₁ – U₁ = 28.57 – 15 = 13.27 m/s.

V_w1 = V₁ = 28.27 m/s.                                                 

The blade outlet angle β₂ = 180⁰ – 165⁰ = 15⁰.

Neglecting frictional losses

V_r1 = V_r2 = 13.27 m/s.                                     

From outlet velocity triangle

V_w2 = U₂ - V_r2 cos β₂ = 15 – 13.27 cos15⁰ = 2.18 m/s.

Power developed P = ρQ(V_w1 - V_w2)U₁

= 1000×1×(28.27 – 2.18)×15 = 3913500 W = 391.35 kW.

Available Power = ρgH

= 1000×9.81×42 = 41202 W = 412.02 kW.    

Turbine efficiency η = Power developed/available power  = 391.35/412.02 = 0.95 = 95%.

Q.10a.

At the section 6 m below the throat, i.e. section 1

Pressure p₁ = 5 atm = 5×10.33 = 51.65m of water, velocity V₁ and datum z₁ = 0.

At the throat, i.e. section 2

Pressure p₂ = 10.33 + 0.20 = 10.53 m of water, velocity V₂ and datum z₁ = 6.

Applying Bernoulli’s equation between sections 1 and 2

51.65 + V₁²/2g + 0 = 10.53 + V₂²/2g + 6 → V₂² - V₁² = 35.12×2×9.81 = 689 (m/s)².

Area of section 1, A₁ = π×0.15²/4 = 0.0177 m²,

Area of section 2, A₂ = π×0.07²/4 = 0.00385 m².

Using the continuity equation, discharge Q = A₁V₁ = A₂V₂.                                

V₁ = Q/A₁ = Q/0.0177 = 56.6Q and V₂ = Q/A₂ = Q/0.00385 = 260.

Hence V₂² - V₁² = (260² -56.6²)Q² = 689 → Q = 0.1034 m³/s.                       

Q.10b.

Let the subscripts m and p denote the model and prototype, respectively.

The inertial and viscous forces are important. Hence, the Reynolds number must be identical in the model and prototype flow.

R_e = (ρVL/μ)_m =  (ρVL/μ)_p

As the fluid is the same ρ and μ of the model and prototype are the same, Hence

(VL)_m =(VL)_p → V_m = V_mL_p/L_m = 60×6 = 360 km/h.                                                  

The non-dimensional term for the drag force F and inertia force ρV²L² is (F/ρV²L²) and would be the same for the model and prototype, i.e. (F/ρV²L²)_p = (F/ρV²L²)_m

Hence, prototype drag F_p = F_m (V_m²/V_p²)( L_m²/L_p²) = 510×(360/60)²(1/6)² = 510 N.

Q.11a.

V_r = (∂ψ/∂θ)/r = V(1 - R²/r²)cosθ , V_θ = -∂ψ/∂r = -V(1 + R²/r²)sinθ.                           

For the stagnation points in the flow V_r = 0 → r = R and V_θ = 0 → θ = 0, π.

Hence the two stagnation points are (R, 0) and (R, π).                                      

The velocity on the surface of the cylinder is V_r = 0 and V_θ = -2Vsinθ.              

As the flow is given to be irrotational, Bernoulli’s equation can be applied between a point on the surface of the cylinder r = R and a point far upstream in the uniform flow where the velocity is V and pressure p_∞. If p is the pressure on a point on the cylinder,

p/ρ + (-2Vsinθ)²/2 = p_∞/ρ + V²/2 → p = p_∞ + ρ(1 - 4sin²θ)V²/2.                                  

Q.11b.

A fully developed laminar flow through a horizontal pipe of radius R is shown in Fig.11b. The axial equilibrium of a cylinder of fluid of radius r and length dx is considered.             

(p + dp) πr² – pπr² = τ 2πrdx → τ = (r/2) dp/dx

According to Newton’s law of viscosity τ = μdu/dr. Hence, μdu/dr = (r/2)dp/dx

→ du/dr = (1/2μ)rdp/dx                                                                                             

On integrating u = (1/4μ)r²dp/dx + C

At r = R, u = 0. Hence, C = - (1/4μ)R²dp/dx.                                                  

Substituting for C, u = (1/4μ)(R² - r²)dp/dx.

This is a parabolic distribution. The maximum velocity is at the centre-line r = 0.

u_max = (1/4μ) R²dp/dx.