SOLUTIONS D-02 APPLIED MECHANICS   (June 2004)

Q.1.     a.         D         If a two force body is in equilibrium, the forces must be equal, opposite and collinear.

            b.         C        

            c.         A         The limiting force of friction F = μR.

            d.         C         The force required to move the load W down is Wtan(α - φ). For α < φ, the load would not move down for zero force.

            e.         B          The moment of inertia of a semicircular section about its diameter   I = (1/2)(πR⁴/4) = 32π cm⁴.

            f.          B          V = dS/dt = 14t +10. At t = 0, V = 10 m/s.

            g.         C         The point of contra-flexure is a point where the beam curvature changes sign and hence bending moment changes sign.

            h.         C         Torsional rigidity of a shaft is given by GJ.

Q.2.

The F.B.D. of the wheel when it is just about to roll over the block is shown in Fig.2. Just when the wheel begins to roll over, there is no force from the ground on the wheel at B. The wheel is subjected to its own weight of 1000 N, the force P and the reaction R of the block.

As the wheel is in equilibrium under three forces, W, P and R only, the forces are concurrent and pass through the point D.

From the geometry of the figure,

cosECA = CE/CA = 15/30 =1/2 → 2θ = 60⁰.

angleCDA = (1/2)angleECA = θ = 30⁰.

From Lame’s theorem,

P/sinθ = W/sin(90-θ) → P = W tanθ = 1000tan30 = 1000/√3 = 577.4 N.

Q.3.

Consider the T section to consist of two rectangular parts 1 and 2 as shown in Fig.3.

Let C(x_C, y_C) be the centroid of the T section. Let A_i be the area and x_i, y_i the cooordiantes of the centroid of the i^th part.

The y coordinate of the centroid is obtained as

y_C = ∑A_iy_i/∑A_i

= [(6×4)×3 + (2×8)×7]/( 6×4 + 2×8)

= 184/40 = 4.6 cm.

The moment of inertia I^C_xx of the T section about an axis parallel to the x axis through C would be

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

= 4×6³/12 + 4×6(4.6 – 3)²

+ 8×2³/12 + 8×2(4.6 – 7)²

= 231 cm⁴.

Q.4.

The F.B.Ds. of the joints E, D and C are shown in Fig.4. The forces are assumed tensile in the members and are assigned positive sign.

Consider the equilibrium of joint E:

∑F_x = - F_CEcos45 = 0 → F_CE = 0, ∑F_y = F_DE - F_CEsin45 = 0 → F_DE = 0.

Consider the equilibrium of joint D:

∑F_y= - F_BDsin45 + F_DE = 0 → F_BD = 0, ∑F_x = -F_CD – F_BDcos45 + 2 = 0 → F_CD = 2 kN(T).

Consider the equilibrium of joint C:

∑F_x = F_CD + F_CEcos45 – F_ACcos45 = 0 → F_AC = 2√2 kN(T).

∑F_y= F_CEsin45 - F_ACsin45 – F_BC = 0 → F_BC = -2 kN, i.e. 2 kN(C).

Q.5.

The F.B.D. of the ladder is shown in Fig.5. The man is at D (AD = d) for the ladder to start slipping. The ladder is subjected to its own weight W at C, the weight of the man W/2 at D, the normal reaction N_A and the limiting friction force N_A/2 from the floor at A and the normal reaction N_B and limiting friction force N_B/3 from the wall at B.

From the force equlibrium equations,

∑F_x = N_B – N_A/2 = 0 and ∑F_y = N_A + N_B/3 – W – W/2 = 0.         

N_A = 9W/7 and N_B = 9W/14.

From moment equilibrium about A,

∑M_A = – N_B×7/√2–(N_B/3)×7/√2+W×3.5/√2+(W/2)d/√2 = 0.

Substituting for N_B in the moment equation, d = 5m.

Q.6.

Let the effort P be given a virtual displacement δ_P downward, then the virtual displacement of the load W would be δ_W = δ_P/2 upward.

If the system is in equilibrium, the sum of the virtual work must be zero, i.e.

Pδ_P – Wδ_W = 0. → P = W/2 = 500 N.

Q.7.

Let a_t be the uniform tangential acceleration of the car. If it starts from rest, the speed V after time t would be,

V = a_tt.

→ a_t = V₆₀/60= (18×1000/3600)/60 = 1/12 = 0.083 m/s².

The speed after 30 s would be V₃₀ = a_t×30 = (1/12)×30 = 2.5 m/s.

The normal acceleration a_n =     V₃₀²/R = 2.5²/250 = 0.025 m/s².

Q.8.

The stress σ = P/A = P/(πd²/4) = 100×10³/(πd²/4) = 100×10⁶.

Hence, d = √(4/1000π) = 0.03568 m = 3.568 cm.

The total elongation δ = (P/E)[L₁/A₁ + L₂/A₂ + L₃/A₃]

= 100×10³/(290×10⁹)[0.1/(π×0.03568²/4) + 0.15/(π×0.1²/4) + 0.15/(π×0.08²/4)]

= 0.0514×10^-3 m = 0.0514 mm.

Q.9.

The F.B.D. of the girder is reproduced in Fig.9. The support reactions are determined by considering its equilibrium,

∑M_A=9R_B–180×4.5-30×6-40×7.5= 0. → R_B = 143.33 kN.

∑F_x = R_A + R_B - 20×9 – 30 -40 = 0. → R_A = 106.67 kN.

The S.F. V and the B.M. M are,

0 ≤ x ≤ 6m:

V = - 106.67 + 20x.

M = 106.67x – 20x²/2.

6 m ≤ x ≤ 7.5m:

V = - 106.67 + 20x + 30.

M = 106.67x – 20x²/2 + 30(x - 6).

7.5 m ≤ x ≤ 9m:

V = 143.33 – 20(9 - x).

M = 143.33(9 – x) – 20(9 - x)²/2.

The S.F. and B.M. diagrams are also shown in Fig.9.

The maximum S.F. V_max = 143.33 kN at the right support, i.e. x = 9 m.

The S.F. changes sign at x = 5.33 m.

The maximum B.M. is at x = 5.33 m

M_max=284.4 kNm.

Q.10.

The F.B.D. of the girder is shown in Fig.10. R_A and R_B are the support reactions. From the equilibrium equations,

∑M_A = R_B×L –P×a = 0. → R_B = Pa/L.

∑F_y = R_A + R_B - P = 0. → R_A = P(L – a)/L.

The expression for the B.M. at any section at a distance x from the origin A using singularity functions is:

M = - R_Ax + P < x - a> = - P(L – a)x/L + P < x - a>.

If v is the deflection of the elastic line of the beam,

EI d²v/dx² = M = - P(L – a)x/L + P < x - a>.

Integrating twice,

EI dv/dx = - P(L – a)x²/2L + P < x - a>²/2 + C₁

EI v = - P(L – a)x³/6L + P < x - a>³/6 + C₁x + C₂

At x = 0, v = 0 → C₂ = 0.

At x = L, v = 0 → C₁ = Pa(L - a)(2L - a)/6L.

Hence, v = - P(L – a)x³/6L + P < x - a>³/6 + Pa(L - a)(2L - a)x/6L.

The deflection v_P under the load P at x = a,

v_P = Pa²(L – a)²/(3EIL)

= 120×10³×4.5² (14 – 4.5)²/(3×210×10⁹×16×10^-4×14) = 1.55×10^-2 m = 1.55 cm.

Q.11.

A shaft of diameter d is subjected to a torque T. The shear modulus of the material is G. The shear stress τ at any radius r and the angle of twist θ over a length L is given by the torsion formula for circular shafts as

τ/r = T/J = Gθ/L,

where the polar moment of  inertia of the cross-section J = πd⁴/32. 

(i) The maximum shear stress τ_max would be in the outer fibers at r_max = d/2.

τ_max= T(d/2)/J = 16T/(πd³) = 16×560/(π×0.03³) = 105.63×10⁶ N/m² =105.63 MPa.

(ii) The angle of twist θ over a length L = 1m would be

θ = TL/GJ = 32TL/(Gπd⁴) = 32×560×1/(82×10⁹×π×0.03⁴) = 0.0859 rad. = 4.92⁰.

(iii) The shear stress τ at r = 0.01m would be

τ = Tr/J = 32Tr/( πd⁴)= 32×560×0.01/(π×0.03⁴) = 70.42×10⁶ N/m² = 70.42 MPa.