(a) A simple beam is loaded as shown in the figure. Using the double-integration method, determine the following : (i) The…

(a) Simply supported beam under a triangular load, by double integration (20 marks)

Reactions. The load intensity is w = w₀(1 − x/L), with x measured from A. The total load is W = w₀L/2, acting at L/3 from A. Moments about B give R_A = W(2L/3)/L = w₀L/3, and R_B = w₀L/6.

Bending moment. For a section at x from A, with sagging moment positive:

M = R_A x − ∫₀ˣ w₀(1 − ξ/L)(x − ξ) dξ = w₀Lx/3 − w₀x²/2 + w₀x³/(6L)

(M = 0 at x = L, as required.)

(i) Elastic curve. Take y positive upward, so EI d²y/dx² = M:

EI y″ = w₀Lx/3 − w₀x²/2 + w₀x³/(6L)

EI y′ = w₀Lx²/6 − w₀x³/6 + w₀x⁴/(24L) + C₁

EI y = w₀Lx³/18 − w₀x⁴/24 + w₀x⁵/(120L) + C₁x + C₂

Boundary conditions: y = 0 at x = 0 gives C₂ = 0. y = 0 at x = L gives w₀L⁴(1/18 − 1/24 + 1/120) + C₁L = 0, that is w₀L⁴/45 + C₁L = 0, so C₁ = −w₀L³/45. Hence

y = (w₀/(360 EI L)) (20L²x³ − 15Lx⁴ + 3x⁵ − 8L⁴x)

which is negative (downward) throughout.

(ii) Slope at A. At x = 0, y′ = C₁/EI:

θ_A = −w₀L³/(45 EI)

This is a slope of magnitude w₀L³/(45EI) = 8w₀L³/(360EI), with the beam falling to the right (clockwise). The slope at B is 7w₀L³/(360EI) in magnitude, and it is smaller because R_B is smaller.

(iii) Deflection at midspan. At x = L/2:

EI y = w₀L⁴(1/144 − 1/384 + 1/3840 − 1/90) = w₀L⁴ (80 − 30 + 3 − 128)/11 520 = −75 w₀L⁴/11 520

y(L/2) = −5w₀L⁴/(768 EI), that is a downward deflection of 5w₀L⁴/(768EI).

(b) Balancing of three eccentric masses in different planes (20 marks)

Data. Three masses of 1 kg at radii 20, 30 and 20 mm, in planes 50 mm apart, with the middle mass (r = 30 mm) in the central plane. Angular positions are 120° apart. The balancing masses A and B are at r = 700 mm, in planes 100 mm either side of the central plane. Take the central plane as reference, and measure angles from mass 1 (the mass on the side of A) in the sense mass 1 → 2 → 3, so mass 1 at 0°, mass 2 at 120° and mass 3 at 240°.

For each mass, mr (kg·mm), position z along the shaft from the central plane (mm) and angle are:

• 1: mr = 20, z = −50, angle 0°.
• 2: mr = 30, z = 0, angle 120°.
• 3: mr = 20, z = +50, angle 240°.
• A: unknown, z = −100.
• B: unknown, z = +100.

Force balance (Σ mr = 0). Components x, y in kg·mm:

• 1: (20, 0)
• 2: (−15, 25.98)
• 3: (−10, −17.32)

Sum of the three = (−5, 8.66), so the balancing masses must satisfy (mr)_A + (mr)_B = (5, −8.66).

Couple balance about the central plane (Σ mr z = 0). The couples of the three masses are 20(−50) = −1000 at 0°, zero, and 20(50) = +1000 at 240°:

−1000∠0° = (−1000, 0) and 1000∠240° = (−500, −866.0), sum (−1500, −866.0).

With (mr)_A at z = −100 and (mr)_B at z = +100: −100 (mr)_A + 100 (mr)_B = (1500, 866.0), so (mr)_B − (mr)_A = (15, 8.66).

Solving. Adding and subtracting the two vector equations:

• (mr)_B = ½[(5, −8.66) + (15, 8.66)] = (10, 0), so (mr)_B = 10 kg·mm at 0°.
• (mr)_A = ½[(5, −8.66) − (15, 8.66)] = (−5, −8.66), so (mr)_A = 10 kg·mm at 240°.

(i) Balancing masses. With r = 700 mm:

• m_A = 10/700 = 0.01429 kg = 14.3 g at 240°, in the plane 100 mm on the mass-1 side of the central plane.
• m_B = 10/700 = 0.01429 kg = 14.3 g at 0°, in the plane 100 mm on the mass-3 side.

(ii) One mass moved to r = 50 mm. The balance conditions involve only the products mr, the planes and the angles, so the product (mr) of the shifted mass must remain 10 kg·mm at the same angle and plane. At r = 50 mm:

m = 10/50 = 0.20 kg = 200 g, an increase of about 186 g (14 times the original 14.3 g).

Other masses. The other balancing mass, and the angular positions and planes of both, do not change, because the force and couple polygons are unchanged.

(c) Force to start the block, with the bar pinned at B (10 marks)

Model. The uniform bar AB (mass m, length ℓ) is pinned at B and rests at A on the block (mass m₀), at angle θ to the horizontal. The contact at A is sliding with μ_k (the block slides under the bar's end), and the block-floor contact is at the point of slipping with μ_s. Let N be the normal force at A, and f = ±μ_k N the friction force on the bar at A.

Bar in equilibrium. Take moments about B. The weight mg acts at the middle, at horizontal distance (ℓ/2) cos θ from B, and N and f act at A, at horizontal distance ℓ cos θ and vertical distance ℓ sin θ:

N ℓ cos θ − f ℓ sin θ = mg (ℓ/2) cos θ

Block in equilibrium. The normal force from the floor is N_f = m₀g + N, and the floor friction is μ_s N_f. The friction from the bar on the block is μ_k N, opposing the motion, so

P = μ_k N + μ_s (m₀g + N) = (μ_k + μ_s) N + μ_s m₀ g

(A) P towards the right. The block slides to the right under A, so the friction on the bar is f = +μ_k N, to the right. Then N (cos θ − μ_k sin θ) = mg cos θ/2:

N = mg / [2 (1 − μ_k tan θ)]

P_A = (μ_k + μ_s) mg / [2 (1 − μ_k tan θ)] + μ_s m₀ g

(B) P towards the left. The friction on the bar is now to the left, f = −μ_k N, so N (cos θ + μ_k sin θ) = mg cos θ/2:

N = mg / [2 (1 + μ_k tan θ)]

P_B = (μ_k + μ_s) mg / [2 (1 + μ_k tan θ)] + μ_s m₀ g

Numerical values. θ = 30° (tan θ = 0.5774), m = m₀ = 3 kg, mg = 29.43 N, μ_s = 0.60, μ_k = 0.50, so μ_k tan θ = 0.2887.

• Case (A): N = 29.43/(2 × 0.7113) = 20.69 N, and P_A = 1.1 × 20.69 + 0.6 × 29.43 = 22.76 + 17.66 = 40.4 N.
• Case (B): N = 29.43/(2 × 1.2887) = 11.42 N, and P_B = 1.1 × 11.42 + 17.66 = 12.56 + 17.66 = 30.2 N.

Pushing towards the wall needs more force, because the friction at A then presses the bar end harder onto the block (the bar wedges), which raises the normal force on the floor. The result holds only while 1 − μ_k tan θ > 0, that is μ_k tan θ < 1; if it fails, the block cannot be moved towards the wall at all (self-locking).