Explanation Given information: The mass of the rod is m . The length of the rod is L . Calculation: Calculate the mass moment of inertia ( I ) as shown below. I = 1 12 m L 2 Sketch the Free Body Diagram as shown in Figure 1. Refer to Figure 1. Calculate the acceleration ( a ) as shown below. a = L 2 α Apply the Equilibrium of moment about A as shown below. ∑ M A = I G α + m a d W × L 2 = I α + m a L 2 Substitute m g for W , m L 2 12 for I , and L 2 α for a . m g L 2 = m L 2 12 α + m L 2 α × L 2 g 2 = L 12 α + L 4 α g 2 L = 4 α 12 α = 3 2 g L Calculate the acceleration at end B ( a B ) as shown below. a B = a n + a B / A = 0 + L α = L α Substitute 3 2 g L for α . a B = L × 3 2 g L = 3 2 g Apply the Equilibrium of forces along y direction as shown below. ∑ F y = m a y A − m g = − m a Substitute L 2 α for a . A − m g = − m × L 2 α Substitute 3 2 g L for α . A − m g = − m × L 2 × 3 2 g L A = m g − 3 4 m g A = 1 4 m g Sketch the Free Body Diagram with portion AJ as shown in Figure 2. Refer to Figure 2. The external forces acting on the portion AJ of the rod are, reaction at A , uniformly distributed load, shear V J , and bending moment M J . Calculate the acceleration ( a J ) as shown below. a J = x α Substitute 3 g 2 L for α . a J = 3 g x 2 L Calculate the inertial terms of the mass center ( m L a J ) as shown below. m L a J = m L × 3 g x 2 L = 3 m g x 2 L 2 Sketch the Free Body Diagram of the portion AJ as shown in Figure 3. Refer to Figure 3. Apply the Equilibrium of forces along y direction as shown below. ∑ F y = m a y m g L x − m g 4 + V J = 1 2 × 3 m g 2 L 2 x × x V J = m g 4 − m g L x + 3 m g 4 L 2 x 2 (1) Apply the Equilibrium of moment about J as shown below. ∑ M J = I G α + m a d ( m g L x ) × x 2 − m g 4 x + M J = 1 2 × 3 m g 2 L 2 x × x × x 3 M J = m g 4 x − m g 2 L x 2 + m g 4 L 2 x 3

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Author: BEER

Publisher: MCGRAW-HILL HIGHER EDUCATION

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Chapter 16.2, Problem 16.152P

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Draw the shear and bending moment diagram of the beam.

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Chapter 16.2, Problem 16.151P

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Bài 1. Cho cơ hệ như hình 1. Hình biểu diễn lược đổ cơ hệ tại vị trí cân bằng tĩnh. Trục tọa độ Oy hướng theo phương chuyển động của vật 1, gốc O đặt tại vị trí cân bằng của vật 1(tức khi lò xo biến dạng tĩnh). Bỏ qua khối lượng của thanh số 3. Vật rắn 2 là pulley 2 tầng đồng chất có bán kính ngoài 21, bán kính trong I, bán kính quán tính đối với trục qua tâm P-1.5, khối lượng m:. Vật rắn 4 là thanh thắng đồng chất có khối lượng m, chiều dài 1. Cho các số liệu: m = 2kg, m= = 5kg, m = 4kg, k=40(N/cm), ! – 0.8(m),r=0.1(m). Điều kiện đầu y; =0.5 cm );j = 10 cm/s) . Giả sử hệ dao động bé, Vật rắn 2 chuyển động lăn không trượt trên mặt phẳng ngang. 1. Viết phương trình chuyển động của hệ. 2. Xác định tần số dao động tự do của hệ. 3. Xác định đáp ứng dao động tự do của hệ. dây dây 1 2r Hình 1 y 3 -2 I k www. -2

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Draw the shear and bending moment diagram of the beam.

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