13.4 Resultants, moments, and torques b z

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1 Homework 13. Chapters 14, 14, 15. Forces, torque and replacement 13.1 Momentsofforcesaboutvariouspoints Consider the following sets of forces. Circle the set(s) in which the following are all equal: ote: All forces have the same magnitude. Forces that are not horizontal or vertical are 3 from vertical. Moment of the set around point Moment of the set around point Moment of the set around point 13.2 Resultants, moments, and torques rove that the couple of forces shown to the right cannot be replaced by just its resultant, no matter where the resultant is applied ptional : Resultants, moments, and torques rove that a set S of coplanar forces (other than a couple) can be replaced by its resultant as long as the resultant s line of action lies along a certain line (called the central axis of S) Resultants, moments, and torques b z b x rove that the set of two forces F x and F z shown to the right cannot be replaced by its resultant, no matter where the resultant is applied. a F x F z b c Copyright c by aul Mitiguy 323 Homework 13

2 13.5 Cantilever beam with end load: Shear force and bending moment diagram. he following figure shows points i (i=,..., 8) uniformly distributed along a relatively light (massless) rigid horizontal beam of length L = 8 m. he beam is loaded with vertically downward forces of magnitude F i (i =2, 4, 6, 8) at points i (i =2, 4, 6, 8), respectively, with F 2 = F 8 = 2 and F 4 = F 6 = 1. he beam is cantilevered to ground (a ewtonian reference frame ) which means that: oint of cannot translate relative to b z b x F 2 F 4 F 6 F 8 eam cannot rotate relative to o Right-handed orthogonal unit vectors b x,, b z are fixed in L with b x horizontally right and vertically downward. At point i of the beam, the beam can be sliced into two sections namely the beam-section containing point i and the beam to the left of i and the beam-section to the right of i.hecutis made to create a planar cross-section normal to b x. Since i cannot translate relative to the right-beam-section and since the left-beam-section cannot rotate relative to the right-beam section, one knows the set S of forces exerted on i s left-beamsection by its right-beam-section can be replaced by an equivalent set consisting of: AforceF i applied to i that is equal to the resultant of S. ote: he vertical component of F i is called a shear force. ote: he horizontal component of F i is called an axial force. A couple whose torque M S/ i is equal to the moment of S about i. ote: he b z component of M S/ i is called a bending moment. (a) Determine F, the resultant of the set S of forces exerted on s left-beam-section by its right-beam-section, and determine M S/, the moment of S about. F = (F 2 + F 4 + F 6 + F 8 ) ( M S/ L = 4 F 2 + L 2 F L ) 4 F 6 + LF 8 b z (b) Determine F 1, the resultant of the set S of forces exerted on 1 s left-beam-section by its right-beam-section, and determine M S/ 1, the moment of S about 1. F 1 = (F 2 + F 4 + F 6 + F 8 ) ( M S/ 1 L = 8 F L 8 F L 8 F L ) 8 F 8 b z (c) Complete the following table with numerical values. For the points marked with ame of Resultant of set S of forces exerted on s left-beamsection Moment of S point by its right-beam-section () about ( m) Copyright c by aul Mitiguy 324 Homework 13

3 (d) Data from the previous table is marked with an X on the following shear-force diagram and bending-moment diagram. Complete these diagrams by properly connecting the data so that shear-force diagram looks like steps whereas the data in the bending-moment diagrams is connected with lines. b z b x F 2 F 4 F 6 F 8 o L 6 5 Shear force () x-distance along beam (m) 3 25 ending moment (*m) x-distance along beam (m) Copyright c by aul Mitiguy 325 Homework 13

4 13.6 Cantilever beam with uniform load: Shear force and bending moment diagram. he following figure shows points i (i=,..., 8) uniformly distributed along a relatively heavy (massive) rigid horizontal beam of length L = 8 m. he beam is uniformly loaded vertically downward due to its own weight. he beam s weight per unit length is W L m where W = 8 kg. he beam is cantilevered to ground (a ewtonian reference frame ), Right-handed orthogonal unit vectors b x,, b z are fixed in with b x horizontally right and vertically downward. b z b x o Uniform load of W/L At point i of the beam, the beam can be sliced into two sections namely the beam-section containing point i and the beam to the left of i and the beam-section to the right of i.hecutis made to create a planar cross-section normal to b x. (a) Determine F, the resultant of the set S of forces exerted on s left-beam-section by its right-beam-section, and determine M S/, the moment of S about. Repeat for point 4. F = W M S/ = WL 2 b z F 4 = W 2 M S/ 4 = WL 8 b z (b) Consider a point F i located a distance x to the right of. Determine F i,theresultantof the set S of forces exerted on i s left-beam-section by its right-beam-section, and determine M S/ 4, the moment of S about 4. Express these results in terms of W, L, andx. ( ) F i = W M S/ i = W ( ) 2 bz 2 L (c) Draw the shear-force diagram and bending-moment diagram. L Shear force () ending moment (*m) x-distance along beam (m) x-distance along beam (m) 13.7 roblems from Sheri Sheppard s textbook. roblem roblem Copyright c by aul Mitiguy 326 Homework 13

5 13.8 Mechanical advantage of pulley systems. his problem investigates the mechanical advantage of several pulley systems. he following figure shows several pulley systems that are attached to a rigid ceiling and support a heavy engine block. he pulleys and rope are relatively light (massless), and the pulley are frictionless (hence the tension Mechanical is the same on either side of the pulley). For each system, draw a suitable number of free-body diagram(s) to determine each pulley system s mechanical System Advantage advantage. Ensure each free-body diagram clearly A 2 Identifies the system being analyzed C Identifies the system s external contact and distance forces D Determine each pulley s mechanical advantage defined as E Mechanical Advantage = utput force Input force = W where is the tension in the pulley s rope and W is the weight of the engine block. A C D E W W W W W Copyright c by aul Mitiguy 327 Homework 13

6 13.9 Static equilibrium of a triple-pendulum. AsystemS consists of 3 frictionless pin-connected rods A,, andc, each of mass m and length L, suspended from a horizontal ceiling. Horizontally-directed forces of magnitude F A, F,andF C are applied to distal (outboard) ends of rods A,, andc, respectively as shown. Knowing that S is in static equilibrium, find three equations that relate m, L, F A, F, F C,andg (the local gravitational constant) with q A, q,andq C (angles between the local vertical and the long-axes of rods A,, andc, respectively). o do this most efficiently, Use only three free-body-diagrams Use replacement and equivalent sets of forces to simplify your results Start with a free-body-diagram of rod C (F A + F C F )cos(q A ) = 2.5 mg sin(q A ) = F C cos(q C ) =.5 mg sin(q C ) F A q A F A q C q C F C Knowing F A =3n, F =2 n, F C =1n, m =1kg, and g =9.8 m/sec 2. determine q A, q and q C. q A = q =34.2 q C = he set of forces exerted by on A across the light (massless) revolute joint (pin) is equivalent to aforcef A/ applied to the proximal (inboard) end of A. his force is written in terms of the unknown force variables Fx A and F y A as shown below. Similarly forces exerted by A on and on C are replaced, resulting in the following expressions. F A/ = F A x n x + F A y n y F /A = F x n x + F y n y F C/ = F C x n x + F C y n y Find these force variables in terms of m, g, F A, F,andF C. Fx A = Fy A = 3mg Fx = F F C Fy = Fx C = Fy C = mg It was easier for me to find the angles/forces (circle one). Copyright c by aul Mitiguy 328 Homework 13

Engineering Mechanics: Statics in SI Units, 12e

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