StaticS Fourteenth edition in si units

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2 EnginEEring MEchanics StaticS Fourteenth edition in si units r.. hibbeler SI onversion b Kai beng Yap hoboken oston columbus san Francisco ne York indianapolis ondon Toronto sdne singapore Toko Montreal Dubai Madrid hong Kong Meico cit Munich Paris amsterdam cape Ton

3 .8 further SimPlification of a force and couple SStem Replace the loading acting on the beam b a single resultant force. Specif here the force acts, measured from end. * 12. Replace the parallel force sstem acting on the plate b a resultant force and specif its location on the - plane Replace the loading acting on the beam b a single resultant force. Specif here the force acts, measured from. 1 m N 700 N 300 N 30 1 m 2 kn 5 kn m 3 m Probs. 121/ N m 1 m 3 kn Prob Replace the loading on the frame b a single resultant force. Specif here its line of action intersects a vertical line along member, measured from Replace the loading on the frame b a single resultant force. Specif here its line of action intersects member, measured from. 00 N 600 N 126. Replace the loading on the frame b a single resultant force. Specif here its line of action intersects member D, measured from end. 300N 1m 2m 3m 250N 5 3 D 00N m 2m 1.5 m N 3m Prob. 123 Probs. 125/126

4 188 chapter force SStem resultants 127. Replace the to renches and the force, acting on the pipe assembl, b an equivalent resultant force and couple moment at point The tube supports the four parallel forces. Determine the magnitudes of forces F and F D acting at and D so that the equivalent resultant force of the force sstem acts through the midpoint of the tube. 100 N m F D 300 N 600 N D F 0.6 m 0.8 m N 00 mm 00 mm 500 N 200 mm 200 mm 180 N m Prob. 127 Prob. 129 * 128. Replace the force sstem b a rench and specif the magnitude of the force and couple moment of the rench and the point here the rench intersects the plane If F = 7 kn and F = 5 kn, represent the force sstem acting on the corbels b a resultant force, and specif its location on the plane Determine the magnitudes of F and F so that the resultant force passes through point of the column N 3 m 6 kn 150 mm 100 mm 650 mm F 600 mm 750 mm F 8 kn 700 mm 100 mm 150 mm Prob. 128 Probs. 130/131

5 .8 further SimPlification of a force and couple SStem 189 * 132. If F = 0 kn and F = 35 kn, determine the magnitude of the resultant force and specif the location of its point of application (, ) on the slab If the resultant force is required to act at the center of the slab, determine the magnitude of the column loadings F and F and the magnitude of the resultant force. * 136. Replace the five forces acting on the plate b a rench. Specif the magnitude of the force and couple moment for the rench and the point P(, ) here the rench intersects the plane. 800 N m 30 kn 2.5 m 0.75 m 0.75 m 3 m F 0.75 m 90 kn 2.5 m F 3 m 0.75 m Probs. 132/ kn m 600 N 300 N Prob N 13. The building slab is subjected to four parallel column loadings. Determine the equivalent resultant force and specif its location (, ) on the slab. Take F 1 = 8 kn and F 2 = 9 kn The building slab is subjected to four parallel column loadings. Determine F 1 and F 2 if the resultant force acts through point (1, 10 m) Replace the three forces acting on the plate b a rench. Specif the magnitude of the force and couple moment for the rench and the point P(, ) here the rench intersects the plate. 12 kn F 1 F 2 F { 300k} N F {200j} N 8 m 1 6 kn m 16 m 6 m 5 m F {00i} N P 3 m Probs. 13/135 Prob. 137

6 190 chapter force SStem resultants p.9 Reduction of a Simple Distributed oading b p p() Sometimes, a bod ma be subjected to a loading that is distributed over its surface. For eample, the pressure of the ind on the face of a sign, the pressure of ater ithin a tank, or the eight of sand on the floor of a storage container, are all distributed loadings. The pressure eerted at each point on the surface indicates the intensit of the loading. It is measured using pascals Pa (or N>m 2 ) in SI units. (a) d df d () (b) oading long a Single is. The most common tpe of distributed loading encountered in engineering practice can be represented along a single ais.* For eample, consider the beam (or plate) in Fig. 8a that has a constant idth and is subjected to a pressure loading that varies onl along the ais. This loading can be described b the function p = p( ) N>m 2. It contains onl one variable, and for this reason, e can also represent it as a coplanar distributed load. To do so, e multipl the loading function b the idth b m of the beam, so that ( ) = p( )b N>m, Fig. 8b. Using the methods of Sec..8, e can replace this coplanar parallel force sstem ith a single equivalent resultant force acting at a specific location on the beam, Fig. 8c. (c) Fig. 8 Magnitude of Resultant Force. From Eq. 17 ( = F ), the magnitude of is equivalent to the sum of all the forces in the sstem. In this case integration must be used since there is an infinite number of parallel forces df acting on the beam, Fig. 8b. Since df is acting on an element of length d, and () is a force per unit length, then df = () d = d. In other ords, the magnitude of df is determined from the colored differential area d under the loading curve. For the entire length, +T = F; = () d = d = ( 19) Therefore, the magnitude of the resultant force is equal to the area under the loading diagram, Fig. 8c. *The more general case of a surface loading acting on a bod is considered in Sec. 9.5.

7 .9 reduction of a SimPle distriuted loading 191 ocation of Resultant Force. ppling Eq. 17 (M R = M ), the location of the line of action of can be determined b equating the moments of the force resultant and the parallel force distribution about point (the ais). Since df produces a moment of df = () d about, Fig. 8b, then for the entire length, Fig. 8c, p p p() a+ (M R ) = M ; - = - () d b Solving for, using Eq. 19, e have () d = () d = d d ( 20) (a) df d () This coordinate, locates the geometric center or centroid of the area under the distributed loading. In other ords, the resultant force has a line of action hich passes through the centroid (geometric center) of the area under the loading diagram, Fig. 8c. Detailed treatment of the integration techniques for finding the location of the centroid for areas is given in hapter 9. In man cases, hoever, the distributed-loading diagram is in the shape of a rectangle, triangle, or some other simple geometric form. The centroid location for such common shapes does not have to be determined from the above equation but can be obtained directl from the tabulation given on the inside back cover. nce is determined, b smmetr passes through point (, 0) on the surface of the beam, Fig. 8a. Therefore, in this case the resultant force has a magnitude equal to the volume under the loading curve p = p() and a line of action hich passes through the centroid (geometric center) of this volume. d (b) (c) Fig. 8 (Repeated) Important Points oplanar distributed loadings are defined b using a loading function = () that indicates the intensit of the loading along the length of a member. This intensit is measured in N>m. The eternal effects caused b a coplanar distributed load acting on a bod can be represented b a single resultant force. This resultant force is equivalent to the area under the loading diagram, and has a line of action that passes through the centroid or geometric center of this area. The pile of brick creates an approimate triangular distributed loading on the board.