2 - P - V = 0 V = - P 2. M + Pax - L 2 b - P 2 x = 0. M = P 1L - x2

Transcription

1 E X P L E 6.1 Draw the shear and moment diagrams for the beam shown in Fig. 6 4a. P P B L L C P L P Fig. 6 4b Support Reactions. The support reactions have been determined, Fig. 6 4d. Shear and oment Functions. The beam is sectioned at an arbitrary distance from the support, etending within region B, and the freebody diagram of the left segment is shown in Fig. 6 4b. The unknowns and are indicated acting in the positive sense on the right-hand face of the segment according to the established sign convention. pplying the equilibrium equations yields +q F y = 0; = P (1) P d+ = 0; = P () P P P P free-body diagram for a left segment of the beam etending a distance within region BC is shown in Fig. 6 4c. s always, and are shown acting in the positive sense. Hence, +q F y = 0; P - P - = 0 = - P (3) = ma PL 4 d+ = 0; + Pa - L b - P = 0 = P 1L - (4) (d) Fig. 6 4 The shear diagram represents a plot of Eqs. 1 and 3, and the moment diagram represents a plot of Eqs. and 4, Fig. 6 4d. These equations can be checked in part by noting that d>d = -w and d>d = in each case. (These relationships are developed in the net section as Eqs. 6 1 and 6.)

2 E X P L E 6. Draw the shear and moment diagrams for the beam shown in Fig. 6 5a. B 0 C L L L 0 Support Reactions. Fig. 6 5d. The support reactions have been determined in Fig. 6 5b Shear and oment Functions. This problem is similar to the previous eample, where two coordinates must be used to epress the shear and moment in the beam throughout its length. For the segment within region B, Fig.6 5b, we have L 0 +q F y = 0; = - 0 L L 0 d+ = 0; nd for the segment within region BC, Fig.6 5c, +q Fy = 0; = - 0 L = - 0 L Fig. 6 5c 0 L 0 L 0 d+ = 0; = 0-0 L = 0 a1 - L b Shear and oment Diagrams. When the above functions are plotted, the shear and moment diagrams shown in Fig. 6 5d are obtained. In this case, notice that the shear is constant over the entire length of the beam; i.e., it is not affected by the couple moment 0 acting at the center of the beam. Just as a force creates a jump in the shear diagram, Eample 6 1, a couple moment creates a jump in the moment diagram. L (d) Fig. 6 5

3 E X P L E 6.3 Draw the shear and moment diagrams for the beam shown in Fig. 6 6a. Fig. 6 6a w Support Reactions. Fig. 6 6c. The support reactions have been computed in wl L w Fig. 6 6b Shear and oment Functions. free-body diagram of the left segment of the beam is shown in Fig. 6 6b.The distributed loading on this segment is represented by its resultant force only after the segment is isolated as a free-body diagram. Since the segment has a length, the magnitude of the resultant force is w. This force acts through the centroid of the area comprising the distributed loading, a distance of / from the right end. pplying the two equations of equilibrium yields +q F wl y = 0; - w - = 0 d+ = 0; = wa L - b - a wl b + 1wa b + = 0 = w 1L - (1) () w These results for and can be checked by noting that d>d = -w. This is indeed correct, since positive w acts downward. lso, notice that d>d =. wl L wl Shear and oment Diagrams. The shear and moment diagrams shown in Fig. 6 6c are obtained by plotting Eqs. 1 and. The point of zero shear can be found from Eq. 1: wl L wl = wa L - b = 0 = L ma = wl 8 From the moment diagram, this value of happens to represent the point on the beam where the maimum moment occurs, since by Eq. 6, the slope = 0 = d>d. From Eq., we have L ma = w cla L b - a L b d Fig. 6 6 = wl 8

4 E X P L E 6.4 Draw the shear and moment diagrams for the beam shown in Fig. 6 7a. w0 L w 0 w 0 w0 L L w 0 L 3 L 3 Fig. 6 7b Fig. 6 7a Support Reactions. The distributed load is replaced by its resultant force and the reactions have been determined as shown in Fig. 6 7b. Shear and oment Functions. free-body diagram of a beam segment of length is shown in Fig. 6 7c. Note that the intensity of the triangular load at the section is found by proportion, that is, w> = w 0 >L or w = w 0 >L. With the load intensity known, the resultant of the distributed loading is determined from the area under the diagram, Fig. 6 7c. Thus, w 0 L 3 w0 L 3 1 w 0 L w = w 0 L +q F y = 0; w 0 L - 1 a w 0 L b - = 0 w 0 = w 0 L 1L - (1) w0 L d+ = 0; w 0 L 3 - w 0 L a w 0 L ba 1 3 b + = 0 = w 0 6L 1-L3 + 3L - 3 () w 0 L 3 w0 L These results can be checked by applying Eqs. 6 1 and 6, that is, = d d = w 0 6L L - 3 = w 0 L 1L - Shear and oment Diagrams. in Fig. 6 7d. w = - d d = -w 0 L 10 - = w 0 L OK OK The graphs of Eqs. 1 and are shown w 0 L 3 (d) Fig. 6 7

5 E X P L E 6.5 kn/m 30 kn 30 kn kn/m 30 kn (kn) 30 (kn m) Fig m 36 kn 36 kn 9 m 1 m 18 m m kn/m kn/m (d) 6 kn/m 4 kn 6 kn/m 4 kn 4 kn/m kn/m { { (m) 4 ma = 163 kn m (m) Draw the shear and moment diagrams for the beam shown in Fig. 6 8a. Support Reactions. The distributed load is divided into triangular and rectangular component loadings and these loadings are then replaced by their resultant forces. The reactions have been determined as shown on the beam s free-body diagram, Fig. 6 8b. Shear and oment Functions. free-body diagram of the left segment is shown in Fig. 6 8c. s above, the trapezoidal loading is replaced by rectangular and triangular distributions. Note that the intensity of the triangular load at the section is found by proportion. The resultant force and the location of each distributed loading are also shown. pplying the equilibrium equations, we have +q F kip - 1 kip>ft - 1 y = 0; 30 kn ( kn/m) 14 (4 kn/m) kip>fta ft b - = 0 18 m d+ = 0; = a b kn kip -30 kip1 + 1 kip>fta b kn() ( kn/m) 14 (4 kn/m) kip>fta ft ba 3 b + = 0 18 m = a () 7 b kn kip # ft m Equation may be checked by noting that d>d =, that is, Eq. 1. lso, w = -d>d = + 9. This equation checks, since when = 0, w kn/m, and when 18 m, w 6 kn/m, Fig. 6 8a. Shear and oment Diagrams. Equations 1 and are plotted in Fig. 6 8d. Since the point of maimum moment occurs when d>d = = 0, then, from Eq. 1, = 0 = Choosing the positive root, m Thus, from Eq., ma = kn m (1)

6 E X P L E 6.6 Draw the shear and moment diagrams for the beam shown in Fig. 6 9a. 15 kn 5( 5) 80 kn m 15 kn B 5 m 5 m 5 kn/m C 80 kn m 5.75 kn 1 Fig. 6 9b Support Reactions. The reactions at the supports have been determined and are shown on the free-body diagram of the beam, Fig. 6 9d. Shear and oment Functions. Since there is a discontinuity of distributed load and also a concentrated load at the beam s center, two regions of must be considered in order to describe the shear and moment functions for the entire beam. 80 kn m 5.75 kn 5 m Fig. 6 9c m, Fig. 6 9b: +q F y = 0; d+ = 0; -80 kn # m kn 1 + = 0 = kn # m 5 m 6 10 m, Fig. 6 9c: +q F y = 0; d+ = 0; 5.75 kn - = 0 = 5.75 kn 5.75 kn - 15 kn - 5 kn>m1-5 m - = 0 = kn -80 kn # m kn + 15 kn1-5 m = kn # m (4) These results can be checked in part by noting that by applying w = -d>d and = d>d. lso, when Eqs. 1 and give and = 80 kn # 1 = 0, = 5.75 kn m; when = 10 m, Eqs. 3 and 4 give = kn and = 0. These values check with the support reactions shown on the free-body diagram, Fig. 6 9d. Shear and oment Diagrams. Equations 1 through 4 are plotted in Fig. 6 9d. (1) () (3) + 5 kn>m1-5 ma - 5 m b + = 0 (kn) (kn m) B 5 m 5 m 5.75 kn 34.5 kn kn m kn (d) Fig kn/m C 34.5 (m) (m)

7 E X P L E 6.7 Draw the shear and moment diagrams for the beam in Fig. 6 13a. P L Support Reactions. Fig. 6 13b. Fig. 6 13a The reactions are shown on a free-body diagram, P P PL Shear Diagram. ccording Fig. to 6 13b the sign convention, Fig. 6 3, at = 0, = +P and at = L, = +P. These points are plotted in Fig. 6 13b. Since w = 0, Fig. 6 13a, the slope of the shear diagram will be zero 1d>d = -w = 0 at all points, and therefore a horizontal straight line connects the end points. P oment Diagram. t = 0, Fig. 6 13c = -PL and at = L, = 0, Fig. 6 13d. The shear diagram indicates that the shear is constant positive and therefore the slope of the moment diagram will be constant positive, d>d = = +P at all points. Hence, the end points are connected by a straight positive sloped line as shown in Fig. 6 13d. PL (d) Fig. 6 13

8 E X P L E 6.8 Draw the shear and moment diagrams for the beam shown in Fig. 6 14a. L Fig. 6 14a Support Reactions. The reaction at the fied support is shown on the free-body diagram, Fig. 6 14b. 0 0 Fig. 6 14b 0 Shear Diagram. The shear = 0 at each end is plotted first, Fig. 6 14c. Since no distributed load eists on the beam the shear diagram will have zero slope, at all points. Therefore, a horizontal line connects the end points, which indicates that the shear is zero throughout the beam. L Fig. 6 14c 0 oment Diagram. The moment at the beam s end points is plotted first, Fig. 6 14d. From the shear diagram the slope of the moment diagram will be zero since = 0. Therefore, a horizontal line connects the end points as shown. 0 (d) Fig. 6 14

9 E X P L E 6.9 Draw the shear and moment diagrams for the beam shown in Fig. 6 15a. w 0 L Fig. 6 15a Support Reactions. The reactions at the fied support are shown on the free-body diagram, Fig. 6 15b. w 0 L w 0 w 0 L Shear Diagram. The shear at Fig. each 6 15b end point, is plotted first, Fig. 6 15c. The distributed loading on the beam is constant positive, and so the slope of the shear diagram will be constant negative 1d>d = -w 0. This requires a straight negative sloped line that connects the end points. w 0 L Constant negative slope = w 0 oment Diagram. The moment Fig. 6 15c at each end point is plotted first, Fig. 6 15d. The shear diagram indicates that is positive and decreases from w 0 L to zero, and so the moment diagram must start with a positive slope of w 0 L and decrease to zero. Specifically, since the shear diagram is a straight sloping line, the moment diagram will be parabolic, having a decreasing slope as shown in the figure. w 0 L Decreasingly positive slope (d) Fig. 6 15

10 E X P L E 6.10 w 0 L Fig Draw the shear and moment diagrams for the beam shown in Fig. 6 16a. Support Reactions. The reactions at the fied support have been calculated and are shown on the free-body diagram, Fig. 6 16b. w 0 L w 0 w 0 L 6 Fig. 6 16b Shear Diagram. The shear at each end point is plotted first, Fig. 6 16c. The distributed loading on the beam is positive yet decreasing. Therefore, the slope of the shear diagram will be negatively decreasing. t = 0, the slope begins at -w 0 and goes to zero at = L. Since the loading is linear, the shear diagram is a parabola having a negatively decreasing slope. w 0 L Negatively decreasing slope 0 Fig. 6 16c oment Diagram. The moment at each end is plotted first, Fig. 6 16d. From the shear diagram, is positive but decreases from w 0 L> at = 0 to zero at = L. The curve of the moment diagram having this slope behavior is a cubic function of, as shown in the figure. w 0 L 6 Positively decreasing slope (d) Fig. 6 16d

11 E X P L E 6.11 Draw the shear and moment diagrams for the beam in Fig. 6 17a. kn/m kn/m 1.5 kn 4.5 m 3 kn 4.5 m Fig (kn) Support Reactions. The reactions have been determined and are shown on the free-body diagram, Fig. 6 17b. Shear Diagram. The end points = 0, 1.5, and 4.5, 3, are plotted first, Fig. 6 17c. From the behavior of the distributed load, the slope of the shear diagram will vary from zero at = 0 to - at 4.5. s a result, the shear diagram is a parabola having the shape shown. The point of zero shear can be found by using the method of sections for a beam segment of length,fig.6 17e. We require that = 0, so that +q F 15 kip - 1 y = 0; 1.5 kn c kip>fta kn/m bd = 0; = mft (kn m) ftm oment Diagram. The end points = 0, = 0 and = 45, = 0 are plotted first, Fig. 6 17d. From the behavior of the shear diagram, the slope of the moment diagram will begin at 1.5, then it becomes decreasingly positive until it reaches zero at.6 m. It then becomes increasingly negative reaching 3 at 4.5 m. Here the moment diagram is a cubic function of. Why? Notice that the maimum moment is at.6, since d>d = = 0 at this point. From the free-body diagram in Fig. 6 17e we have d+ = 0; -15 kip16.0 ft mft.6 mft c kip>fta bd16.0 (.6 m) fta6.0 b + = ft m ( 3 ) 1.5 kn(.6 m) kn/m ( ) ( ).6 kn m Slope = 0 Slope = kn Fig. 6 17c Fig. 6 17b Increasingly negative slope.6 m Decreasingly Fig. 6 17d positive slope Slope = 0 (d).6.6 m Fig. 6 17e (e) Slope = Slope = (m) Increasingly negative slope (m)

12 E X P L E 6.1 Draw the shear and moment diagrams for the beam shown in Fig. 6 18a. 8 kn 8 kn 6 m m m B C D Fig. 6 18b 8 kn 8 kn 6 m m m B C D 4.8 kn 11. kn (kn) (kn m) (d) 8 kn 6 m 8.8 kn m (e) 3. kn 4.8 kn Fig (m) (m) Fig. 6 18a Support Reactions. diagram, Fig. 6 18b. The reactions are indicated on the free-body Shear Diagram. t = 0, = +4.8 kn, and at = 10, D = -11. kn, Fig. 6 18c. t intermediate points between each force, the slope of the shear diagram will be zero.why? Hence, the shear retains its value of +4.8 up to point B.T B, the shear is discontinuous, since there is a concentrated force of 8 kn there.the value of the shear just to the right of B can be found by sectioning the beam at this point, Fig. 6 18e, where for equilibrium = -3. kn. Use the method of sections and show that the diagram jumps again at C, as shown, then closes to the value of -11. kn at D. It should be noted that based on Eq. 6 5, = -F, the shear diagram can also be constructed by following the load on the freebody diagram. Beginning at the 4.8-kN force acts upward, so = +4.8 kn. No distributed load acts between and B, so the shear remains constant 1d>d = 0. t B the 8-kN force is down, so the shear jumps down 8 kn, from +4.8 kn to -3. kn. gain, the shear is constant from B to C (no distributed load), then at C it jumps down another 8 kn to -11. kn. Finally, with no distributed load between C and D, it ends at -11. kn. oment Diagram. The moment at each end of the beam is zero, Fig. 6 18d. The slope of the moment diagram from to B is constant at Why? The value of the moment at B can be determined by using statics, Fig. 6 18c, or by finding the area under the shear diagram between and B, that is, B = 14.8 kn16 m = 8.8 kn # m. Since = 0, then B = + B = kn # m = 8.8 kn # m. From point B, the slope of the moment diagram is -3. until point C is reached. gain, the value of the moment can be obtained by statics or by finding the area under the shear diagram from B to C, that is, BC = 1-3. kn1 m = -6.4 kn # m, so that kn # m =.4 kn # C = 8.8 kn # m m. Continuing in this manner, verify that closure occurs at D.

13 E X P L E 6.13 Draw the shear and moment diagrams for the overhanging beam shown in Fig. 6 19a. 8 kn kn/m 4 m B C 6 m 4 m Fig. 6 19a Support Reactions. The free-body diagram with the calculated support reactions is shown in Fig. 6 19b. Shear Diagram. s usual we start by plotting the end shears 4.40 kn, and D = 0, Fig. 6 19c. The shear diagram will have zero slope from to B. It then jumps down 8 kn to 3.60 kn. It then has a slope that is increasingly negative.the shear at C can be determined from the area under the load diagram, C B BC 3.60 kn (1/)(6 m)( kn/m) 9.60 kn. It then jumps up 17.6 kn to 8 kn. Finally, from C to D, the slope of the shear diagram will be constant yet negative, until the shear reaches zero at D. oment Diagram. The end moments = 0 and D = 0 are plotted first, Fig. 6 19d. Study the diagram and note how the slopes and therefore the various curves are established from the shear diagram using d>d =. erify the numerical values for the peaks using the method of sections and statics or by computing the appropriate areas under the shear diagram to find the change in moment between two points. In particular, the point of zero moment can be determined by establishing as a function of, where, for convenience, etends from point B into region BC, Fig. 6 19e. Hence, d+ = 0; D ( ) kip14 ft + + kip1 + 1 a kip>ft kn/m 4.40 kn(4 m ) 8 kn() b1a 6 mft 3 b + = 0 = a b kn kip # ft m = m Reviewing these diagrams, we see that because of the integration process for region B the load is zero, shear is constant, and moment is linear; for region BC the load is linear, shear is parabolic, and moment is cubic; and for region CD the load is constant, the shear is linear, and the moment is parabolic. It is recommended that Eamples 6.1 through 6.6 also be solved using this method. B C D 4 m 6 m 4 m 4.40 kn 17.6 kn 4.40 (d) (kn) (kn m) kn Slope = 4.40 (e) Slope = 3.60 Slope = Slope = 8 Fig. 6 19b 4.40 kn 8 kn B 4 m Fig () w = 6 kn/m (m) (m)

14 E X P L E 6.14 beam has a rectangular cross section and is subjected to the stress distribution shown in Fig. 6 7a. Determine the internal moment at the section caused by the stress distribution using the fleure formula, by finding the resultant of the stress distribution using basic principles. 60 mm 0 Pa N 600 mm 60 mm 0 Pa Fig. 6 7 Part. The fleure formula is s ma = c>i. From Fig. 6 7a, c 60 mm and ma 0 Pa. The neutral ais is defined as line N, because the stress is zero along this line. Since the cross section has a rectangular shape, the moment of inertia for the area about N is determined from the formula for a rectangle given on the inside front cover; i.e., Therefore, I = 1 1 bh3 = (60 in.11 mm)(10 in.3 mm) = in864( ) mm 4 s ma = c I ; 0 N/mm (60 mm) 864(10 4 ) mm 4 88(10 4 ) N mm.88 kn m ns. Continued

15 0 Pa z d F N F F y 60 mm d dy y 60 mm 60 mm 0 Pa Fig. 6 7b 60 mm 60 mm 40 mm 40 mm 60 mm Fig. 6 7 Part. First we will show that the resultant force of the stress distribution is zero. s shown in Fig. 6 7b, the stress acting on the arbitrary element strip d (60 mm) dy, located y from the neutral ais, is The force created by this stress is df = s d, and thus, for the entire cross section, 60 6 in. mm F R = s d = ca -y 6 in. b1 y [ kip>in (0 d16 N/mmin. ) dy in. mm 60 mm ] = ( N/mm kip>in ) y ` ( ) s = a -y y 6 in. b1 (0 kip>in N/mm ) 60 mm ( ) (60 mm) dy in. mm in. mm = 0 0 The resultant moment of the stress distribution about the neutral ais (z ais) must equal. Since the magnitude of the moment of df about this ais is d = y df, and d is always positive, Fig. 6 7b, then for the entire area, 60 6 in. mm = y df = yca y 6 in. b1 y y [( kip>in d16 in. dy in. mm 60 mm ) (0 N/mm ) ] (60 mm) dy = a in. mm 0 ( kip>in N/mm by ) 3 ` in. mm 88(10 4 ) N mm.88 kn m ns. The above result can also be determined without the need for integration. The resultant force for each of the two triangular stress distributions in Fig. 6 7c is graphically equivalent to the volume contained within each stress distribution. Thus, each volume is 1 F (60 mm)(0 N/mm )(60 mm) 36(10 3 ) N 36 kn These forces, which form a couple, act in the same direction as the stresses within each distribution, Fig. 6 7c. Furthermore, they act through 1 the centroid of each volume, i.e., 3 (60 mm) 0 mm from the top and bottom of the beam. Hence the distance between them is 80 mm as shown. The moment of the couple is therefore 36 kn (80 mm) 880 kn mm.88 kn m ns.

16 E X P L E 6.15 The simply supported beam in Fig. 6 8a has the cross-sectional area shown in Fig. 6 8b. Determine the absolute maimum bending stress in the beam and draw the stress distribution over the cross section at this location. Fig. 6 8a a 5 kn/m 3 m 6 m (kn m) (m) Fig. 6 8 aimum Internal oment. The maimum internal moment in the beam, =.5 kn # m, occurs at the center as shown on the bending moment diagram, Fig. 6 8c. See Eample 6.3. Section Property. By reasons of symmetry, the centroid C and thus the neutral ais pass through the midheight of the beam, Fig. 6 8b. The area is subdivided into the three parts shown, and the moment of inertia of each part is computed about the neutral ais using the parallel-ais theorem. (See Eq. 5 of ppendi.) Choosing to work in meters, we have I = 1I + d = c m10.00 m m10.00 m m d 0 mm 0 mm N B C 0 mm 50 mm 150 mm 150 mm D + c m m3 d = m 4 Fig. 6 8c Continued

17 1.7 Pa B 11. Pa =.5 kn m B 1.7 Pa 11.Pa D =.5 kn m D 1.7 Pa 1.7 Pa (d) 11. Pa 1.7 Pa B 11. Pa D 1.7 Pa (e) Fig. 6 8d Bending Stress. pplying the fleure formula, with c = 170 mm, the absolute maimum bending stress is s ma = c I ; ns. Two-and-three-dimensional views of the stress distribution are shown in Fig. 6 8d. Notice how the stress at each point on the cross section develops a force that contributes a moment d about the neutral ais such that it has the same direction as. Specifically, at point B, y B = 150 mm, and so s B = y B I ; s ma =.5 kn # m m m 4 = 1.7 Pa s B =.5 kn # m m m 4 = 11. Pa The normal stress acting on elements of material located at points B and D is shown in Fig. 6 8e.

18 E X P L E 6.16 The beam shown in Fig. 6 9a has a cross-sectional area in the shape of a channel, Fig. 6 9b. Determine the maimum bending stress that occurs in the beam at section a a. Internal oment. Here the beam s support reactions do not have to be determined. Instead, by the method of sections, the segment to the left of section a a can be used, Fig. 6 9c. In particular, note that the resultant internal aial force N passes through the centroid of the cross section. lso, realize that the resultant internal moment must be computed about the beam s neutral ais at section a a. To find the location of the neutral ais, the cross-sectional area is subdivided into three composite parts as shown in Fig. 6 9b. Since the neutral ais passes through the centroid, then using Eq. of ppendi, we have y = y [0.100 m]10.00 m m + [0.010 m]10.0 m10.50 m = m m m10.50 m.6 kn = m = mm This dimension is shown in Fig. 6 9c. pplying the moment equation of equilibrium about the neutral ais, we have y =59.09 mm d+ N = 0; 4 kn1 m kn m - = 0 N = kn # m Section Property. The moment of inertia about the neutral ais is determined using the parallel-ais theorem applied to each of the three composite parts of the cross-sectional area. Working in meters, we have 15 mm a m a 1 m Fig. 6 9a Fig b mm 0 mm C 00 mm 15 mm I = c m10.00 m m10.00 m m m d + c m10.00 m m10.00 m m m d = m 4 aimum Bending Stress. The maimum bending stress occurs at points farthest away from the neutral ais. This is at the bottom of the beam, c = 0.00 m m = m. Thus, s ma = c ns. I = kn # m m m 4 = 16. Pa Show that at the top of the beam the bending stress is s =6.79 Pa. Note that in addition to this effect of bending, the normal force of N = 1 kn and shear force =.4 kn will also contribute additional stress on the cross section. The superposition of all these effects will be discussed in a later chapter..4 kn 1.0 kn m m Fig. 6 9 C N

19 E X P L E mm 40 N m 30 mm N y 10 mm 60 mm Fig. 6 30a 40 N m 10 mm Fig y 5 mm The member having a rectangular cross section, Fig. 6 30a, is designed to resist a moment of 40 N # m. In order to increase its strength and rigidity, it is proposed that two small ribs be added at its bottom, Fig. 6 30b. Determine the maimum normal stress in the member for both cases. Without Ribs. Clearly the neutral ais is at the center of the cross section, Fig. 6 30a, so y = c = 15 mm = m. Thus, I = 1 1 bh3 = m10.03 m3 = m 4 Therefore the maimum normal stress is ns. With Ribs. From Fig. 6 30b, segmenting the area into the large main rectangle and the bottom two rectangles (ribs), the location y of the centroid and the neutral ais is determined as follows: y = y [0.015 m] m m + [0.035 m] m m = m m m m = m s ma = c I This value does not represent c. Instead = 140 N # m m m 4 = 4.44 Pa c = m m = m Using the parallel-ais theorem, the moment of inertia about the neutral ais is I = c m m m m m m d + c m m m m m m d = m 4 Therefore, the maimum normal stress is s ma = c I = 40 N # m m m 4 = 4.65 Pa ns. This surprising result indicates that the addition of the ribs to the cross section will increase the normal stress rather than decrease it, and for this reason they should be omitted.

20 E X P L E 6.18 The rectangular cross section shown in Fig. 6 35a is subjected to a bending moment of = 1 kn # m. Determine the normal stress developed at each corner of the section, and specify the orientation of the neutral ais. Internal oment Components. By inspection, it is seen that the y and z aes represent the principal aes of inertia since they are aes of symmetry for the cross section. s required, we have established the z ais as the principal ais for maimum moment of inertia. The moment is resolved into its y and z components, where y = kn # m = KN # m z = kn # m = 7.0 kn # m 0. m B 0.1 m y 0.1 m 0. m C E D z = 1 kn m Section Properties. Bending Stress. s = - z y I z Thus, + y z I y The moments of inertia about the y and z aes are I y = m10. m3 = m 4 I z = m10.4 m3 = m 4 Fig s B = N # m10. m m N # m1-0.1 m m 4 =.5 Pa s C = N # m10. m m N # m10.1 m m 4 = Pa s D = N # m1-0. m m N # m10.1 m m 4 = -.5 Pa s E = N # m1-0. m m N # m1-0.1 m m 4 = 4.95 Pa ns. ns. ns. ns. The resultant normal-stress distribution has been sketched using these values, Fig. 6 35b. Since superposition applies, the distribution is linear as shown. Continued

21 = 1 kn m.5 Pa B N z 0. m C 4.95 Pa E 4.95 Pa.5 Pa D E B D θ = 53.1 α = 79.4 α C z N y Fig. 6 35b Fig. 6 35c Orientation of Neutral is. The location z of the neutral ais (N), Fig. 6 35b, can be established by proportion. long the edge BC, we require.5 Pa z z = 4.95z In the same manner, this is also the distance from D to the neutral ais in Fig. 6 35b. We can also establish the orientation of the N using Eq. 6 19, which is used to specify the angle a that the ais makes with the z or maimum principal ais.ccording to our sign convention, u must be measured from the +z ais toward the +y ais. By comparison, in Fig. 6 35c, tan (or ).Thus, tan a = I z I y tan u = 4.95 Pa 10. m - z z = m tan a = m m 4 tan a = ns. This result is shown in Fig. 6 35c. Using the value of z calculated above, verify, using the geometry of the cross section, that one obtains the same answer.

22 E X P L E 6.19 T-beam is subjected to the bending moment of 15 kn # m as shown in Fig. 6 36a. Determine the maimum normal stress in the beam and the orientation of the neutral ais. z Fig. 6 36a 30 mm 100 mm = 15 kn m mm 40 mm 80 mm y Internal oment Components. The y and z aes are principal aes of inertia. Why? From Fig. 6 36a, both moment components are positive. We have Section Properties. meters, we have z = z = m With reference to Fig. 6 36b, working in units of [0.05 m] m10.04 m + [0.115 m]10.03 m10.00 m = m10.04 m m10.00 m Using the parallel-ais theorem of ppendi, I = I + d, principal moments of inertia are thus I z = m10.04 m m10.00 m3 = m 4 = m 4 y = 115 kn # m cos 30 = 1.99 kn # m z = 115 kn # m sin 30 = 7.50 kn # m 0.03 m I y = c m m m10.04 m m m d + c m10.03 m m10.03 m m m d the m 0.0 m m m z Fig m z y Continued

23 Fig. z 6 36c z Fig. 6 36d m 7.50 kn m m B 1.99 kn m y α = 68.6 θ = m y C N 0.0 m aimum Bending Stress. The moment components are shown in Fig. 6 36c. By inspection, the largest tensile stress occurs at point B, since by superposition both moment components create a tensile stress there. Likewise, the greatest compressive stress occurs at point C. Thus, s = - z y I z s B = kn # m m m kn # m m m 4 = 74.8 Pa + y z I y s C = kn # m m m kn # m m m 4 = Pa ns. By comparison, the largest normal stress is therefore compressive and occurs at point C. Orientation of Neutral is. When applying Eq. 6 19, it is important to be sure the angles a and u are defined correctly. s previously stated, y must represent the ais for minimum principal moment of inertia, and z must represent the ais for maimum principal moment of inertia. These aes are properly positioned here since I y 6 I z. Using this setup, u and a are measured positive from the +z ais toward the +y ais. Hence, from Fig. 6 36a, u = +60. Thus, tan a = a m b tan 60 4 m a = 68.6 ns. The neutral ais is shown in Fig. 6 36d. s epected, it lies between the y ais and the line of action of. (d)

24 E X P L E 6.0 The Z-section shown in Fig. 6 37a is subjected to the bending moment of = 0 kn # m. Using the methods of ppendi (see Eample.4 or.5), the principal aes y and z are oriented as shown, such that they represent the minimum and maimum principal moments of inertia, I and I z = m 4 y = m 4, respectively. Determine the normal stress at point P and the orientation of the neutral ais. For use of Eq. 6 19, it is important that the z ais be the principal ais for the maimum moment of inertia, which it is because most of the area is located furthest from this ais. Internal oment Components. From Fig. 6 37a, Bending Stress. The y and z coordinates of point P must be determined first. Note that the y, z coordinates of P are 1-0. m, 0.35 m. Using the colored and shaded triangles from the construction shown in Fig. 6 37b, we have pplying Eq. 6 17, we have s P = - z y P I z y P = sin cos 3.9 = m z P = 0.35 cos sin 3.9 = m = kn # m m m kn # m m m 4 = 3.76 Pa y = 0 kn # m sin 57.1 = kn # m z = 0 kn # m cos 57.1 = kn # m + y z P I y ns. Orientation of Neutral is. The angle u = 57.1 is shown in Fig. 6 37a. Thus, N 400 mm 100 mm 3.9 P P 100 mm 300 mm 0.00 m z Fig. 6 37az 3.9 z Fig z θ = m 3.9 y α = 85.3 z = 0 kn m y y y y tan a = c m d tan m a = 85.3 The neutral ais is located as shown in Fig. 6 37b. ns.

25 E X P L E 6.1 composite beam is made of wood and reinforced with a steel strap located on its bottom side. It has the cross-sectional area shown in Fig. 6 40a. If the beam is subjected to a bending moment of = kn # m, determine the normal stress at points B and C. Take E w = 1 GPa GPa and E st = 00 GPa. GPa. 9 mm 150 mm B = kn m B _ y 150 mm 0 mm C Fig N 150 mm 0 mm Section Properties lthough the choice is arbitrary, here we will transform the section into one made entirely of steel. Since steel has a greater stiffness than wood 1E st 7 E w, the width of the wood must be reduced to an equivalent width for steel. Hence n must be less than one. For this to be the case, n = E w >E st, so that The transformed section is shown in Fig. 6 40b. The location of the centroid (neutral ais), computed from a reference ais located at the bottom of the section, is y = y b st = nb w = 1 GPa 1150 mm = 9 mm 00 GPa = [0.01 m]10.0 m m + [0.095 m] m m 0.0 m m m m The moment of inertia about the neutral ais is therefore C 150 mm Fig. 6 40b I N = c m10.0 m m10.0 m m m d = m + c m m m m m m d = m 4 Continued

26 Fig. 6 40c B 8.6 Pa Fig. 6 40d B 1.71 Pa 3.50 Pa 0.10 Pa 3.50 Pa C 7.78 Pa = kn m C 7.78 Pa (d) = kn m Normal Stress. and C is pplying the fleure formula, the normal stress at B s B = kn # m m m m 4 = 8.6 Pa s C = kn # m m m 4 = 7.78 Pa ns. The normal-stress distribution on the transformed (all steel) section is shown in Fig. 6 40c. The normal stress in the wood, located at B in Fig. 6 40a, is determined from Eq. 6 1; that is, s B = ns B = 1 GPa Pa = 1.71 Pa 00 GPa ns. Using these concepts, show that the normal stress in the steel and the wood at the point where they are in contact is s st = 3.50 Pa and s w = 0.10 Pa, respectively. The normal-stress distribution in the actual beam is shown in Fig. 6 40d.

27 E X P L E 6. In order to reinforce the steel beam, an oak board is placed between its flanges as shown in Fig. 6 41a. If the allowable normal stress for the steel is ( allow ) st 168 Pa, and for the wood ( allow ) w 1 Pa, determine the maimum bending moment the beam can support, with and without the wood reinforcement. E st 00 GPa, E w 1 GPa. The moment of inertia of the steel beam is I z mm 4, and its cross-sectional area is Fig. 6 41a mm. 18 mm 100 mm z 100 mm 5 mm N c z 5 mm y 30 mm 105 mm Fig Without Board. Here the neutral ais coincides with the z ais. Direct application of the fleure formula to the steel beam yields 1s allow st = c I z 4 kip>in (105 mm) in. 168 N/mm = 7.93( in 6 ) 4 mm kn m ns. With Board. Since now we have a composite beam, we must transform the section to a single material. It will be easier to transform the wood to an equivalent amount of steel. To do this, n = E w >E st. Why? Thus, the width of an equivalent amount of steel is b st = nb w = (10 3 ) Pa ksi (300 in. mm) = in. mm 00(10 3 ) ksi Pa Continued

28 The transformed section is shown in Fig. 6 41b. The neutral ais is at y = y' = [0]18.79 [0]( in mm + [.0 ) [55 in.]14 mm](100 in mm)(18 in. mm) in mm (18) in mm mm nd the moment of inertia about the neutral ais is I = [0.3 [7.93(10 in 46 ) + mm in mmin. (13.57 ] + mm) ] [ ] 1 (18 mm)(100 mm) 3 (18 mm)(100 mm)(55 mm mm) (10 6 ) mm 4 The maimum normal stress in the steel will occur at the bottom of the beam, Fig. 6 41b. Here c 105 mm mm mm. The maimum moment based on the allowable stress for the steel is therefore 1s allow st = c I 4 kip>in ( in. mm) 168 N/mm = in mm kn m The maimum normal stress in the wood occurs at the top of the beam, Fig. 6 41b. Here c 105 mm mm mm. Since s w = ns st, the maimum moment based on the allowable stress for the wood is c 1s allow w = n I 3 kip>in = c (10 3 ) Pa ksi (91.43 in. mm) 9110 d 00(10 3 ) Pa ksi in mm 4 1 N/mm [ ] kn m By comparison, the maimum moment is limited by the allowable stress in the steel. Thus, kn m ns. Note also that by using the board as reinforcement, one provides an additional 51% moment capacity for the beam.

29 Fig. 6 43a E X P L E mm The reinforced concrete beam has the cross-sectional area shown in Fig. 6 43a.If it is subjected to a bending moment of 60 kn m, determine the normal stress in each of the steel reinforcing rods and the maimum normal stress in the concrete. Take E st 00 GPa and E conc 5 GPa. 60 kn m 450 mm Since the beam is made from concrete, in the following analysis we will neglect its strength in supporting a tensile stress. 5-mm-diameter bars 50 mm Section Properties. The total area of steel, st [ (1.5 mm) ] 98 mm will be transformed into an equivalent area of concrete, Fig. 6 43b. Here 00(10 n st ) Pa (98 mm ) 7856 mm 5(10 3 ) Pa 300 mm We require the centroid to lie on the neutral ais. Thus y ' = 0, or h h 400 mm 3001 mm in.1h mm in 116in. (400 mm - h = h ) 0 0 N C h 5.37h Solving for the positive root, = 7856 mm h mm Using this value for h, the moment of inertia of the transformed section, Fig. 6 43b computed about the neutral ais, is I = c in in. (300 mm)(10.90 in.3 mm) mm 1 300in mm(10.90 in.a mm) ( b d mm in 116 (400in. mm in. mm) ) mm 4 Normal Stress. pplying the fleure formula to the transformed section, the maimum normal stress in the concrete is 60 kn m (1000 mm/m)(10.90 mm)(1000 N/kN) ( conc ) ma 9.0 Pa ns mm Pa The normal stress resisted by the concrete strip, which replaced the steel, is 60 kn m (1000 mm/m)(1000 N/kN)(400 mm 10.9 mm) mm conc 1.3 Pa mm 4 The normal stress in each of the two reinforcing rods is therefore Pa Pa 00(10 3 ) Pa st n conc ( 1.3 Pa Pa ns. 5(10 3 ) Pa ) Fig The normal-stress distribution is shown graphically in Fig. 6 43c.

30 E X P L E 6.4 steel bar having a rectangular cross section is shaped into a circular arc as shown in Fig. 6 45a. If the allowable normal stress is allow 140 Pa, determine the maimum bending moment that can be applied to the bar. What would this moment be if the bar was straight? 0 mm dr 0 mm r o = 110 mm r i = 90 mm r O Fig Internal oment. Since tends to increase the bar s radius of curvature, it is positive. Section Properties. The location of the neutral ais is determined using Eq From Fig. 6 45a, we have d r 110 in. mm 1 in. (0 dr mm) dr 11 in. = = 1 in. (0 ln rmm) ` ln = r in mm r r 90 in. mm 9 in. 110 mm 90 mm This same result can of course be obtained directly from Table 6. Thus, R = (0 1 in.1 mm)(0 in.mm) = = in. mm d in. mm r Continued

31 Fig. 6 45b 1.5 Pa 140 Pa It should be noted that throughout the above calculations, R must be determined to several significant figures to ensure that 1r - R is accurate to at least three significant figures. It is unknown if the normal stress reaches its maimum at the top or at the bottom of the bar, and so we must compute the moment for each case separately. Since the normal stress at the bar s top is compressive, 140 Pa, s = 1R - r o r o 1r - R -0 kip>in in. - in. 140 N/mm ( mm 110 mm) = (0 mm)(0 1 in.1 mm)(110 in.111 in.110 mm)(100 in. -mm in. mm) N mm kn m Likewise, at the bottom of the bar the normal stress will be tensile, so 140 Pa. Therefore, s = 1R - r i r i 1r - R 0 kip>in in. - 9 in. 140 N/mm ( mm 90 mm) = (0 mm)(0 1 in.1 mm)(90 in.19 in.110 mm)(100 in. mm in. mm) N mm kn m ns. By comparison, the maimum moment that can be applied is kn m and so maimum normal stress occurs at the bottom of the bar. The compressive stress at the top of the bar is then N kip mm( # in in. mm in. mm) s = (0 mm)(0 1 in.1 mm)(110 in.111 in.110 mm)(100 in. mm in. mm) 1.5 N/mm The stress distribution is shown in Fig. 6 45b. If the bar was straight, then s = c I 0 kip>in 11 in. 140 N/mm (10 mm) = 1 1 (0 1 in.1 mm)(0 in.3 mm) N mm kn m ns. This represents an error of about 7% from the more eact value determined above.

32 E X P L E 6.5 The curved bar has a cross-sectional area shown in Fig. 6 46a. If it is subjected to bending moments of 4 kn # m, determine the maimum normal stress developed in the bar. 4 kn m 00 mm O 4 kn m 50 mm 80 mm B 00 mm r 50 mm 50 mm 30 mm Fig Internal oment. Each section of the bar is subjected to the same resultant internal moment of 4 kn # m. Since this moment tends to decrease the bar s radius of curvature, it is negative. Thus, = -4 kn # m. Section Properties. Here we will consider the cross section to be composed of a rectangle and triangle. The total cross-sectional area is The location of the centroid is determined with reference to the center of curvature, point O, Fig. 6 46a. r = r' = m m10.03 m = m = [0.5 m]10.05 m10.05 m + [0.60 m] m m m = m Continued

33 We can compute d>r for each part using Table 6. For the rectangle, d r 0.50 m = 0.05 maln 0.00 m b = m nd for the triangle, d r = m10.80 m m m aln 0.80 m b m = m 0.50 m Thus the location of the neutral ais is determined from R = d>r = m m m = m Note that R 6 r as epected. lso, the calculations were performed with sufficient accuracy so that ( r R) m m m is now accurate to three significant figures. Normal Stress. The maimum normal stress occurs either at or B. pplying the curved-beam formula to calculate the normal stress at B, r B = 0.00 m, we have 4 kn m s B = 1R - r B r B 1r - R = = -116 Pa 1-4 kn # m m m m m m t point, r = 0.80 m and the normal stress is 116 Pa B 19 Pa s = 1R - r r 1r - R = 1-4 kn # m m m m m m = 19 Pa ns. By comparison, the maimum normal stress is at. two dimensional representation of the stress distribution is shown in Fig. 6 46b. Fig. 6 46b

34 E X P L E 6.6 The transition in the cross-sectional area of the steel bar is achieved using shoulder fillets as shown in Fig. 6 51a. If the bar is subjected to a bending moment of 5 kn # m, determine the maimum normal stress developed in the steel. The yield stress is s Y = 500 Pa. 5 kn m 5 kn m Fig. 6 51b 10 mm r = 16 mm 80 mm 5 kn m 340 Pa 5 kn m 0 mm 340 Pa Fig The moment creates the largest stress in the bar at the base of the fillet, where the cross-sectional area is smallest. The stressconcentration factor can be determined by using Fig From the geometry of the bar, we have r = 16 mm, h = 80 mm, w = 10 mm. Thus, r 16 mm = h 80 mm = 0. w 10 mm = h 80 mm = 1.5 These values give K = pplying Eq. 6 6, we have c s ma = K I = kn # m10.04 m [ m10.08 m3 ] 5 kn m = 340 Pa This result indicates that the steel remains elastic since the stress is below the yield stress (500 Pa). The normal-stress distribution is nonlinear and is shown in Fig. 6 51b. Realize, however, that by Saint-enant s principle, Sec. 4.1, these localized stresses smooth out and become linear when one moves (approimately) a distance of 80 mm or more to the right of the transition. In this case, the fleure formula gives s ma = 34 Pa, Fig. 6 51c.lso note that the choice of a larger-radius fillet will significantly reduce s ma, since as r increases in Fig. 6 48, K will decrease. Fig. 6 51c 34 Pa 34 Pa 5 kn m

35 E X P L E 6.7 The steel wide-flange beam has the dimensions shown in Fig. 6 56a. If it is made of an elastic perfectly plastic material having a tensile and compressive yield stress of Y 50 Pa, determine the shape factor for the beam. In order to determine the shape factor, it is first necessary to compute the maimum elastic moment Y and the plastic moment p. aimum Elastic oment. The normal-stress distribution for the maimum elastic moment is shown in Fig. 6 56b. The moment of inertia about the neutral ais is I = c (1.5 in.19 mm)(5 in.3 d mm) [ + ] 3 c 1 18 (00 in.10.5 mm)(1.5 in.3 + mm) 8 in in mm(1.5 in. mm)( d = 11.0 in mm) [ 4 1 ] mm 4 pplying the fleure formula, we have 36 kip>in = Y15 in. s ma = c I ; 50 N/mm Y (15 mm) in mm 4 Y kn m Plastic oment. The plastic moment causes the steel over the entire cross section of the beam to yield, so that the normal-stress distribution looks like that shown in Fig. 6 56c. Due to symmetry of the cross-sectional area and since the tension and compression stress strain diagrams are the same, the neutral ais passes through the centroid of the cross section. In order to determine the plastic moment, the stress distribution is divided into four composite rectangular blocks, and the force produced by each block is equal to the volume of the block. Therefore, we have C 1 T 1 50 N/mm (1.5 mm)(11.5 mm) kn C T 50 N/mm (1.5 mm)(00 mm) 65 kn These forces act through the centroid of the volume for each block. Computing the moments of these forces about the neutral ais, we obtain the plastic moment: p [56.5 mm)( kn)] [( mm)(65 kn)] 188 kn m Shape Factor. pplying Eq gives k = p = kn kip # m in. ns. Y kip # = kn in. m This value indicates that a wide-flange beam provides a very efficient section for resisting an elastic moment.ost of the moment is developed in the flanges, i.e., in the top and bottom segments, whereas the web or vertical segment contributes very little. In this particular case, only 14% additional moment can be supported by the beam beyond that which can be supported elastically. Fig. 6 56a N N 50 Pa 1.5 mm 00 mm Fig. 6 56b 50 Pa Fig mm 1.5 mm 5 mm 50 Pa 50 Pa C C 1 T 1 T Y p

36 E X P L E mm T-beam has the dimensions shown in Fig. 6 57a. If it is made of an elastic perfectly plastic material having a tensile and compressive yield stress of s Y = 50 Pa, determine the plastic moment that can be resisted by the beam. Fig. 6 57b 100 mm 15 mm 10 mm p 50 Pa N C C 1 T d 15 mm (10 mm d ) 15 mm Fig mm The plastic stress distribution acting over the beam s cross-sectional area is shown in Fig. 6 57b. In this case the cross section is not symmetric with respect to a horizontal ais, and consequently, the neutral ais will not pass through the centroid of the cross section. To determine the location of the neutral ais, d, we require the stress distribution to produce a zero resultant force on the cross section. ssuming that d 10 mm, we have s d = 0; T - C 1 - C = 0 50 Pa m1d - 50 Pa m10.10 m - d - 50 Pa m m = 0 d = m m OK Using this result, the forces acting on each segment are T = 50 N>m m m = 41.5 kn C 1 = 50 N>m m m = 37.5 kn C = 50 N>m m m = 375 kn Hence, the resultant plastic moment about the neutral ais is p = 41.5 kna m p = 9.4 kn # m b kna 0.01 m b kna0.01 m m b ns.

37 E X P L E 6.9 The beam in Fig. 6 58a is made of an alloy of titanium that has a stress strain diagram that can in part be approimated by two straight lines. If the material behavior is the same in both tension and compression, determine the bending moment that can be applied to the beam that will cause the material at the top and bottom of the beam to be subjected to a strain of mm/mm. σ (Pa) 3 cm σ = 105(10 ) σ = cm (mm/mm) Fig. 6 58a I By inspection of the stress strain diagram, the material is said to ehibit elastic-plastic behavior with strain hardening. Since the cross section is symmetric and the tension compression s P diagrams are the same, the neutral ais must pass through the centroid of the cross section. The strain distribution, which is always linear, is shown in Fig. 6 58b. In particular, the point where maimum elastic strain (0.010 mm/mm) occurs has been determined by proportion, such that 0.05/1.5 cm 0.010/y or y 0.3 cm 3 mm. The corresponding normal-stress distribution acting over the cross section is shown in Fig. 6 58c. The moment produced by this distribution can be calculated by finding the volume of the stress blocks. To do so, we will subdivide this distribution into two triangular blocks and a rectangular block in both the tension and compression regions, Fig. 6 58d. Since the beam is cm wide, the resultants and their locations are determined as follows: y = 0.3 cm cm Strain distribution Fig Continued

38 1330 Pa y= 0.3 cm 1050 Pa 1050 Pa 1.5 cm 1330 Pa Stress distribution T 1 = C 1 = (1 mm)(80 in.140 kip>in N/mm1 )(0 in. mm) = 48 kip N 33.6 kn y 1 = 0.3 cm in (1. in. cm) = 1.10 in. cm 11.0 mm 3 T = C = (1 11. mm)(1050 in.1150 kip>in N/mm 1 )(0 in. mm) = 360 kip 500 N 5 kn y = 0.3 cm in (1. in. cm) = 0.90 in. cm 9 mm T 3 = C 3 = (3 mm)(1050 in.1150 ksi1 N/mmin. )(0 = mm) 45 kip N 31.5 kn y 3 = 10.3 (0.3 in. cm) = 0. in. cm mm 3 y 1 y C 1 C y 3 T T 1 C 3 T 3 Fig. 6 58c 0.3 cm 0. cm The moment produced by this normal-stress distribution about the neutral ais is therefore [33.6 kn (110 mm) 5 kn (9 mm) 31.5 kn ( mm)] kn mm 5.40 kn m ns. II 1050 Pa 80 Pa (d) Fig. 6 58d Rather than using the above semigraphical technique, it is also possible to compute the moment analytically. To do this, we must epress the stress distribution in Fig. 6 58c as a function of position y along the beam. Note that s = f1p has been given in Fig. 6 58a. lso, from Fig. 6 58b, the normal strain can be determined as a function of position y by proportional triangles; i.e., cm P= y 0 y 1.5 in. cm 15 mm Substituting this into the s -Pfunctions shown in Fig. 6 58a gives N σ y dy s = 500y 350y s = 3.33y 33.33y y 0.3 cm in. 3 mm cm in. y 1.5 in. cm 15 mm 11 1 From Fig. 6 58e, the moment caused by s acting on the area strip d = dy is d = y1s d = ys1 (0 dy dy) (e) Fig Using Eqs. 1 and, the moment for the entire cross section is thus = c 0 350y 500y dy + 0 (3.33y 133.3y + dy 140y 980y) dy ddy ] [ (10 3 ) N mm 5.40 kn m ns.

39 E X P L E 6.30 The steel wide-flange beam shown in Fig. 6 60a is subjected to a fully plastic moment of p. If this moment is removed, determine the residual-stress distribution in the beam. The material is elastic perfectly plastic and has a yield stress of Y 50 Pa. The normal-stress distribution in the beam caused by p is shown in Fig. 6 60b. When p is removed, the material responds elastically. Removal of p requires applying p in its reverse direction and therefore leads to an assumed elastic stress distribution as shown in Fig. 6 60c. The modulus of rupture s r is computed from the fleure formula. Using P 188 kn m and I mm 4 from Eample 6.7, we have 1.5 mm 1.5 mm 5 mm 1.5 mm 00 mm s ma = c I ; ( N mm)(15 mm) allow mm 4 Fig. 6 60a 85.1 N/mm 85.1 Pa s epected, s r 6 s Y. Superposition of the stresses gives the residual-stress distribution shown in Fig. 6 60d. Note that the point of zero normal stress was determined by proportion; i.e., from Fig. 6 60b and 6 60c, we require that Pa ksi kis Pa = 15 5 mm in. y y y mm 50 Pa σ r = 85.1 Pa 35.1 Pa 15 mm 15 mm p 15 mm 15 mm y 50 Pa p mm mm 50 Pa Plastic moment applied (profile view) Fig. 6 60b Plastic moment reversed (profile view) Fig. 6 60c Fig σ r = 85.1 Pa Residual stress distribution (d) 35.1 Pa