Determine the resultant internal loadings acting on the cross section at C of the beam shown in Fig. 1 4a.

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1 E X M P L E 1.1 Determine the resultant internal loadings acting on the cross section at of the beam shown in Fig. 1 a. 70 N/m m 6 m Fig. 1 Support Reactions. This problem can be solved in the most direct manner by considering segment of the beam, since then the support reactions at do not have to be computed. Free-ody Diagram. Passing an imaginary section perpendicular to the longitudinal axis of the beam yields the free-body diagram of segment shown in Fig. 1 b. It is important to keep the distributed loading exactly where it is on the segment until after the section is made. Only then should this loading be replaced by a single resultant force. Notice that the intensity of the distributed loading at is found by proportion, i.e., from Fig.1 a, w>6 m = 170 N>m>9m, w = 180 N>m. The magnitude of the resultant of the distributed load is equal to the area under the loading curve (triangle) and acts throughthe centroid of this area.thus, F = N>m16 m = 0 N, which acts 1>16 m = m from as shown in Fig. 1 b. Equations of Equilibrium. pplying the equations of equilibrium we have : + F x = 0; -N = 0 N = 0 +q F y = 0; V - 0 N = 0 V = 0 N d+ M = 0; -M - 0 N1 m = 0 M = N # m The negative sign indicates that M acts in the opposite direction to that shown on the free-body diagram. Try solving this problem using segment, by first obtaining the support reactions at, which are given in Fig. 1 c. N M 180 N/m V 6 N m 90 N/m 11 N 0 N m m Fig. 1 b 0 N 1 N 1 m 1. m 0. m Fig. 1 c 180 N/m M V N

2 E X M P L E 1. Determine the resultant internal loadings acting on the cross section at of the machine shaft shown in Fig. 1 a. The shaft is supported by bearings at and, which exert only vertical forces on the shaft. 800 N/m N (800 N/m)(0.10 m) = 10 N N 18.7 N 00 mm 0 N 0.0 m 0.0 m Fig. 1 c 0 mm V M 100 mm 100 mm N 0 mm D Fig. 1 y 0.7 m Fig. 1 b 0.1 m y m We will solve this problem using segment of the shaft. Support Reactions. free-body diagram of the entire shaft is shown in Fig. 1 b. Since segment is to be considered, only the reaction at has to be determined. Why? d+ M = 0; - y m + 10 N10.1 m - N m = 0 y = N The negative sign for y indicates that y acts in the opposite sense to that shown on the free-body diagram. Free-ody Diagram. Passing an imaginary section perpendicular to the axis of the shaft through yields the free-body diagram of segment shown in Fig. 1 c. Equations of Equilibrium. : + F x = 0; N = 0 +q F y = 0; N - 0 N - V = 0 V = -8.8 N d+ M = 0; M + 0 N10.0 m N10.0 m = 0 M = -.69 N # m What do the negative signs for V and M indicate? s an exercise, calculate the reaction at and try to obtain the same results using segment D of the shaft.

3 E X M P L E 1. The hoist in Fig. 1 6a consists of the beam and attached pulleys, the cable, and the motor. Determine the resultant internal loadings acting on the cross section at if the motor is lifting the 000 N ( 00 kg) load W with constant velocity. Neglect the weight of the pulleys and beam. 1 mm 1 m 1. m 0. m 1 mm D W 1 m 1.1 m 000 N Fig. 1 6a M V N Fig N The most direct way to solve this problem is to section both the cable and the beam at and then consider the entire left segment. Free-ody Diagram. See Fig. 1 6b. Equations of Equilibrium. : + F x = 0; lb N + N = 0 N = lbn +q F y = 0; lb N - V = 0 V = lbn d+ M = 0; lb N( ftm) lb 10. N(0.1 ft + m) M 1 = M0 0 M = lb N # ft m s an exercise, try obtaining these same results by considering just the beam segment, i.e., remove the pulley at from the beam and show the 000-N force components of the pulley acting on the beam segment. lso, this problem can be worked by first finding the reactions at, ( x 0, y 000 N, M 7000 N m) and then considering segment.

4 E X M P L E 1. Determine the resultant internal loadings acting on the cross section at G of the wooden beam shown in Fig. 1 7a. ssume the joints at,,, D, and E are pin connected. F = 600 N 100 N G D E 1. m 100 N 1. m E x = 600 N E y = 00 N 600 N/m 1 m 1 m m m ( m) = m 1 ( m)(600 N/m) = 900 N F = 770 N F D = 60 N Fig. 1 7c 100 N 770 N N G G 1 m M G V G (d) Fig. Fig. 1 7d N Fig. 1 7b Support Reactions. Here we will consider segment G for the analysis. free-body diagram of the entire structure is shown in Fig. 1 7b. Verify the computed reactions at E and. In particular, note that is a two-force member since only two forces act on it. For this reason, the reaction at must be horizontal as shown. Since and D are also two-force members, the free-body diagram of joint is shown in Fig. 1 7c. gain, verify the magnitudes of the computed forces F and F D. Free-ody Diagram. Using the result for F, the left section G of the beam is shown in Fig. 1 7d. Equations of Equilibrium. pplying the equations of equilibrium to segment G, we have : + F lb x = 0; 770 N + N G = 0 N G = Nlb +q F -100 lb + lb y = 0; 100 N 770 N - V G = 0 V G = 10 Nlb d+ M M G lb G = 0; (770 N) 1 (1 ft m) + (100 lb N)(1 1 ft m) = 0 M G = lb N # ft m s an exercise, compute these same results using segment GE.

5 E X M P L E 1. Determine the resultant internal loadings acting on the cross section at of the pipe shown in Fig. 1 8a. The pipe has a mass of kg/m and is subjected to both a vertical force of 0 N and a couple moment of 70 N # m at its end. It is fixed to the wall at. The problem can be solved by considering segment, which does not involve the support reactions at. Free-ody Diagram. The x, y, z axes are established at and the free-body diagram of segment is shown in Fig. 1 8b. The resultant force and moment components at the section are assumed to act in the positive coordinate directions and to pass through the centroid of the cross-sectional area at. The weight of each segment of pipe is calculated as follows: W D = 1 kg>m10. m19.81 N>kg = 9.81 N W D = 1 kg>m11. m19.81 N>kg =. N These forces act through the center of gravity of each segment. Equations of Equilibrium. pplying the six scalar equations of equilibrium, we have* F x = 0; F y = 0; 1F x = 0 1F y = 0 F z = 0; 1F z N -. N - 0 N = 0 1F z = 8. N 1M x = 0; 1M x + 70 N # m - 0 N 10. m -. N 10. m N 10. m = 0 1M x = -0. N # m 1M y = 0; 1M y +. N 10.6 m + 0 N 11. m = 0 1M y = N # m 1M z = 0; 1M z = 0 What do the negative signs for 1M x and 1M y indicate? Note that the normal force N = 1F y = 0, whereas the shear force is V = lso, the torsional moment is T = 1M y = 77.8 N# = 8. N. m and the bending moment is M = = 0. N # m. *The magnitude of each moment about an axis is equal to the magnitude of each force times the perpendicular distance from the axis to the line of action of the force. The direction of each moment is determined using the right-hand rule, with positive moments (thumb) directed along the positive coordinate axes. x 70 N m 70 N m 0 N 0 N Fig. 1 8a z (F ) y (M ) z 9.81 N (F ) z (M ) y (M ) x 0. m 0. m (F ) x. N Fig m 0.7 m 1. m 0.6 m 0. m D y

6 E X M P L E 1.6 The bar in Fig. 1 16a has a constant width of mm and a thickness of 10 mm. Determine the maximum average normal stress in the bar when it is subjected to the loading shown. 9 kn kn D 1 kn kn mm 9 kn kn 1 kn P = 1 kn 1 kn 9 kn 9 kn P = 0 kn P D = kn kn P(kN) 0 1 x Fig. 1 16a 10 mm mm (d) 8.7 MPa 0 kn Internal Loading. y inspection, the internal axial forces in regions,, and D are all constant yet have different magnitudes. Using the method of sections, these loadings are determined in Fig. 1 16b; and the normal force diagram which represents these results graphically is shown in Fig. 1 16c. y inspection, the largest loading is in region, where P = 0 kn. Since the cross-sectional area of the bar is constant, the largest average normal stress also occurs within this region of the bar. Fig verage Normal Stress. pplying Eq. 1 6, we have s = P = 0110 N 10.0 m m = 8.7 MPa The stress distribution acting on an arbitrary cross section of the bar within region is shown in Fig. 1 16d. Graphically the volume (or block ) represented by this distribution of stress is equivalent to the load of 0 kn; that is, 0 kn = 18.7 MPa1 mm110 mm.

7 E X M P L E 1.7 The 80-kg lamp is supported by two rods and as shown in Fig. 1 17a. If has a diameter of 10 mm and has a diameter of 8 mm, determine the average normal stress in each rod. F y F 60 x 60 80(9.81) = 78.8 N Fig. 1 17b Fig Internal Loading. We must first determine the axial force in each rod. free-body diagram of the lamp is shown in Fig. 1 17b.pplying the equations of force equilibrium yields : + F x = 0; F - F cos 60 = 0 +q F y = 0; F + F sin N = 0 F = 9. N, F = 6. N y Newton s third law of action, equal but opposite reaction, these forces subject the rods to tension throughout their length. 8.0 MPa 8.0 MPa verage Normal Stress. pplying Eq. 1 6, we have s = F = s = F = 9. N = 7.86 MPa p10.00 m 6. N = 8.0 MPa p10.00 m 6. N The average normal stress distribution acting over a cross section of rod is shown in Fig. 1 17c, and at a point on this cross section, an element of material is stressed as shown in Fig. 1 17d. (d) Fig. 1 17c

8 E X M P L E 1.8 z The casting shown in Fig. 1 18a is made of steel having a specific weight of st 80 kn/m. Determine the average compressive stress acting at points and. W st 00 mm 800 mm 800 mm 00 mm 100 mm 00 mm y P 6 kn/m x Fig Fig. 1 18b Internal Loading. free-body diagram of the top segment of the casting where the section passes through points and is shown in Fig. 1 18b.The weight of this segment is determined from W st = g st V st. Thus the internal axial force P at the section is +q F z = 0; P (80 kn/m )(0.8 m) (0. m) 0 P 8.0 kn verage ompressive Stress. The cross-sectional area at the section is (0. m), and so the average compressive stress becomes s = P = kn lb p10.7 (0. m) ft = kn/m psi P - W st = 0 The stress shown on the volume element of material in Fig. 1 18c is representative of the conditions at either point or. Notice that this stress acts upward on the bottom or shaded face of the element since this face forms part of the bottom surface area of the cut section, and on this surface, the resultant internal force P is pushing upward.

9 E X M P L E 1.9 Member shown in Fig. 1 19a is subjected to a vertical force of kn. Determine the position x of this force so that the average compressive stress at the smooth support is equal to the average tensile stress in the tie rod. The rod has a cross-sectional area of 00 mm and the contact area at is 60 mm. kn F kn x x 00 mm Fig Internal Loading. The forces at and can be related by considering the free-body diagram for member, Fig. 1 19b. There are three unknowns, namely, F, F, and x. To solve this problem, we will work in units of newtons and millimeters. +q F y = 0; F + F N = 0 (1) d+ M = 0; -000 N1x + F 100 mm = 0 () verage Normal Stress. necessary third equation can be written that requires the tensile stress in the bar and the compressive stress at to be equivalent, i.e., s = F 00 mm = F 60 mm F = 1.6F Substituting this into Eq. 1, solving for F, then solving for F, we obtain F = 11 N F = 187 N The position of the applied load is determined from Eq., x = 1 mm Note that 0 6 x 6 00 mm, as required. 00 mm Fig. 1 19b F

10 E X M P L E 1.10 The bar shown in Fig. 1 a has a square cross section for which the depth and thickness are 0 mm. If an axial force of 800 N is applied along the centroidal axis of the bar s cross-sectional area, determine the average normal stress and average shear stress acting on the material along section plane a a and section plane b b. a b 0 mm 800 N a Fig. 1 a b 60 0 mm 00 kpa 800 N P = 800 N 00 kpa Fig. 1 Part Internal Loading. The bar is sectioned, Fig. 1 b, and the internal resultant loading consists only of an axial force for which P = 800 N. verage Stress. The average normal stress is determined from Eq.1 6. s = P = 800 N 10.0 m10.0 m = 00 kpa No shear stress exists on the section, since the shear force at the section is zero. t avg = 0 The distribution of average normal stress over the cross section is shown in Fig. 1 c.

11 y y x 0 x V 800 N N (d) N Fig. 1 d Part Internal Loading. If the bar is sectioned along b b, the free-body diagram of the left segment is shown in Fig. 1 d. Here both a normal force (N) and shear force (V) act on the sectioned area. Using x, y axes, we require : + F x = 0; +q F y = 0; -800 N + N sin 60 + V cos 60 = 0 V sin 60 - N cos 60 = 0 or, more directly, using x, y axes, +Ω F x = 0; + F y = 0; N N cos 0 = 0 V N sin 0 = 0 Solving either set of equations, verage Stresses. In this case the sectioned area has a thickness and depth of 0 mm and 0 mm>sin 60 = 6.19 mm, respectively, Fig. 1 a. Thus the average normal stress is s = N = N = 69.8 N V = 00 N 69.8 N 10.0 m m = 7 kpa 7 kpa and the average shear stress is t avg = V = 00 N 10.0 m m = 17 kpa (e) 17 kpa 7 kpa The stress distribution is shown in Fig. 1 e. Fig. 1 e

12 E X M P L E 1.11 b a c d 0 mm 0 mm The wooden strut shown in Fig. 1 a is suspended from a 10-mmdiameter steel rod, which is fastened to the wall. If the strut supports a vertical load of kn, compute the average shear stress in the rod at the wall and along the two shaded planes of the strut, one of which is indicated as abcd. Internal Shear. s shown on the free-body diagram in Fig. 1 b, the rod resists a shear force of kn where it is fastened to the wall. free-body diagram of the sectioned segment of the strut that is in contact with the rod is shown in Fig. 1 c. Here the shear force acting along each shaded plane is. kn. force of strut on rod kn Fig. 1 a verage Shear Stress. For the strut, t avg = V = For the rod, 000 N = 6.7 MPa p10.00 m kn V = kn t avg = V = 00 N 10.0 m10.0 m =.1 MPa Fig. 1 b The average-shear-stress distribution on the sectioned rod and strut segment is shown in Figs. 1 d and 1 e, respectively. lso shown with these figures is a typical volume element of the material taken at a point located on the surface of each section. Note carefully how the shear stress must act on each shaded face of these elements and then on the adjacent faces of the elements. V =. kn V =. kn c kn b 6.7 MPa a d force of rod on strut kn kn.1 MPa (d) (e) Fig. 1 d Fig. 1 Fig. 1 e

13 E X M P L E 1.1 The inclined member in Fig. 1 6a is subjected to a compressive force of 000 N. Determine the average compressive stress along the smooth areas of contact defined by and, and the average shear stress along the horizontal plane defined by ED. 000 N 000 N F mm E D Fig. 1 6 Internal Loadings. The free-body diagram of the inclined member is shown in Fig. 1 6b. The compressive forces acting on the areas of contact are : + F F lb x = 0; F 000 N = 0 F = lbn +q F F lb y = 0; F 000 N = 0 F = lbn lso, from the free-body diagram of the top segment of the bottom member, Fig. 1 6c, the shear force acting on the sectioned horizontal plane ED is : + F x = 0; V = lbn verage Stress. The average compressive stresses along the horizontal and vertical planes of the inclined member are lb N s = = 0 psi 1.80 N/mm 11 in.11. in. ( mm)(0 mm) lb N s = = 160 psi 1.0 N/mm 1 (0 in.11. mm)(0 in. mm) These stress distributions are shown in Fig. 1 6d. The average shear stress acting on the horizontal plane defined by ED is lb N t avg = = 80 psi 0.60 N/mm 1 (7 in.11. mm)(0 in. mm) This stress is shown distributed over the sectioned area in Fig. 1 6e. 0 mm 7 mm 0 mm V 1.80 N/mm Fig. 1 6e F Fig. 1 6b Fig. 1 6c (d) Fig. 1 6d 1800 N 1.0 N/mm 1800 N 0.60 N/mm (e) 000 N

14 E X M P L E 1.1 The two members are pinned together at as shown in Fig. 1 1a. Top views of the pin connections at and are also given in the figure. If the pins have an allowable shear stress of allow 90 MPa and the allowable tensile stress of rod is ( t ) allow 11 MPa, determine to the nearest mm the smallest diameter of pins and and the diameter of rod necessary to support the load. 6 kn m 1 m Fig. 1 1 Recognizing to be a two-force member, the free-body diagram of member along with the computed reactions at and is shown in Fig. 1 1b. s an exercise, verify the computations and notice that the resultant force at must be used for the design of pin, since this is the shear force the pin resists..68 kn 0.6 kn 6 kn 6.67 kn. kn m 1 m Fig. 1 1b ontinued

15 .68 kn.8 kn Pin at.8 kn 6.67 kn Pin at 6.67 kn Fig. 1 1c Diameter of the Pins. From Fig. 1 1a and the free-body diagrams of the sectioned portion of each pin in contact with member, Fig. 1 1c, it is seen that pin is subjected to double shear, whereas pin is subjected to single shear. Thus, V T allow V T allow.8 kn m kpa 6.67 kn m kpa d ( ) d ( ) lthough these values represent the smallest allowable pin diameters, a fabricated or available pin size will have to be chosen.we will choose a size larger to the nearest millimeter as required. d 6. mm d 9.7 mm d 7 mm d 10 mm Diameter of Rod. midsection is thus, = P 1s t allow = The required diameter of the rod throughout its kip kn 16. kip>in = 0.08 in = pa d 10 6 m b kpa d ( ) d 8.9 mm We will choose d 9 mm

16 E X M P L E 1.1 The control arm is subjected to the loading shown in Fig. 1 a. Determine to the nearest mm the required diameter of the steel pin at if the allowable shear stress for the steel is allow MPa. Note in the figure that the pin is subjected to double shear. F 00 mm 00 mm 0 mm 7 mm 1 kn x kn Internal Shear Force. free-body Fig. 1 a diagram of the arm is shown in Fig. 1 b. For equilibrium, we have d+ M = 0; ( ) F (0. m) 1 kn(0.07 m) kn (0.1 m) 0 y 0 mm 7 mm 1 kn Fig. 1 b kn : + F x = 0; +q F y = 0; F 1 kn 1 - kn kip - x + kip x kn = 0 x = 1 kip kn y - kip - kip y 1 kn kn = 0 y = 60 kip kn 0.1 kn The pin at resists the resultant force at. Therefore, F = 11 ( kip kn) + (0 16 kip kn) = kip kn Since the pin is subjected to double shear, a shear force of 1.0 kn acts over its cross-sectional area between the arm and each supporting leaf for the pin, Fig. 1 c. Required rea. We have = V kip kn = = 0.80 t in m allow 8 kip>in 10 kn/m pa d b = mm in Use a pin having a diameter of d = mm in. d = d in. = 0.70 mm in. 1.0 kn 1.0 kn Pin at Fig. 1

17 E X M P L E 1.1 The suspender rod is supported at its end by a fixed-connected circular disk as shown in Fig. 1 a. If the rod passes through a 0-mmdiameter hole, determine the minimum required diameter of the rod and the minimum thickness of the disk needed to support the 0-kN load. The allowable normal stress for the rod is s allow = 60 MPa, and the allowable shear stress for the disk is t allow = MPa. 0 mm t 0 mm allow d 0 kn 0 kn Fig. 1 Fig. 1 b Diameter of Rod. y inspection, the axial force in the rod is 0 kn. Thus, the required cross-sectional area of the rod is So that = = pa d b = m d = m = 0.6 mm Thickness of Disk. s shown on the free-body diagram of the core section of the disk, Fig. 1 b, the material at the sectioned area must resist shear stress to prevent movement of the disk through the hole. If this shear stress is assumed to be distributed uniformly over the sectioned area, then, since V = 0 kn, we have = P s allow = V t allow = Since the sectioned area = p10.0 m1t, the required thickness of the disk is t = m p10.0 m 0110 N N>m = m 0110 N N>m = m = m =. mm

18 E X M P L E 1.16 n axial load on the shaft shown in Fig. 1 a is resisted by the collar at, which is attached to the shaft and located on the right side of the bearing at. Determine the largest value of P for the two axial forces at E and F so that the stress in the collar does not exceed an allowable bearing stress at of 1s b allow = 7 MPa and the average normal stress in the shaft does not exceed an allowable tensile stress of 1s t allow = MPa. P F E Fig. 1 a 60 mm P xial Load 0 mm 80 mm P P Fig. 1 b Fig. 1 P To solve the problem we will determine P for each possible failure condition. Then we will choose the smallest value. Why? Normal Stress. Using the method of sections, the axial load within region FE of the shaft is P, whereas the largest axial load, P, occurs within region E, Fig. 1 b. The variation of the internal loading is clearly shown on the normal-force diagram, Fig. 1 c. Since the crosssectional area of the entire shaft is constant, region E will be subjected to the maximum average normal stress. pplying Eq. 1 11, we have earing Stress. s shown on the free-body diagram in Fig. 1 d, the collar at must resist the load of P, which acts over a bearing area of b = [p10.0 m - p10.0 m ] = m. Thus, = P s allow ; y comparison, the largest load that can be applied to the shaft is P = 1.8 kn, since any load larger than this will cause the allowable normal stress in the shaft to be exceeded. P P s allow = P N>m P = p10.0 m P = 1.8 kn N>m P = m P =.0 kn Position P (d) Fig. 1 d

19 E X M P L E 1.17 Steel P 0.7 m luminum F m Fig. 1 a P 0.7 m 1. m Fig. 1 F The rigid bar shown in Fig. 1 a is supported by a steel rod having a diameter of 0 mm and an aluminum block having a crosssectional area of 1800 mm. The 18-mm-diameter pins at and are subjected to single shear. If the failure stress for the steel and aluminum is 1s st fail = 680 MPa and 1s al fail = 70 MPa, respectively, and the failure shear stress for each pin is t fail = 900 MPa, determine the largest load P that can be applied to the bar.pply a factor of safety of F.S. =. Using Eqs. 1 9 and 1 10, the allowable stresses are 1s st allow = 1s st fail F.S. 1s al allow = 1s al fail F.S. t allow = t fail F.S. = 680 MPa = = 0 MPa 70 MPa = = MPa 900 MPa = 0 MPa The free-body diagram for the bar is shown in Fig. 1 b. There are three unknowns. Here we will apply the equations of equilibrium so as to express F and F in terms of the applied load P. We have d+ M = 0; d+ M = 0; P11. m - F 1 m = 0 F 1 m - P10.7 m = 0 (1) () We will now determine each value of P that creates the allowable stress in the rod, block, and pins, respectively. Rod. This requires F = 1s st allow 1 = N>m [p10.01 m ] = kn Using Eq. 1, lock. P = In this case, kn1 m 1. m = 171 kn F = 1s al allow = N>m [1800 mm m >mm ] = 6.0 kn Using Eq., 16.0 kn1 m P = = 168 kn 0.7 m Pin or. Here V = F = t allow = N>m [p m ] = 11. kn From Eq. 1, 11. kn1 m P = = 18 kn 1. m y comparison, when P reaches its smallest value (168 kn), it develops the allowable normal stress in the aluminum block. Hence, P = 168 kn