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Solution Manual for Structures, 7th Edition

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Solution Manual for Structures, 7th Edition - Page 1 preview imageSolution Manualto accompanySTRUCTURESSeventh EditionDaniel L. SchodekMartin Bechthold
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Solution Manual for Structures, 7th Edition - Page 2 preview image
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Solution Manual for Structures, 7th Edition - Page 3 preview imageviiSchodek, BechtholdStructuresChapter 2Chapter 3Chapter 4Chapter 5Chapter 6Chapter 7Chapter 8Chapter 9Chapter 10Chapter 11Chapter 12Chapter 16Table of ContentsMechanicsAnalysisTrussesCables and ArchesBeamsColumnsContinuous BeamsRigid FramesStructural Plates and GridsStructural Membranes and NetsStructural ShellsConnections1-1213-1415-4445-6465-7879-8485-8889-9697-104105-106107-112113-114
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Solution Manual for Structures, 7th Edition - Page 4 preview imageSchodek, BechtholdStructuresChapter 21= PX/P= PX= 0.26P= PY/P= PY= 0.97PQuestion 2.2: The components of a force on the x and y axes are 0.50P and 1.50P, respectively.What are the magnitude and direction of the resultant force?RR2Rtanθθθ= resultant force= (0.50P)2+ (1.50P)2= 1.58P= 1.50P/0.50P= tan-11.50P/0.50P= 71.6°Question 2.3: The following three forces act concurrently through a point: a force P acting to theright atθX= 30° to the horizontal, a force P acting to the right atθX= 45° to the horizontal, anda force P acting to the right atθX= 60° to the horizontal. Find the single resultant force that isequivalent to this three-force system.Step 1: Find the horizontal and verticalcomponents of each force and the net horizontaland vertical force.P1xP1xP1yP1yP2xP2xP2yP2yP3xP3xP3yP3yRX= RYRX= RY= P * cos 30°= .87P= P * sin 30°= .50P= P * cos 45°= .71P= P * sin 45°= .71P= P * cos 60°= .50P= P * sin 60°= .87P= .71P + .50P + .87P= 2.08Pcos 75°P * cos 75°PXsin 75°P * sin 75°PYChapter 2Question 2.1: A force of P defined by the angleθX= 75° to the horizontal acts through a point. Whatare the components of this force on the x and y axes?PY= ?PPX= ?75°RY= 1.50PR = ?θ= ?71.6°1.50P1.58PP1y= .50PP1x= .87P30°P2y= .71PP2x= .71P45°60°P3x= .50PP3y= .87PP3P2P1P1P2P3RX= 0.50P0.50P
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Solution Manual for Structures, 7th Edition - Page 5 preview imageSchodek, BechtholdStructuresChapter 22Question 2.3 (continued):R2Rtanθtanθθθ= (2.08P)2+ (2.08P)2= 2.93P= RY/RX= 2.08P/2.08P= tan-11= 45°Question 2.4: The following three forces act through a point: P atθx= 45°, 2P atθx= 180°, andP atθx= 270°. Find the equivalent resultant force.Step 1: Find the horizontal and verticalcomponents of each force.= P * cos 45°= .71P= P * sin 45°= .71P= -2P= 0= 0= -P= .71P - 2P= -1.29P= .71P - P= -0.29P= (-1.29P)2+ (-0.29P)2= 1.32P= -0.29P / -1.29P= 12.7°= 1.33P acting at 192.7°F1xF1xF1yF1yF2xF2yF3xF3yRXRXRYRYR2Rtanθθresultant force = RQuestion 2.6: Determine the reactions for the structure shown in Figure 2.59(Q6).SummaryStep 3: Find the magnitude and direction of theresultant force.Step 2: Find the net horizontal and verticalforce.RY= 2.08PR = ?RX= 2.08Pθ= ?Step 2: Find the magnitude and direction of theresultant force.45°F1y= ?F1x= ?F2= 2PF3= PF1= Pθ= ?RX= -1.29PRY= -0.29PR = ?RA V= ?RB V= ?LLL2PPM0- (2PL) + (R2L) - (P3L)0RABB∑=∗∗∗=2L2PL + 3PLR5PL/2LB∗==R5P/2F0R+ R- 2P - P0BYAB=↑==∑R3P - RR3P - 5ABA==PP/2RP/2A=↑Sum rotational moments about point A. Assume thata counter-clockwise rotational effect is positive.
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Solution Manual for Structures, 7th Edition - Page 6 preview imageSchodek, BechtholdStructuresChapter 23Question 2.8: Determine the reactions for the structure shown in Figure 2.59(Q8).Question 2.10: Determine the reactions for the structure shown in Figure 2.59(Q10).Question 2.12: Determine the reactions for the structure shown in Figure 2.59(Q12).L/3L/3L/34P2PRA= ?RB= ?wL/3 = equivalent point loadRB= ?L/62L/3L/6RA= ?45°RBx= ?RBy= ?RBRAy= ?L/2RB= ?L/2RAx= ?wL = equiv pt loadwM0- (4PL/3) - (2P2L/3)0(RL)AB∑=∗∗=∗=0RL(4PL/3) + (2P2L/3)B∗=∗∗RL4PL/3 + 4PL/3R8P/3F0BBY∗==↑=∑RR+ R- 4P - 2P0R6P - RABAB==R6P - 8P/3RAA==110P/3↑M0- (wL/3L/6)(RL)0RLwABB∑=∗∗+∗=∗=L/3L/6RLwL /18B2×∗∗=RwL/18F0R+ R- wL/30BYAB=↑==∑RwL/3 - RR6wL/18ABA==- wL/18R5wL/18A=↑M0(RL)(wLL/2)0RLwLL/2AByBy∑=∗∗=∗=∗-RwL/2RRByByBx=↑=tan/45D11RRRRRByBxBxBy==/BBwL/2x=←YAyByAyByF0RRwL0RwLR=+==∑--RwLwL/2RwL/2F0RAyAyXA==↑=∑-xxBxAxBAxR0RRRwL/2x+===→Sum moments about A. Assume that counter-clockwisemoments are positive. Convert the uniform loadwintoan equivalent concentrated load for purposes offindingreactions.The angle of the roller on the right determinesthe direction of the reactive force at B whichis then considered in terms of its components.The fact that the reaction at B is inclinedmeans that the reaction at A must also beinclined (the horizontal components ofthe reactions must sum to zero because ofequilibrium in the x direction).An equivalent point load of(w)(L)is used to model theuniform load ofwacting overthe length of the beam.
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Solution Manual for Structures, 7th Edition - Page 7 preview imageSchodek, BechtholdStructuresChapter 24M0(PL/2) - (Rh)0BAx=∗∗=∑(Rh)PL/2RPL/2hAxAx∗==→Question 2.13: Determine the reactions for the four beams shown in Figure 2.59(Q13).Step 2: Figure 2.33(e)-2Step 1: Figure 2.33(e)-1Step 3: Figure 2.33(e)-3RBy= ?RAy= ?RAx= ?PL/2L/2LRAx= ?RAy= ?RBy= ?Step 4: Figure 2.33(e)-4L/2L/2LPRBy= ?RAy= ?RBy= ?RBx= ?RAx= ?L/2L/2PPL/2L/2LM0(RL)(PL/2)0RLPL/2AByBy∑=∗∗=∗=-RP/2By=↑F0R0XAx∑==F0RRP0YAyBy=+=∑-RRPRRPP/2RP/2AyByAyAy===↑--RLPL/2RPByBy∗==//2F0Y↑=∑RP - RRP - P/2RAyByAyAy===P/2↑M0(RL)(PL/2)0RLPL/2AByBy∑=∗∗=∗=-RP/2By=↑F0R0XAx∑==F0RRP0YAyBy=+=∑-RRPRRPP/2RP/2AyByAyAy===↑--Notice that the three inclined members are identical except for the type of end conditions present. Note howchanging the support types radically alters the nature of the reactive forces.FX∑∑=+==0RR0RRRAxBxBxAx-BBxPL/2h=←YByByF0R- P0RP===↑∑
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Solution Manual for Structures, 7th Edition - Page 8 preview imageSchodek, BechtholdStructuresChapter 25LLLLLLQuestion 2.15: Draw shear and moment diagrams for the beam analyzed in Question 2.6 [Figure 2.59(Q6)]. What is the maximum shear force present? What is the maximum bending moment present?= P/2 (upward)= 5P/2 (upward)= P/2= P/2 - 2P= -3P/2= P/2 - 2P + 5P/2= P= (P/2)x= (P/2)L= PL/2= (P/2)x - (2P)(x - L)= (P/2)x - (2P)(x - L)= Px/2 - 2Px + 2PL= -3Px/2 + 2PL= 2PL= 2PL(2/3P)= 4L/3= (P/2)2L - (2P)(2L - L)= PL - 2PL= -PL= (P/2)x - (2P)(x - L)+ (5P/2)(x - 2L)= (P/2)3L - (2P)(3L - L)+ (5P/2)(3L - 2L)= 3PL/2 - (2P)(2L) + (5P/2)L= 3PL/2 - 4PL + 5PL/2= 8PL/2 - 4PL= 0= -3P/2= -PLRARBFor 0 < x < L:VXFor L < x < 2L:VXVXFor 2L < x < 3L:VXVXFor 0 < x < L:MXWhen x = L:MLMLFor L < x < 2L:MXWhen MX= 0:0003Px/2(2/3P)3Px/2xWhen x = 2L:M2LM2LM2LFor 2L < x < 3L:MXCheck: when x = 3L:M3LM3LM3LM3LM3LVMAXMMAXStep 1: Find the reactions (see Question 2.6).Step 2: Draw the shear diagram.Step 3: Draw the moment diagram.Vx= P/2Vx= -3P/2Vx= PM3L= 0M2L= -PLML= PL/2x = 4L/3When the shear is positive, the slope to the momentdiagram is positive and vice-versa. Also notethat when the shear diagram passes through zerothe bending moment values are critical. Sinceonly concentrated loads are present, the momentdiagram consists of linearly sloped lines only(uniform loadings produce curved lines). The pointof zero moment on the bending moment diagramcorresponds to a "point of inflection" (reversecurvature) on the deflected shape of the structure (seeSection 2.4.4).Summary
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Solution Manual for Structures, 7th Edition - Page 9 preview imageSchodek, BechtholdStructuresChapter 26L/3L/3L/3L/3L/3L/3RARBFor 0 < x < L/3:VXFor L/3 < x < 2L/3:VXVXFor 2L/3 < x < L:VXVXFor 0 < x < L/3:MXWhen x = L/3:ML/3ML/3For L/3 < x < 2L/3:MXWhen x = 2L/3:M2L/3M2L/3M2L/3M2L/3For 2L/3 < x < L:MXCheck: when x = L:MLMLMLMLVMAXMMAX= 10P/3 (upward)= 8P/3 (upward)= 10P/3= 10P/3 - 4P= -2P/3= 10P/3 - 4P - 2P= -8P/3= (10P/3)x= (10P/3)(L/3)= 10PL/9= (10P/3)x - 4P(x - L/3)= (10P/3)(2L/3) - 4P(L/3)= 20PL/9 - 4PL/3= 20PL/9 - 12PL/9= 8PL/9= (10P/3)x - 4P(x - L/3)- 2P(x - 2L/3)= (10P/3)L - 4P(L - L/3)- 2P(L - 2L/3)= 10PL/3 - 4P(2L/3)- 2P(L/3)= 10PL/3 - 8PL/3 - 2PL/3= 0= +10P/3= +10PL/9Question 2.17: Draw shear and moment diagrams for the beam analyzed in Question 2.8 [Figure2.59]. What is the maximum shear force present? What is the maximum bending momentpresent?Step 1: Find the reactions (see Question 2.8).Step 2: Draw the shear diagram.Step 3: Draw the moment diagram.M0= 0ML/3=10PL/9M2L/3=8PL/9ML= 0Vx= 10P/3Vx= -8P/3Vx= -2P/3Summary
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Solution Manual for Structures, 7th Edition - Page 10 preview imageSchodek, BechtholdStructuresChapter 27RARBFor 0 < x < L/3:VXWhen x = 0:VXVXWhen VX= 0:0wxxFor L/3 < x < L:VXVXVXFor 0 < x < L/3:MXMXWhen x = 5L/18 (Vx= 0):M5L/18M5L/18M5L/18M5L/18When x = L/3:ML/3ML/3ML/3ML/3ML/3ML/3For L/3 < x < L:MxVMAXMMAXStep 1: Find the reactions (see Question2.7).= 5wL/18 (upward)= wL/18 (upward)= 5wL/18 - wx= 5wL/18 - wx= 5wL/18= 5wL/18 - wx= 5wL/18= 5L/18= 5wL/18 - w * L/3= 5wL/18 - 6wL/18= -wL/18= (5wL/18)x - wx(x/2)= 5wxL/18 - wx2/2= (5wL/18)(5L/18)- w(5L/18)2/2= 25wL2/324 - 25wL2/648= 25wL2/648= 0.039 wL2= (5wL/18)(L/3) - w(L/3)2/2= 5wL2/54 - wL2/18= 5wL2/54 - 3wL2/54= 2wL2/54= wL2/27= 0.037wL2= (5wL/18)x - (wL/3)(x - L/6)= +5wL/18= +25wL2/648Question 2.19: Draw shear and moment diagrams for the beam analyzed in Question 2.10 [Figure2.59]. What is the maximum shear force present? What is the maximum bending moment present?Step 2: Draw the shear diagram.Step 3: Draw the moment diagram.Check: when x = L:MLMLMLML= (5wL/18)L- (wL/3)(L - L/6)= 5wL2/18- (wL/3)(5L/6)= 5wL2/18 - 5wL2/18= 0L/32L/35L/18V0=5wL/18V5L/18= 0VX= -wL/18L/32L/35L/18M5L/18=0.039wL2ML/3= 0.037wL2ML= 0Summary
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Solution Manual for Structures, 7th Edition - Page 11 preview imageSchodek, BechtholdStructuresChapter 28Question 2.21: Draw shear and moment diagrams for the beam analyzed in Question 2.12[Figure 2.59]. What is the maximum shear force present? What is the maximum bendingmoment present?RAxRAyRBxRByFor 0 > x > L:VXWhen x = 0:V0When x = L:VLVLWhen VX= 0:0wxxFor 0 > x > L:MXMXWhen x = 0:MXWhen x = L/2:MXMXMXCheck: when x = L:MLMLMLVMAXMMAX= wL/2 (to the right)= wL/2 (upward)= wL/2 (to the left)= wL/2 (upward)= wL/2 - wx= wL/2= wL/2 - wL= -wL/2= wL/2 - wx= wL/2= L/2= (wL/2)x - wx(x/2)= wxL/2 - wx2/2= 0= (wL/2)(L/2) - w(L/2)(L/4)= wL2/4 - wL2/8= wL2/8= (wL/2)L - wL2/2= wL2/2 - wL2/2= 0= ±wL/2= +wL2/8Step 1: Find the reactions (see Question 2.12).Step 2: Draw the shear diagram.SummaryStep 3: Draw the moment diagram.L/2L/2V0= wL/2VL= -wL/2L/2L/2ML/2= wL2/8
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Solution Manual for Structures, 7th Edition - Page 12 preview imageSchodek, BechtholdStructuresChapter 29L/2L/2RAxRAyRByFor 0 < x < L/2:VXFor L/2 < x < L:VXVXFor 0 < x < L/2:MXWhen x = L/2:MXMXFor L/2 < x < L:MXRAxRAyRBycos 45°longitudinal axislongitudinal axisPYPYRAy= RByRAy= RByFor 0 < x < .71L:VXFor .71L < x < 1.41L:VXVX= 0= P/2 (upward)= P/2 (upward)= P/2= P/2 - P= -P/2= (P/2)x= P/2 * L/2= PL/4= P/2(x) - P(x - L/2)= 0= P/2 (upward)= P/2 (upward)= L/longitudinal axis= L/cos 45°= 1.41L= P * sin 45°= 0.71 P= P/2 * sin 45°= 0.35 P= 0.35P= 0.35P - 0.71P= -0.35PQuestion 2.22: Draw shear and moment diagrams for the four beams in Question 13 [Figure 2.59].For the inclined members, the shear and moment diagrams should be drawn with respect to thelongitudinal axes of the members. Transverse components of the applied and reactive forces shouldthus be considered in determining shears and moments. Compare the maximum moments developedin all four beams.Beam 2.59(Q13a)Step 1: Find the reactions (see Question 2.13).Step 2: Draw the shear diagram.Step 3: Draw the moment diagram.Beam 2.59(Q13b)Step 1: Find the reactions (see Question 2.13).Step 2: Calculate the longitudinal axis of themember.Step 3: Calculate the transverse components ofapplied and reactive forces.Step 4: Draw the shear diagram.VX= P/2VX= -P/2L/2L/2L/2L/2ML/2= PL/4LL45°longitudinal axis = 1.41LPXPYPRA= P/2RAxRAyRByRBxRB= P/2Vx= 0.35PVx= -0.35P
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Solution Manual for Structures, 7th Edition - Page 13 preview imageSchodek, BechtholdStructuresChapter 210For 0 < x < .71L:MXWhen x = .71L:M.71LM.71LM.71LFor .71L < x < 1.41L:MXMXMXRARB1RB2longitudinal axisPYRAyRAyRB1yRB1yRB2yRB2yRBy(net reaction)Question 2.22 (continued):Beam 2.59(Q13b) (continued).Step 5: Draw the moment diagram= 0.35Px= 0.35P * .71L= 0.25PL= PL/4= 0.35Px - 0.71P(x - .71L)= 0.35Px - 0.71Px + 0.50PL= - 0.35Px + 0.50PL= P/2= P/2= P= 1.41L= 0.71 P= P/2 * sin 45°= 0.35P (upward)= P/2 * sin 45°= -0.35P (downward)= P * sin 45°= 0.707P (upward)= 0.35P (upward)Beam 2.59(Q13c)Step 1: Find the reactions (see Question 2.13).Step 2: Calculate the transverse components ofapplied and reactive forces.The formulas and diagrams will be the same as those for Beam2.59(Q13b).Step 3: Draw the shear and moment diagrams.RARBPYRAy= RByRAy= RBy= P/2= P/2= 0.71 P= P/2 * sin 45°= 0.35PThe formulas and diagrams will be the same as those for Beam2.59(Q13b).MMAX= PL/4 (for all four beams)L/2L/2ML/2= PL/4PXPPYRAyRAxRB2xRB2yRB2= PRB1= P/2RB1yRB1xRA= P/2Step 3: Draw the shear and moment diagrams.SummaryBeam 2.59(Q13d)Step 1: Find the reactions (see Question 2-13).Step 2: Calculate the transverse components ofapplied and reactive forces.PXPYPRA= P/2RByRBxRB= P/2RAxRAy
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Solution Manual for Structures, 7th Edition - Page 14 preview imageSchodek, BechtholdStructuresChapter 211ƒ/ εεεε= E= ƒ / E= (10,000 lb/in2)/(11.3 * 106lb/in2)= 0.000885 in/inQuestion 2.24: What is the unit strain present in an aluminum specimen loaded to 10,000 lb/in2?Assume that Ea= 11.3 * 106lb/in2.= modulus of elasticity= 10,000 lb/in2= 11.3 * 106lb/in2= ?stress / strainƒ (stress)E (modulus of elasticity)ε(strain)ƒ/ εεεεQuestion 2.25: What is the unit strain present in a steel specimen loaded to 24,000 lb/in2?Assume that Es= 29.6 * 106lb/in2.stress / strainƒ (stress)E (modulus of elasticity)ε(strain)= modulus of elasticity= 24,000 lb/in2= 29.6 * 106lb/in2= ?= E= ƒ / E= (24,000 lb/in2)/(29.6 * 106lb/in2)= 0.000811 in/in∆L∆L∆L= PL/AE= (16,000 lb * 240 in)/(4 in2* 29.6 * 106lb/in2)= 0.032 inQuestion 2.26: A 2 in square steel bar is 20 ft long and carries a tension force of 16,000 lb. Howmuch does the bar elongate? Assume that Es= 29.6 * 106lb/in2.A (cross-sectional area)AL (member length)LP (load)E (modulus of elasticity)∆L (elongation)= 2 in * 2 in= 4 in2= 20 ft. * 12 in/1 ft= 240 in= 16,000 lb.= 29.6 * 106lb/in2= ?Question 2.27: A steel bar that is 20 mm in diameter is 5 m long and carries a tension force of 20kN.How much does the bar elongate? Assume that ES= 0.204 * 106N/mm2.∆L∆L∆L= PL/AE= (20 000 N * 5000 mm)/(314 mm2* 0.204 * 106N/mm2)= 1.56 mmA (cross-sectional area)AAL (member length)LP (load)PE (modulus of elasticity)∆L (elongation)=πr2=π(10 mm)2= 314 mm2= 5 m * 1000 mm/1 m= 5000 mm= 20 kN *1000 N/1 kN= 20 000 N= 0.204 * 106N/mm2= ?
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Solution Manual for Structures, 7th Edition - Page 15 preview image
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Solution Manual for Structures, 7th Edition - Page 16 preview imageSchodek, BechtholdStructuresChapter 313Chapter 3Question 3.5: For thefloor system shown infigure 3.13, assume a combined live and dead load of80 lb/ft2. Assume a beam span of 16 ft and a beam spacing of 3 ft. Determine the reactions for beamsA, B, and C in thefloor system.Step 1: Determine the load per unit length foreach beam.Beam 1, Beam 3: 1.5 ft x 80 lb/ft2= 120 lb/ftBeam 2:3 ft x 80 lb/ft2= 240 lb/ftStep 2: Find the reactions using moment equilib-rium around one of the supports. Since the beamsare symmetrical the reactions on both sides areequal.Beam 1, Beam 3:Beam 2:∑== −∗∗∗⇒=∑== −()()()MlbftftftRftRlbsMlABBA0120168169600240/bbftftftRftRlbsBB/∗∗∗⇒=()()()168161920
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