Solution Manual to accompany STRUCTURES Seventh Edition Daniel L. Schodek Martin Bechthold vii Schodek, Bechthold Structures Chapter 2 Chapter 3 Chapter 4 Chapter 5 Chapter 6 Chapter 7 Chapter 8 Chapter 9 Chapter 10 Chapter 11 Chapter 12 Chapter 16 Table of Contents Mechanics Analysis Trusses Cables and Arches Beams Columns Continuous Beams Rigid Frames Structural Plates and Grids Structural Membranes and Nets Structural Shells Connections 1-12 13-14 15-44 45-64 65-78 79-84 85-88 89-96 97-104 105-106 107-112 113-114 Schodek, Bechthold Structures Chapter 2 1 = P X /P = P X = 0.26P = P Y /P = P Y = 0.97P Question 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? R R 2 R tan θ θ θ = resultant force = (0.50P) 2 + (1.50P) 2 = 1.58P = 1.50P/0.50P = tan -1 1.50P/0.50P = 71.6° Question 2.3: The following three forces act concurrently through a point: a force P acting to the right at θ X = 30° to the horizontal, a force P acting to the right at θ X = 45° to the horizontal, and a force P acting to the right at θ X = 60° to the horizontal. Find the single resultant force that is equivalent to this three-force system. Step 1: Find the horizontal and vertical components of each force and the net horizontal and vertical force. P 1x P 1x P 1y P 1y P 2x P 2x P 2y P 2y P 3x P 3x P 3y P 3y R X = R Y R X = R Y = 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 = 2.08P cos 75° P * cos 75° P X sin 75° P * sin 75° P Y Chapter 2 Question 2.1: A force of P de fi ned by the angle θ X = 75° to the horizontal acts through a point. What are the components of this force on the x and y axes? P Y = ? P P X = ? 75° R Y = 1.50P R = ? θ = ? 71.6° 1.50P 1.58P P 1y = .50P P 1x = .87P 30° P 2y = .71P P 2x = .71P 45° 60° P 3x = .50P P 3y = .87P P 3 P 2 P 1 P 1 P 2 P 3 R X = 0.50P 0.50P Schodek, Bechthold Structures Chapter 2 2 Question 2.3 (continued): R 2 R tan θ tan θ θ θ = (2.08P) 2 + (2.08P) 2 = 2.93P = R Y /R X = 2.08P/2.08P = tan -1 1 = 45° Question 2.4: The following three forces act through a point: P at θ x = 45°, 2P at θ x = 180°, and P at θ x = 270°. Find the equivalent resultant force. Step 1: Find the horizontal and vertical components 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° F 1x F 1x F 1y F 1y F 2x F 2y F 3x F 3y R X R X R Y R Y R 2 R tan θ θ resultant force = R Question 2.6: Determine the reactions for the structure shown in Figure 2.59(Q6). Summary Step 3: Find the magnitude and direction of the resultant force. Step 2: Find the net horizontal and vertical force. R Y = 2.08P R = ? R X = 2.08P θ = ? Step 2: Find the magnitude and direction of the resultant force. 45° F 1y = ? F 1x = ? F 2 = 2P F 3 = P F 1 = P θ = ? R X = -1.29P R Y = -0.29P R = ? R A V = ? R B V = ? L L L 2P P M 0 - (2P L) + (R 2L) - (P 3L) 0 R A B B ∑ = ∗ ∗ ∗ = 2L 2PL + 3PL R 5PL/2L B ∗ = = R 5P/2 F 0 R + R - 2P - P 0 B Y A B = ↑ = = ∑ R 3P - R R 3P - 5 A B A = = P P/2 R P/2 A = ↑ Sum rotational moments about point A. Assume that a counter-clockwise rotational effect is positive. Schodek, Bechthold Structures Chapter 2 3 Question 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/3 L/3 L/3 4P 2P R A = ? R B = ? wL/3 = equivalent point load R B = ? L/6 2L/3 L/6 R A = ? 45° R Bx = ? R By = ? R B R Ay = ? L/2 R B = ? L/2 R Ax = ? wL = equiv pt load w M 0 - (4P L/3) - (2P 2L/3) 0 (R L) A B ∑ = ∗ ∗ = ∗ = 0 R L (4P L/3) + (2P 2L/3) B ∗ = ∗ ∗ R L 4PL/3 + 4PL/3 R 8P/3 F 0 B B Y ∗ = = ↑ = ∑ R R + R - 4P - 2P 0 R 6P - R A B A B = = R 6P - 8P/3 R A A = = 1 10P/3 ↑ M 0 - (w L/3 L/6) (R L) 0 R L w A B B ∑ = ∗ ∗ + ∗ = ∗ = L/3 L/6 R L wL /18 B 2 × ∗ ∗ = R wL/18 F 0 R + R - wL/3 0 B Y A B = ↑ = = ∑ R wL/3 - R R 6wL/18 A B A = = - wL/18 R 5wL/18 A = ↑ M 0 (R L) (wL L/2) 0 R L wL L/2 A By By ∑ = ∗ ∗ = ∗ = ∗ - R wL/2 R R By By Bx = ↑ = tan / 45 D 1 1 R R R R R By Bx Bx By = = / B B wL/2 x = ← Y Ay By Ay By F 0 R R wL 0 R wL R = + = = ∑ - - R wL wL/2 R wL/2 F 0 R Ay Ay X A = = ↑ = ∑ - x x Bx Ax B Ax R 0 R R R wL/2 x + = = = → Sum moments about A. Assume that counter-clockwise moments are positive. Convert the uniform load w into an equivalent concentrated load for purposes of fi nding reactions. The angle of the roller on the right determines the direction of the reactive force at B which is then considered in terms of its components. The fact that the reaction at B is inclined means that the reaction at A must also be inclined (the horizontal components of the reactions must sum to zero because of equilibrium in the x direction). An equivalent point load of ( w)(L) is used to model the uniform load of w acting over the length of the beam. Schodek, Bechthold Structures Chapter 2 4 M 0 (P L/2) - (R h) 0 B Ax = ∗ ∗ = ∑ (R h) PL/2 R PL/2h Ax Ax ∗ = = → Question 2.13: Determine the reactions for the four beams shown in Figure 2.59(Q13). Step 2: Figure 2.33(e)-2 Step 1: Figure 2.33(e)-1 Step 3: Figure 2.33(e)-3 R By = ? R Ay = ? R Ax = ? P L/2 L/2 L R Ax = ? R Ay = ? R By = ? Step 4: Figure 2.33(e)-4 L/2 L/2 L P R By = ? R Ay = ? R By = ? R Bx = ? R Ax = ? L/2 L/2 P P L/2 L/2 L M 0 (R L) (P L/2) 0 R L PL/2 A By By ∑ = ∗ ∗ = ∗ = - R P/2 By = ↑ F 0 R 0 X Ax ∑ = = F 0 R R P 0 Y Ay By = + = ∑ - R R P R R P P/2 R P/2 Ay By Ay Ay = = = ↑ - - R L PL/2 R P By By ∗ = = / /2 F 0 Y ↑ = ∑ R P - R R P - P/2 R Ay By Ay Ay = = = P/2 ↑ M 0 (R L) (P L/2) 0 R L PL/2 A By By ∑ = ∗ ∗ = ∗ = - R P/2 By = ↑ F 0 R 0 X Ax ∑ = = F 0 R R P 0 Y Ay By = + = ∑ - R R P R R P P/2 R P/2 Ay By Ay Ay = = = ↑ - - Notice that the three inclined members are identical except for the type of end conditions present. Note how changing the support types radically alters the nature of the reactive forces. F X ∑ ∑ = + = = 0 R R 0 R R R Ax Bx Bx Ax - B Bx PL/2h = ← Y By By F 0 R - P 0 R P = = = ↑ ∑ Schodek, Bechthold Structures Chapter 2 5 L L L L L L Question 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 = -PL R A R B For 0 < x < L: V X For L < x < 2L: V X V X For 2L < x < 3L: V X V X For 0 < x < L: M X When x = L: M L M L For L < x < 2L: M X When M X = 0: 0 0 0 3Px/2 (2/3P)3Px/2 x When x = 2L: M 2L M 2L M 2L For 2L < x < 3L: M X Check: when x = 3L: M 3L M 3L M 3L M 3L M 3L V MAX M MAX Step 1: Find the reactions (see Question 2.6). Step 2: Draw the shear diagram. Step 3: Draw the moment diagram. V x = P/2 V x = -3P/2 V x = P M 3L = 0 M 2L = -PL M L = PL/2 x = 4L/3 When the shear is positive, the slope to the moment diagram is positive and vice-versa. Also note that when the shear diagram passes through zero the bending moment values are critical. Since only concentrated loads are present, the moment diagram consists of linearly sloped lines only (uniform loadings produce curved lines). The point of zero moment on the bending moment diagram corresponds to a "point of in fl ection" (reverse curvature) on the de fl ected shape of the structure (see Section 2.4.4). Summary Schodek, Bechthold Structures Chapter 2 6 L/3 L/3 L/3 L/3 L/3 L/3 R A R B For 0 < x < L/3: V X For L/3 < x < 2L/3: V X V X For 2L/3 < x < L: V X V X For 0 < x < L/3: M X When x = L/3: M L/3 M L/3 For L/3 < x < 2L/3: M X When x = 2L/3: M 2L/3 M 2L/3 M 2L/3 M 2L/3 For 2L/3 < x < L: M X Check: when x = L: M L M L M L M L V MAX M MAX = 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/9 Question 2.17: Draw shear and moment diagrams for the beam analyzed in Question 2.8 [Figure 2.59]. What is the maximum shear force present? What is the maximum bending moment present? Step 1: Find the reactions (see Question 2.8). Step 2: Draw the shear diagram. Step 3: Draw the moment diagram. M 0 = 0 M L/3 = 10PL/9 M 2L/3 = 8PL/9 M L = 0 V x = 10P/3 V x = -8P/3 V x = -2P/3 Summary Schodek, Bechthold Structures Chapter 2 7 R A R B For 0 < x < L/3: V X When x = 0: V X V X When V X = 0: 0 wx x For L/3 < x < L: V X V X V X For 0 < x < L/3: M X M X When x = 5L/18 (V x = 0): M 5L/18 M 5L/18 M 5L/18 M 5L/18 When x = L/3: M L/3 M L/3 M L/3 M L/3 M L/3 M L/3 For L/3 < x < L: M x V MAX M MAX Step 1: Find the reactions (see Question 2.7). = 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 - wx 2 /2 = (5wL/18)(5L/18) - w(5L/18) 2 /2 = 25wL 2 /324 - 25wL 2 /648 = 25wL 2 /648 = 0.039 wL 2 = (5wL/18)(L/3) - w(L/3) 2 /2 = 5wL 2 /54 - wL 2 /18 = 5wL 2 /54 - 3wL 2 /54 = 2wL 2 /54 = wL 2 /27 = 0.037wL 2 = (5wL/18)x - (wL/3)(x - L/6) = +5wL/18 = +25wL 2 /648 Question 2.19: Draw shear and moment diagrams for the beam analyzed in Question 2.10 [Figure 2.59]. 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: M L M L M L M L = (5wL/18)L - (wL/3)(L - L/6) = 5wL 2 /18 - (wL/3)(5L/6) = 5wL 2 /18 - 5wL 2 /18 = 0 L/3 2L/3 5L/18 V 0 = 5wL/18 V 5L/18 = 0 V X = -wL/18 L/3 2L/3 5L/18 M 5L/18 = 0.039wL 2 M L/3 = 0.037wL 2 M L = 0 Summary Schodek, Bechthold Structures Chapter 2 8 Question 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 bending moment present? R Ax R Ay R Bx R By For 0 > x > L: V X When x = 0: V 0 When x = L: V L V L When V X = 0: 0 wx x For 0 > x > L: M X M X When x = 0: M X When x = L/2: M X M X M X Check: when x = L: M L M L M L V MAX M MAX = 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 - wx 2 /2 = 0 = (wL/2)(L/2) - w(L/2)(L/4) = wL 2 /4 - wL 2 /8 = wL 2 /8 = (wL/2)L - wL 2 /2 = wL 2 /2 - wL 2 /2 = 0 = ±wL/2 = +wL 2 /8 Step 1: Find the reactions (see Question 2.12). Step 2: Draw the shear diagram. Summary Step 3: Draw the moment diagram. L/2 L/2 V 0 = wL/2 V L = -wL/2 L/2 L/2 M L/2 = wL 2 /8 Schodek, Bechthold Structures Chapter 2 9 L/2 L/2 R Ax R Ay R By For 0 < x < L/2: V X For L/2 < x < L: V X V X For 0 < x < L/2: M X When x = L/2: M X M X For L/2 < x < L: M X R Ax R Ay R By cos 45° longitudinal axis longitudinal axis P Y P Y R Ay = R By R Ay = R By For 0 < x < .71L: V X For .71L < x < 1.41L: V X V X = 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 = -0.35P Question 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 the longitudinal axes of the members. Transverse components of the applied and reactive forces should thus be considered in determining shears and moments. Compare the maximum moments developed in 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 the member. Step 3: Calculate the transverse components of applied and reactive forces. Step 4: Draw the shear diagram. V X = P/2 V X = -P/2 L/2 L/2 L/2 L/2 M L/2 = PL/4 L L 45° longitudinal axis = 1.41L P X P Y P R A = P/2 R Ax R Ay R By R Bx R B = P/2 V x = 0.35P V x = -0.35P Schodek, Bechthold Structures Chapter 2 10 For 0 < x < .71L: M X When x = .71L: M .71L M .71L M .71L For .71L < x < 1.41L: M X M X M X R A R B1 R B2 longitudinal axis P Y R Ay R Ay R B1y R B1y R B2y R B2y R By (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 of applied and reactive forces. The formulas and diagrams will be the same as those for Beam 2.59(Q13b). Step 3: Draw the shear and moment diagrams. R A R B P Y R Ay = R By R Ay = R By = P/2 = P/2 = 0.71 P = P/2 * sin 45° = 0.35P The formulas and diagrams will be the same as those for Beam 2.59(Q13b). M MAX = PL/4 (for all four beams) L/2 L/2 M L/2 = PL/4 P X P P Y R Ay R Ax R B2x R B2y R B2 = P R B1 = P/2 R B1y R B1x R A = P/2 Step 3: Draw the shear and moment diagrams. Summary Beam 2.59(Q13d) Step 1: Find the reactions (see Question 2-13). Step 2: Calculate the transverse components of applied and reactive forces. P X P Y P R A = P/2 R By R Bx R B = P/2 R Ax R Ay Schodek, Bechthold Structures Chapter 2 11 ƒ / ε ε ε ε = E = ƒ / E = (10,000 lb/in 2 )/ (11.3 * 10 6 lb/in 2 ) = 0.000885 in/in Question 2.24: What is the unit strain present in an aluminum specimen loaded to 10,000 lb/in 2 ? Assume that E a = 11.3 * 10 6 lb/in 2 . = modulus of elasticity = 10,000 lb/in 2 = 11.3 * 10 6 lb/in 2 = ? stress / strain ƒ (stress) E (modulus of elasticity) ε (strain) ƒ / ε ε ε ε Question 2.25: What is the unit strain present in a steel specimen loaded to 24,000 lb/in 2 ? Assume that E s = 29.6 * 10 6 lb/in 2 . stress / strain ƒ (stress) E (modulus of elasticity) ε (strain) = modulus of elasticity = 24,000 lb/in 2 = 29.6 * 10 6 lb/in 2 = ? = E = ƒ / E = (24,000 lb/in 2 )/ (29.6 * 10 6 lb/in 2 ) = 0.000811 in/in ∆ L ∆ L ∆ L = PL/AE = (16,000 lb * 240 in)/ (4 in 2 * 29.6 * 10 6 lb/in 2 ) = 0.032 in Question 2.26: A 2 in square steel bar is 20 ft long and carries a tension force of 16,000 lb. How much does the bar elongate? Assume that E s = 29.6 * 10 6 lb/in 2 . A (cross-sectional area) A L (member length) L P (load) E (modulus of elasticity) ∆ L (elongation) = 2 in * 2 in = 4 in 2 = 20 ft. * 12 in/1 ft = 240 in = 16,000 lb. = 29.6 * 10 6 lb/in 2 = ? 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 E S = 0.204 * 10 6 N/mm 2 . ∆ L ∆ L ∆ L = PL/AE = (20 000 N * 5000 mm)/ (314 mm 2 * 0.204 * 10 6 N/mm 2 ) = 1.56 mm A (cross-sectional area) A A L (member length) L P (load) P E (modulus of elasticity) ∆ L (elongation) = π r 2 = π (10 mm) 2 = 314 mm 2 = 5 m * 1000 mm/1 m = 5000 mm = 20 kN *1000 N/1 kN = 20 000 N = 0.204 * 10 6 N/mm 2 = ? Schodek, Bechthold Structures Chapter 3 13 Chapter 3 Question 3.5: For the fl oor system shown in fi gure 3.13, assume a combined live and dead load of 80 lb/ft 2 . Assume a beam span of 16 ft and a beam spacing of 3 ft. Determine the reactions for beams A, B, and C in the fl oor system. Step 1: Determine the load per unit length for each beam. Beam 1, Beam 3: 1.5 ft x 80 lb/ft 2 = 120 lb/ft Beam 2: 3 ft x 80 lb/ft 2 = 240 lb/ft Step 2: Find the reactions using moment equilib- rium around one of the supports. Since the beams are symmetrical the reactions on both sides are equal. Beam 1, Beam 3: Beam 2: ∑ = = − ∗ ∗ ∗ ⇒ = ∑ = = − ( )( ) ( ) M lb ft ft ft R ft R lbs M l A B B A 0 120 16 8 16 960 0 240 / b b ft ft ft R ft R lbs B B / ∗ ∗ ∗ ⇒ = ( )( ) ( ) 16 8 16 1920