Meriam, Kraige & Bolton — Distributed Forces: Centroids and Centers of Gravity: Problem 5_93

⚡ Mecademy AIENG정역학 · ch5  Problem Statement Calculate the support reactions at and for the loaded beam. Problem 5/93 (a) Vertical support reaction at 1. Formula: Sum of moments about point must be zero for equilibrium: 2. Substitution: The total length of the beam is . The loading consists of two triangular distributions: Resultant , acting at from . Resultant , acting at from . 3. Calculation: 4. Result: ● Final Conclusion: The vertical reaction at support is acting upwards. (b) Vertical support reaction at AB A B M = ∑ B 0⟹R ⋅ A L− ( F ⋅ ∑ i d )= i 0 L= + 3 l = 3 2l l F = 1 ⋅ 2 1 ⋅ ( 3 l )w = 0 6 w l 0 d = 1 L− = 3 1 ( 3 l )l− = 9 l 9 8l B F = 2 ⋅ 2 1 ⋅ ( 3 2l )w = 0 3 w l 0 d = 2 = 3 1 ( 3 2l ) 9 2l B R ⋅ A l− − ( 6 w l 0 )( 9 8l ) = ( 3 w l 0 )( 9 2l )0 R ⋅ A l= + 54 8w l 0 2 27 2w l 0 2 R ⋅ A l= + 27 4w l 0 2 27 2w l 0 2 R ⋅ A l= = 27 6w l 0 2 9 2w l 0 2 R = A w l 9 2 0 AA = y w l 9 2 0 B 1. Formula: Sum of vertical forces must be zero for equilibrium: 2. Substitution: 3. Calculation: To combine, use a common denominator of 18: 4. Result: ● Final Conclusion: The vertical reaction at support is acting upwards. (c) Horizontal support reaction at 1. Formula: Sum of horizontal forces must be zero for equilibrium: 2. Substitution: There are no applied horizontal loads. 3. Calculation: No additional calculation required as no other horizontal components exist. 4. Result: F = ∑ y 0⟹R + A R − B F − 1 F = 2 0 + 9 2w l 0 R − B − 6 w l 0 = 3 wl 0 0 R = B + 6 w l 0 − 3 w l 0 9 2w