on a piece of paper, integrate M(x) from Part 3 once to obtain a beam slope...

 
 
 
 

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on a piece of paper, integrate M(x) from Part 3 once to obtain a beam slope function, which can be written as: El dv/dx = [ M(x) terms from Part 3 integrated once ] + C₁ where dv/dx is the beam slope and C₁ is a constant of integration. Integrate a second time to obtain a beam deflection function, which can be written as: El v = [ M(x) terms from Part 3 integrated twice ] + C₁ x + C₂ where v is the beam deflection and C₂ is a constant of integration. Apply appropriate boundary conditions for this case and use the relationships from Parts 1 and 2 to calculate the constants C₁ and C₂, and the reaction forces Ay, Cy, and Dy. Enter the reaction forces below. Positive values for the reactions are indicated by the directions of the red arrows on the figure shown in Part 3. Answers: Ay= Cy= Dy = i i i kips kips kips Determine the beam deflection at B. Your answer must be consistent with the sign convention. A downward deflection is negative, and an upward deflection is positive. Answer: VB = i in. For the beam shown, assume that El = 150,000 kip-ft2, P= 75 kips, and w = 6.5 kips/ft. Use discontinuity functions to determine (a) the reactions at A, C, and D. (b) the beam deflection at B. Assume LAB LBC = 7.0 ft, LCD = 14.0 ft. A B LAB LAB LBC LCD Sum the forces in the y direction to find an expression that includes the reaction forces Ay, Cy, and Dy acting on the beam. Positive values for the reactions are indicated by the directions of the red arrows shown on the free-body diagram below. (Note: Since Dx = 0, it has been omitted from the free-body diagram.) Enter the correct value in the box below to produce the correct expression. P B LBC Answer: Ay+ Cy+ Dy= i W C W LCD D kips X Sum the moments about support A to find an expression that includes the reaction forces Cy and Dy acting on the beam. Positive values for the reactions are indicated by the directions of the red arrows shown on the free-body diagram below. (Note: Since Dx= O, it has been omitted from the free-body diagram.) Enter the correct value in the box below to produce the correct expression. Assume LAB = LBC = 7.0 ft, LCD = 14.0 ft. A Ay LAB B LBC C Cy Answer: (14.0 ft) Cy+ (28.0 ft) Dy = i W LCD D kip-ft For this loading, select the correct moment expression M(x) from those listed below, where each expression is based on discontinuity functions. Assume LAB LBC= 7.0 ft, LCD= 14.0 ft. A V Answer: LAB B LBC C Cy W LCD Dy D O (e) W M(x) = Ay(x - Oft)¹ - P(x - 7.0 ft)' + Cy(x - 14.0 ft)' - (x - 14.0 ft)²+(x - 28.0 ft)² + Dy(x - 28.0 ft)¹ O (d) M(x) = Ay(x − 0ft)² – P(x - 7.0 ft)² + Cy(x - 14.0 ft)²-(x - 14.0 ft)³+ 2 O (b) W M(x): = Ay(x − 0 ft)º – P(x − 7.0 ft)⁰ + Cy(x − 14.0 ft) - (x - 14.0 ft)² + W (x - 28.0 ft)³ + Dy(x - 28.0 ft)² W (x - 28.0 ft)² + D¸(x − 28.0 ft)º O (a) w M(x) = Ay(x - Oft)¹ - P(x − 7.0 ft)¹ + Cy(x − 14.0 ft)' + (x - - 14.0 ft)²-(x - 28.0 ft)² + Dy(x − 28.0 ft)¹ 2 O (c) w W M(x) = Ay(x - Oft)° - P(x − 7.0 ft)º + Cy(x − 14.0 ft)° – (x - 14.0 ft) + (x - 28.0 ft)' + Dy(x − 28.0 ft)º 2 on a piece of paper, integrate M(x) from Part 3 once to obtain a beam slope function, which can be written as: El dv/dx = [ M(x) terms from Part 3 integrated once ] + C₁ where dv/dx is the beam slope and C₁ is a constant of integration. Integrate a second time to obtain a beam deflection function, which can be written as: El v = [ M(x) terms from Part 3 integrated twice ] + C₁ x + C₂ where v is the beam deflection and C₂ is a constant of integration. Apply appropriate boundary conditions for this case and use the relationships from Parts 1 and 2 to calculate the constants C₁ and C₂, and the reaction forces Ay, Cy, and Dy. Enter the reaction forces below. Positive values for the reactions are indicated by the directions of the red arrows on the figure shown in Part 3. Answers: Ay= Cy= Dy = i i i kips kips kips Determine the beam deflection at B. Your answer must be consistent with the sign convention. A downward deflection is negative, and an upward deflection is positive. Answer: VB = i in. For the beam shown, assume that El = 150,000 kip-ft2, P= 75 kips, and w = 6.5 kips/ft. Use discontinuity functions to determine (a) the reactions at A, C, and D. (b) the beam deflection at B. Assume LAB LBC = 7.0 ft, LCD = 14.0 ft. A B LAB LAB LBC LCD Sum the forces in the y direction to find an expression that includes the reaction forces Ay, Cy, and Dy acting on the beam. Positive values for the reactions are indicated by the directions of the red arrows shown on the free-body diagram below. (Note: Since Dx = 0, it has been omitted from the free-body diagram.) Enter the correct value in the box below to produce the correct expression. P B LBC Answer: Ay+ Cy+ Dy= i W C W LCD D kips X Sum the moments about support A to find an expression that includes the reaction forces Cy and Dy acting on the beam. Positive values for the reactions are indicated by the directions of the red arrows shown on the free-body diagram below. (Note: Since Dx= O, it has been omitted from the free-body diagram.) Enter the correct value in the box below to produce the correct expression. Assume LAB = LBC = 7.0 ft, LCD = 14.0 ft. A Ay LAB B LBC C Cy Answer: (14.0 ft) Cy+ (28.0 ft) Dy = i W LCD D kip-ft For this loading, select the correct moment expression M(x) from those listed below, where each expression is based on discontinuity functions. Assume LAB LBC= 7.0 ft, LCD= 14.0 ft. A V Answer: LAB B LBC C Cy W LCD Dy D O (e) W M(x) = Ay(x - Oft)¹ - P(x - 7.0 ft)' + Cy(x - 14.0 ft)' - (x - 14.0 ft)²+(x - 28.0 ft)² + Dy(x - 28.0 ft)¹ O (d) M(x) = Ay(x − 0ft)² – P(x - 7.0 ft)² + Cy(x - 14.0 ft)²-(x - 14.0 ft)³+ 2 O (b) W M(x): = Ay(x − 0 ft)º – P(x − 7.0 ft)⁰ + Cy(x − 14.0 ft) - (x - 14.0 ft)² + W (x - 28.0 ft)³ + Dy(x - 28.0 ft)² W (x - 28.0 ft)² + D¸(x − 28.0 ft)º O (a) w M(x) = Ay(x - Oft)¹ - P(x − 7.0 ft)¹ + Cy(x − 14.0 ft)' + (x - - 14.0 ft)²-(x - 28.0 ft)² + Dy(x − 28.0 ft)¹ 2 O (c) w W M(x) = Ay(x - Oft)° - P(x − 7.0 ft)º + Cy(x − 14.0 ft)° – (x - 14.0 ft) + (x - 28.0 ft)' + Dy(x − 28.0 ft)º 2