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a pin at the left side, and a roller at the right side. a) Draw a Free Body Diagram and determine the reaction forces at each end of the beam. b) Cut the beam at the mid-span (5 meters from the left support). Draw a Free Body Diagram of the cut beam, and solve for the internal reactions at the cut. 2 kN/m 10 kN A B C D 5 m 3 m 2 m

a pin at the left side, and a roller at the right side. a) Draw a Free Body Diagram and determine the reaction forces at each end of the beam. b) Cut the beam at the mid-span (5 meters from the left support). Draw a Free Body Diagram of the cut beam, and solve for the internal reactions at the cut. 2 kN/m 10 kN A B C D 5 m 3 m 2 m

Steel Design (Activate Learning with these NEW titles from Engineering!)
Steel Design (Activate Learning with these NEW titles from Engineering!)
6th Edition
ISBN:9781337094740
Author:Segui, William T.
Publisher:Segui, William T.
Chapter10: Plate Girders
Section: Chapter Questions
Problem 10.7.9P
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Textbook: Strength of Materials

 

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The beam below supports a distributed load and a point load. The beam is supported by a pin at the left side, and a roller at the right side. a) Draw a Free Body Diagram and determine the reaction forces at each end of the beam. b) Cut the beam at the mid-span (5 meters from the left support). Draw a Free Body Diagram of the cut beam, and solve for the internal reactions at the cut. 2 kN/m 10 kN A B D 5 m 3 m 2 m
Transcribed Image Text:The beam below supports a distributed load and a point load. The beam is supported by a pin at the left side, and a roller at the right side. a) Draw a Free Body Diagram and determine the reaction forces at each end of the beam. b) Cut the beam at the mid-span (5 meters from the left support). Draw a Free Body Diagram of the cut beam, and solve for the internal reactions at the cut. 2 kN/m 10 kN A B D 5 m 3 m 2 m
Formula Sheet Centre of Gravity and Moment of Inertia ΣΑ ΣΑ EA, ΣΑ I, = I, + Ad² I¸ =E(!.)+ E(Ad²) 1, = E(!.)+ y Axial Stress and Internal Forces ΣΕ, -0 Σ , =0 ΣΜ=0 o = P/ A T =V | A S.F. = o alow /o. allow actual Strain PL 8 = AE · E \ateral V = E = E = L E axial
Transcribed Image Text:Formula Sheet Centre of Gravity and Moment of Inertia ΣΑ ΣΑ EA, ΣΑ I, = I, + Ad² I¸ =E(!.)+ E(Ad²) 1, = E(!.)+ y Axial Stress and Internal Forces ΣΕ, -0 Σ , =0 ΣΜ=0 o = P/ A T =V | A S.F. = o alow /o. allow actual Strain PL 8 = AE · E \ateral V = E = E = L E axial
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