U_-_Relationships_Between_Load__Shear__a

U_-_Relationships_Between_Load__Shear__a - INTERNAL FORCES...

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Unformatted text preview: INTERNAL FORCES FORCES Relationships Between Load, Shear and Bending Moment and LEARNING OBJECTIVES LEARNING Be able to identify the relationship Be between the externally applied load, and the internally generated shear force and bending moment force PRE-REQUISITE KNOWLEDGE PRE-REQUISITE Units of measurement Trigonometry concepts Rectangular component concepts Equilibrium of forces in 2-D Internal force concepts SHEAR FORCE AND BENDING MOMENT DIAGRAMS (SFD AND BMD) Shear force diagram (SFD) is a graphic representation of the vertical shear force (V) along a structural member Bending moment diagram (BMD) is a graphic representation of the bending moment (M) along a structural member Sign convention of shear force (V) and bending moment (M) in the diagrams can be defined as follows SIGN CONVENTION OF SHEAR FORCE AND BENDING MOMENT The internal shear force is considered positive if it causes clockwise rotation of the structural member under consideration as shown below; Negative otherwise Positive Negative SIGN CONVENTION OF SHEAR FORCE AND BENDING MOMENT The internal bending moment is considered positive if it causes compression of the top fiber of the structural member under consideration as shown below. Negative other wise. Tension at Bottom Tension at top Positive Moment Negative Moment STEPS OF CONSTRUCTING SFD AND BMD 1. Draw FBD and solve for support reactions 2. Calculate the shear force along the member from left to right “Shear force due to distributed loading can be calculated as the area underneath the load. 3. Draw SFD based on the calculated shear force from left to right 4. Calculate the bending moment along the member from left to right “Bending moment can be calculated as the area underneath the SFD” 5. Draw BMD based on the calculated bending moment from left to right EXAMPLE 1 EXAMPLE If P = 600 lb, a = 5 ft and b = 7 ft, draw the If SFD and BMD SFD Solve for support reactions 600 lb AX SOLUTION 1 SOLUTION R AY 5 ft 7 ft BY ∑ MA= 0; RBY(12) – 600(5) = 0; R R R = 250 lb SOLUTION 1 -Contd. SOLUTION Draw SFD and BMD 600 lb 5 ft 350 lb 350 lb SFD – 250 lb 0 + 5(350) = 1750 lb-ft BMD 7 ft 250 lb 0 lb 1750 – 7(250) = 0 lb-ft EXAMPLE 2 EXAMPLE Draw SFD and BMD for the beam SOLUTION 2 SOLUTION Solve for support reactions AX R AY BY R R ∑ MA = 0; RBY(20) – 50(20)(10) – 200 = 0; R = 510 lb SOLUTION 2 -Contd. SOLUTION 490 lb 510 lb 490 lb SFD 0 lb 490/50 = 9.8 ft 490 – 50(20) = – 510 lb 0.5(9.8)(490) = 2401 lb-ft BMD 2401 – 0.5(10.2)(510) = – 200 lb-ft QUESTION Which statement is correct relative to the relationship between the SFD and the BMD? BMD? A) The SFD represents the area underneath A) the BMD the B) The BMD represents the area underneath B) the SFD the C) Both of the above D) None of the above QUESTION The values of the BMD indicate The A) External moment B) Internal moment C) Sum of external and internal moments ...
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This note was uploaded on 10/31/2010 for the course CEE CE 221 taught by Professor Baladi during the Fall '10 term at Michigan State University.

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