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 PREREQUISITE KNOWLEDGE PREREQUISITE Units of measurement Trigonometry concepts Rectangular component concepts Equilibrium of forces in 2D 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 lbft BMD 7 ft 250 lb 0 lb 1750 – 7(250) = 0 lbft 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 lbft BMD 2401 – 0.5(10.2)(510) = – 200 lbft 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 ...
View
Full Document
 Fall '10
 Baladi
 Force, 200 lb, 490 lb, 9.8 ft, 510 lb

Click to edit the document details