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Unformatted text preview: INTERNAL FORCES INTERNAL FORCES Internal Forces Developed in Structural Internal Forces Developed in Structural Members Members LEARNING OBJECTIVES LEARNING OBJECTIVES Be able to use the method of sections for Be able to use the method of sections for determining internal forces in 2D load case determining internal forces in 2D load case PREREQUISITE KNOWLEDGE PREREQUISITE KNOWLEDGE Units of measurement Units of measurement Trigonometry concepts Trigonometry concepts Rectangular component concepts Rectangular component concepts Equilibrium of forces in 2D Equilibrium of forces in 2D STRESSES AND STRAINS STRESSES AND STRAINS Stress is defined as load divided by the Stress is defined as load divided by the crosssectional area on which it acts. crosssectional area on which it acts. Strain is defined as the change in length divided by the original length. Stress = = Q/A Strain = = L/L STRESSES AND STRAINS STRESSES AND STRAINS Stress = = Q/A Q Q A = r 2 L L Strain = = L/L STRESSES AND STRAINS STRESSES AND STRAINS Stress = = Q/A Q Q A = r 2 L L Strain = = L/L CALCULATION OF INTERNAL FORCES The internal forces developed in the beam at any section such as C can be determined using the following steps: CALCULATION OF INTERNAL FORCES Step 1 Draw FBD of the structure (beam) and calculate the reactions at the support. CALCULATION OF INTERNAL FORCES Step 2 Draw FBD of either the left or the right section of the structure (beam), this is similar to the method of sections. CALCULATION OF INTERNAL FORCES Step 3 Calculate the internal forces (shear = V C , Normal = N C and moment = M C ) using the equations of equilibrium TYPE OF INTERNAL FORCES SIGN CONVENTION FOR SHEAR FORCES The internal shear force is considered positive if it causes clockwise rotation of the structural member under consideration and negative otherwise as shown below. Positive Negative SIGN CONVENTION FOR BENDING MOMENT The internal bending moment is considered positive if it causes compression in the top fiber of the structural member under consideration as shown below. Positive Moment Negative Moment Tension at Bottom Tension at top EXAMPLE 1  EXAMPLE 1  Simply Supported Beam Simply Supported Beam 200 lb 5 ft 5 ft x y Determine the internal forces at any section along the beam EXAMPLE 1 EXAMPLE 1Simply Supported BeamSimply Supported Beam Step 1 Draw FBD and determine the reactions at the supports 200 lb 5 ft 5 ft x y A B 200 lb R AY R BY EXAMPLE 1 ...
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 Fall '10
 Baladi

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