Axially loaded member Axial load and normal stress
Axially loaded member Axial load and normal stress under equilibrium load, Elastic Deformation 1
Saint-Venant’s Principle • Localized deformation occurs at each end, and the deformations decrease as measurements are taken further away from the ends • At section c-c, stress reaches almost uniform value as compared to a-a, b-b • c-c is sufficiently far enough away from P so that localized deformation “vanishes”, i. e. , minimum distance 2
Saint-Venant’s Principle • • General rule: min. distance is at least equal to largest dimension of loaded xsection. For the bar, the min. distance is equal to width of bar This behavior discovered by Barré de Saint-Venant in 1855, this the name of the principle Saint-Venant Principle states that localized effects caused by any load acting on the body, will dissipate/smooth out within regions that are sufficiently removed from location of load Thus, no need to study stress distributions at that points near application loads or support reactions 3
Elastic Deformation of an Axially Loaded Member Deformation can be calculated using SIGN CONVENTION When the member is tension When the member is compression 4
Elastic Deformation of an Axially Loaded Member Total deformation: The Above Figure:
Elastic Deformation of an Axially Loaded Member 1) Internal Forces 2) Displacement calculation
Example 1 Composite A-36 steel bar shown made from two segments AB and BD. Area AAB = 600 mm 2 and ABD = 1200 mm 2. Determine the vertical displacement of end A and displacement of B relative to C. A-36 Steel: E = 210 GPa 7
Internal axial load:
Displacement of point A Displacement of point B relative to C
Displacement of point A Displacement of point B relative to C
Solve it! The copper shaft is subjected to the axial loads shown. Determine the displacement of end A with respect to end D. The diameter of each segment are d. AB = 75 mm, d. BC= 50 mm and d. CD = 25 mm. Take Ecu = 126 GPa.
1) Internal Forces PAB=30 k. N Tension P=20 k. N PAB=30 k. N 2) Displacement calculation PBC=10 k. N PCD=-5 k. N P=5 k. N Total deformation Compression
Solve it! The assembly consist of three titanium (Ti-6 A 1 -4 V) rods and a rigid bar AC. The cross sectional area for each rod is given in the figure. If a force 30 k. N is applied to the ring F, determine the angle of tilt of bar AC.
FDC 1) Equilibrium Equation FAB F= 30 k. N 2) Based on FAB and FDC, FDC is predicted to deform more. d. DC 3) Angle of tilt d. AB Lecture 1 14
Example 2 The assembly shown in the figure consists of an aluminum tube AB having a cross-sectional area of 400 mm 2. A steel rod having a diameter of 10 mm is attached to a rigid collar and passes through the tube. If a tensile load of 80 k. N is applied to the rod, determine the displacement of the end C of the rod. Take Est = 200 GPa, Eal = 70 GPa. 15
Find the displacement of end C with respect to end B. Displacement of end B with respect to the fixed end A, Since both displacements are to the right, 16
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