average normal stress in each rod and the average shear stress in the pin A between the members V Strain 5.1 Deformation F2 F1 F3 F4 V Strain F2 5.2 Strain F3 F1 5.2.1 Normal strain  AB AB  AB  AB F4 V Strain 5.2.2 Shear Strain     V Strain 5.2.3 Cartesian strain The approximate lengths of the sides of the parallelepiped are 1   x x 1   y 1   z y z The approximate angles between sides, again originally defined by the sides Δx, Δy and Δz are    xy    yz    xz V Strain 5.2.3 Cartesian strain Notice that the normal strains cause a change in volume of rectangular element, whereas the shear strain cause a change in shape V Strain Example: Due to a loading, the plate is deformed into the dashed shape shown in fig a Determine (a) the average normal strain along the side AB, and (b) the average shear strain in the plate at A relative to the and y axes V Strain Solutions Part (a) • Line AB, coincident with the y axis, becomes line after deformation, thus the length of this line is AB '  250  22  32  248.018 mm • The average normal strain for AB is therefore  AB avg AB ' AB 248.018  250    7.93 10 3 mm/mm (Ans) AB 240   • The negative sign indicates the strain causes a contraction of AB V Strain Solutions Part (b) • As noted, the once 900 angle BAC between the sides of the plate, referenced from the x, y axes, changes to θ’ due to the displacement of B to B’ • Since  xy     ' then  xy is the angle shown in the figure • Thus,  xy    arctg    0.121 rad (Ans)  250   V Strain Problem: The rigid beam is supported by a pin at A and wires BD and CE If the load P on the beam causes the end C to be displaced 10 mm downward, determine the normal strain developed in wires CE and BD ... The approximate lengths of the sides of the parallelepiped are 1   x x 1   y 1   z y z The approximate angles between sides, again originally defined by the sides Δx, Δy and Δz are... thus the length of this line is AB '  250  22  32  248. 018 mm • The average normal strain for AB is therefore  AB avg AB ' AB 248. 018  250    7.93 10 3 mm/mm (Ans) AB 240   • The. ..  ' then  xy is the angle shown in the figure • Thus,  xy    arctg    0 .12 1 rad (Ans)  250   V Strain Problem: The rigid beam is supported by a pin at A and wires BD and CE If the