The pin displacements of the truss in Figure P260 were compu

The pin displacements of the truss in Figure P2.60 were computed by the finite-element method. The displacements in x and y directions given by u and v are given in Table P2.60. Determine the axial strains in members AB, BF, FG, and GB.

Solution

For AB,

uA = 0, uB = 12.6 mm => del_u = |12.6 - 0| mm = 0.0126 m (tensile as length is increasing)

vA = 0, vB = -24.48 mm => del_v = |24.48 -0| mm = -0.02448 m

Axial direction for AB is x.

Hence axial strain in AB = del_u / len_AB = 0.0126 m/ 2 m = 0.0063.

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For BF,

uB = 12.6 mm, uF = -8.4 mm => del_u = |(-8.4) - 12.6| mm= 21 mm

vB = -24.48mm, vF = -28.68 mm => del_v = |-28.68 - (-24.48) | = 4.2 mm = 0.0042 m (tensile)

BF axial direction is y.

Hence, axial strain in BF = del_v/len_BF = 0.0042 / 2 = 0.0021.

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For FG, axial direction is x.

uG = 0, uF = -8.4 mm => del_u = |-8.4 - 0| mm = 0.0084 m (Compressive as length is decreasing)

Hence, axial strain in FG = - del_u/ len_FG = -0.0084/2 = - 0.0042 (Compressive and hence negative)

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For GB, the axial direction is 45 deg to x-y.

uG = 0 mm, uB = 12.6 mm => del_u = |12.6 - 0| mm = 0.0126 m (Tensile)

vG = 0 mm, vB = -24.48 mm = > del_v = |-24.48 - 0| MM = 0.02448 M (Compressive)

Along axis, delta displacement is given by

del_u/cos(45) + del_v/sin(45) = 1.414 * (0.0126 - 0.02448) (taking sign for tension/compression)

= - 0.01188*1.414

Hence, axial strain in GB is -0.01188*1.414 / (2*1.414) = -0.00594 (Compressive)