Question

The 225 x 225 x 1,150-mm oak [E = 12 GPa) block (2) shown is reinforced by bolting two 5 x 225 x 1,150 mm steel [E = 200 GPa) plates (1) to opposite sides of the block. A concentrated load of 320 KN is applied to a rigid cap. Assume P=320 kN, L=1.15 m. Determine (a) the normal stresses in the steel plates (1) and the oak block (2). (b) the shortening of the block when the load is applied. Calculate the cross-sectional area of each steel plate, A1, and the cross-sectional area of the oak block, A.

Answer 1

I FBD of eigid cap i 320 KN Efy=0. - 2F, -f - 320=0 - 2f - f = 320 - ①

Now axial deformation of members are S1 F, L, 4 s. fol, - A, E, A E₂ The elongation of both members are equal .: F.L, - fu la - A, E, A, E Also L=4 = 150 mm .: f f Az. E А, Е, :. Substituting this in 0 - 2f, - f, Az. E₂ - 320 A, E, البدل A = 5x225 = 1125 mm² 4 area A = 225x225=50625 mm². 4 € = 200 GPa & E 2 12 Gla. 320. 50625 * 12 1125 200 -2f-2.7f= 320. -4.75 = 320 =-68.085 (KN) 0

- 2x(-68.085) - f = 320. 136-17-320= fa f = -183.83 (KN) : Nound stresses in each member is, 6,= -68.085 x 103 U25 6,= -60.52 (Hlmm²) 6. a 60.52 Compa) Negative means Compressive 6 = 60.52 (MPa) (c) And. .. Stress A, - 183.83 x103 50625 6.-3.63 (Nima?) 6 = -3.63 (MPa) Negative means 6 3 .63 (MPa) Compressive stress (c)

b The shortening of block is equal to elongation be the contraction of member (2) A, E, (-183.83x103) x (1150) (50625) x ( 12 x103 Nimm?) dl= -0.348 mm. Negative means shortening Shortenis is 0.348 mm. I shortening
Using stress and elongation formula to get the result.