3 Determine the maximum stresses in the a wood and b steel u

3. Determine the maximum stresses in the (a) wood and (b) steel using the transformed section method if Mall 40 ft-k Wood: Ew = 2 x 106 psi Steel: E, 30 x 10 psi 2\" Steel 1\" 8\" Wood 0.5\" Steel 6\"

Solution

Let us convert the entire section into equivalent wood section

modular ratio =n= E_s/E_w = 30*10⁶ / 2*10⁶ = 15

equivalent width of steel bar at top = 2*15=30\"

equivalent width of steel at bottom = 6*15=90\"

Now let us determine the depth of neutral axis of the transformed section

let the depth of neutral axis be at d measured from topmost face

d = [(30*1*0.5)+(6*8*5)+(90*0.5*9.25)]/[(30*1)+(8*6)+(0.5*90)]=5.46 in

moment of inertia of the cross section about neutral axis can be determined using parallel axes theorem

moment of inertia=I=(30*1³/12) +30*(5.46-0.5)² + 6*8³/12 + (6*8)*(5.46-5)² + 90*0.5³/12 + (90*0.5*)*(9.25-5.46)²

= 1654 in⁴

bending stress is given by My/I where y is the distnace from the neutral axis

M=40 kip-ft = 480 kip-in

maximum bending stress in wood = 480*(5.46-1)/1654=1.29 ksi

maximum bending stress in steel = n*My/I = 15*480*5.46/1654=23.77 ksi