Answer:
the minimum width is b= 0.1414m = 141mm
Explanation:
]given,
L= 4.25
w₀ = 5.5kN/m,
allowable bending stress = 7MPa
allowable shear stress = 875kPa
h/b = 0.67
b = ?
for a linearly distributed load, with maximum load intensity, w₀ of 5.5kN/m,
the maximum moment, M exerted by the timber is =
given h/b = 0.67, i.e h=0.67b
allowable bending stress = = 7000kPa
7000 = (6*6.37)/ (b *(0.67b)² ) = 38.21/0.449b³
3080b³=38.21
b³ = 38.21/3080 = 0.0124
b = 0.232m
h=0.67b = 0.67* 0.232 = 0.155m
for allowable shear stress = (3V)/(2bh)
875 = (3V)/(2bh) = (3x 7.79)/(2xbx0.67b)
875 = 23.375/1.34b²
1172.5 b²= 23.375
b² =0.0199
b= 0.1414m
h=0.67b = 0.67* 0.1414 = 0.095m
the minimum width is b= 0.1414m = 141mm
