A timber beam 100mm wide and 200mm deep is strengthened by steel plate 100 mm wide and 10mm thick at the top. The beam is fixed at one end and free at another end and carries a UDL of 8 KN/m over the length of 2m. calculate the maximum stress in timber and steel. Take Es=20 Et

1. The 100mm wide timver strip is replaced by an equivalent

$=\frac{E_t}{E_s}*100=\frac{1}{20}*100=50mm$

($E_s=20E_t i.e \frac{E_s}{E_t}=20$ given)

Properties of the equivalent steel section.

Now, $\bar y_t=\frac{\sum ay}{a}=\frac{1105*10^3}{11*10^3}=100.45mm$ ( from (1) - (1) )

2. Moment of inertia of section about axis (1)-(1)

$I_{1-1}=\sum I_{self}+\sum_{ay^2}$

=$33341666.63+121025*10^3$

$I_{1-1}=154.366*10^6 mm^4$

$I_{x-x}=I_{1-1}-A y_{-2}=154.366*10^6-11*10^3*(100.45)^2=43.374*10^6 mm^4$

3. Since, the beam is fixed at one end and free at another end and carries a UDL of 8KN/m over the length of 2m

Now, bending moment M at A is $=8*2*)\frac{2}{2})=16KNm=16*10^6$

4. a) Maximum bending stress in steel$=\frac{16*10^6}{43.37*10^6}*100.45=37.054N/mm^2$

$(\frac{M}{I}=\frac{f}{y} i.e f=\frac{M}{I}*y)$

b) Maximum bending stress in timber$=\frac{10*10^6}{43.374*10^6}*-(100.45-10)*\frac{1}{20} $

$(f=\frac{M}{I}*y*y*m) where m=\frac{E_b}{E_s}$

c) Maximum bending stress in timber=

$=\frac{10*10^6}{43.374*10^6}*109.55*\frac{1}{20}=2.02N/mm^2 (tensile)$

i.e maximum bending stress in timber is $2.02 N/mm^2$ which is tensile

Note : Since equivalent section is in steel, so while calculating the stress for timber always multiply by the modular ratio value m.