Answer:
Support at left = 5.99 N Support at right =15.65 N
Maximum bending moment=33.619 KNm at 9.65 m from left
Minimum height, h=0.491304 m
Explanation:
Assuming a simply supported beam as attached
The sum of upward and downward forces are equal hence to obtain support reactions
Let reaction at the the beginning of dimension a be Ra and reaction at the end of dimension c be Rc
Ra+Rc=4+(3.6*4.9)=4+17.64=21.64
Taking sum of moments at the extreme left end support
(4.1+4.5+4.9)*Rc=4*4.1+(3.6*4.9)*(0.5*4.9+4.5+4.1)
13.5Rc=16.4+ 17.64*11.05=16.4+194.922=211.322
13.5Rc=211.322
Rc=15.65348148 rounded off as 15.65 N
Since Ra+Rc=21.64 as initially found, Ra=21.64-Rc=21.64-15.65=5.99
Maximum moment occurs when shear is zero
Equation for shear [tex]Ra-4-3.6x^{2}/2[/tex] where x is the distance from the left point where UDL starts
5.99-4=[tex]3.6x^{2}/2[/tex]
2*1.99=[tex]3.6x^{2}[/tex]
[tex]x^{2}[/tex]=2*1.99/3.6=3.98/3.6=1.105555556
x=[tex]\sqrt {1.105555556}[/tex]
x=1.05145402
x is approximately 1.05m from the start of UDL from left.
The point of maximum shear is at 4.1+4.5+1.05m=9.65m
Maximum moment is Ra(9.65)-4(9.65-4.1)-[tex]3.6*1.05^{2}/2[/tex] and substituting Ra=5.99
Maximum moment 5.99*9.65-(4*5.55)-[tex]3.6*1.05^{2}/2[/tex]=33.619 KNm
Maximum moment occurs at 9.65m from extreme left and is 33.619KNm
To get maximum stress
[tex]\sigma_{max}=\frac {M_{max}y}{I}[/tex] where I=[tex]\frac {bh^{3}}{12}[/tex] and y=h/2
[tex]\sigma_{max}=\frac {M_{max}h/2}{\frac {bh^{3}}{12} }=\frac {6M_{max}}{bh^{2}}[/tex]
[tex]h^{2}=\frac {6M_{max}}{b\sigma_{max}}[/tex] and b is given as 8.7cm
[tex]h^{2}=\frac {6*33.619*10^{3}}{8.7*10^{-3}*96*10^{6}}[/tex]
h=[tex]\sqrt \frac {6*33.619*10^{3}}{8.7*10^{-3}*96*10^{6}}[/tex]= 0.491304 m
h=0.491304 m
