A circular arched rib of 20m spanwith central rise of 4m is hinged at the crown and springings. It carries a point load of 100kN at 5m from the left hand hing. Calculate:

(1) Reaction

(2) Horizontal thrust

(3) The maximum B.M: positive & negative

(1)$rx^n$

$\sum M_A=0$

$100\times 5-V_B\times 20=0$

$V_B=25kN (\uparrow)$

$\sum F_y=0(\uparrow +ve)$

$V_A-100+25=0$

$V_A=75kN(\uparrow)$

C is an internal hinge

Consider part CB

$B.M_c=0$

$25\times 10-H\times 4=0$

$H=62.5kN(\leftarrow)$

$H_A=H_B=62.5kN$

(2) Bending moment calculation:

Consider portion CA

$B.M_D=75\times 5-62.5\times y$

Note:- In circular arch

$y=\sqrt{R^2-x^2}-R+h$

$R=\frac{l^2}{8h}+\frac{h}{2}$

$y=\sqrt{R^2-x^2}-R+h$ -------------considering as origin

$R=\frac{l^2}{8h}+\frac{h}{2}=\frac{20^2}{8\times 4}+\frac{4}{2}=14.5m$

$y=\sqrt{14.5^2-5^2}-(14.5-4)$

$=3.11m$

At x=5m from C, y=3.11m

$B.M_D=75\times 5-62.5\times 3.11$

$=180.63kNm$

Max positive B.M=180.63kN.m

Consider portion CB

$BM_E=25\times(10-x)-62.5\times y$

$=-62.5y+25(10-x)$

$=-62.5[\sqrt{210.25-x^2}-10.5]+25[10-x]$

$=-62.5\sqrt{210.25-x^2}-10.5]+656.25+250-25x$ ------(1)

$M_x$ to be maximum $\frac{d}{dx}M_x=0$

$\frac{-62.5}{2\sqrt{210.25-x^2}}\times {-2x}+0-25=0$

By solving

$x=5.38m$

Put in equation (1)

$=-62.5\sqrt{210.25-5.38^2}+656.25+250-25\times 5.38$

$B.M_E=-69.811kN.m$

Max negative B.M=69.811kN.m

(3)B.M.D:-