Determine

1) The horizontal thrust

2) The moment,radial shear and normal thrust at a section 6m from the left hand support

3) Draw B.M.D

Note: Since the arch is symmetrical h=10m

1) $Rx^n$ and horizontal thrust

$\sum M_A = 0 (\circlearrowleft+ve)$

$45 \times 10 \times \frac{10}{2} - V_B \times 20 = 0$

$\boxed{V_B = 112.5 KN}$

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

$V_A - 45 \times 10 + 112.5 = 0$

$\boxed{V-A = 337.5 KN}$

Since C is an internal hinge

Consider part CB

$BM_c = 0$

$112.5 \times 10 - H \times 10 =0$

$\boxed{H = 112.5KN}$

2) B.M, Radial shear and normal thrust:

Consider portion AC:

$BM_D = 337.5 \times 6 - 112.5xy - 45 \times 6 \times \frac{6}{2}$ ------- (1)

But,

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

Considering C as origin:

$R = \frac{l^2}{8h} + \frac{h}{2} = 10$

$x = 4m$ From C

$y = \sqrt{10^2 - 4^2} - 10+10$

$y = 9.16m$

Putting in eqn (1):

$BM_D = 337.5 \times 6 - 112.5 \times 9.16 - 45 \times 6 \times \frac{6}{2}$

$\boxed{B.M_D = 184.5 KN}$ [Sagging]

Consider portion BC

$B.M_E = 112.5 \times (10 - x) - 112.5 xy$ - (1)

$M_E$ to be Maximum $\frac{d}{dx} M_E = 0$

$y = \sqrt{10^2 - x^2} - 10 + 10$

$1125 - 112.5x - 112.5 \times (\sqrt{10^2 - x^2} - 10 + 10)$

$-112.5 + \frac{112.5}{\sqrt{10^2 - x^2}} = 0$

By solving x = 7.07 m from C

Put x = 7.07 m in equation (1)

$112.5 \times (10 - x) - 112.5 xy$

$112.5 \times (10 - 7.07) - 112.5 \sqrt{10^2 - 7.07^2}$

$ = -465.99 KN.m$

$BM_E = 465.99 KN.m$ - [Hogging]

3) Radial Shear and Normal thrust:

$N_T = H cos\theta + v sin\theta$

$N_T = 112.5 cos(23.65) + 67.5 sin(23.65)$

$N_T = 130 KN$

$F = Vcos\theta - Hsin\theta$

$F = 67.5 cos(23.65) - 112.5 sin(23.65)$

$F = 16.8 KN$

$V = 337.5 - 45 \times 6 = 67.5 KN$

$H = 112.5 KN$

$tan\theta = \frac{dy}{dx} = \frac{x}{y} = \frac{4}{9.12}$

$tan\theta = 0.438$

$\theta = tan^{-1}(0.438) = 23.65^\circ$

4) B.M.D