Question: A 3 hinged parabolic arch with rise $H_m$ and span 20m carries UDL of 20KN/m from the left half.It also carries a point load of 50 KN on the crown cal
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1) $rn^{n}$ and horizontal thrust

2) B.M at $H_m$ from left hand

3) Normal trust (N) and radial shear (F) at $H_m$ from left hand

4) Max +ve and -ve B.M

5) Draw B.M.D

Subject : Structural Analysis 1

Topic : Hinge Arches

Difficulty : Medium

sa1(74) • 826 views
 modified 12 months ago by Sanket Shingote ♦♦ 250 written 15 months ago by
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1. Support $rx^{n}$:

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

50 x 10 + (20 x 10) x 5 - $V_B$ x 20 = 0

20$V_B$ = 50 x 10 + 200 x 5

$\boxed{V_B = 75 KN}$

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

$V_A$ - 50 - 200 + 75 = 0

$\boxed{V_A = 175 KN}$

C is an internal hing consider part (CB)

$B.M_c = 0$

75 x 10 - $H_B$ x H = 0

$H_B = \frac{750}{4} = 187.5 KN$

$\boxed{H_A = H_B = 187.5 KN}$

2. B.M @ $4 m$ from A

With A as origin {D is situated at $4m$ from A}

$y = \frac{4h}{l^{2}}(lx - x^2)$

$y = \frac{4 x 4}{20^2}(20 x 4 - 4^2)$

$\boxed{y = 2.56m}$

consider part DA

$B.M_D$

$B.M_D = 175 \times 4 - 187.5 \times 2.56 - 20 \times 4 \times \frac{4}{2}$

$B.M_D = 60 KN.m$

B.M @ $4m$ from A = 60 KN.m

3. N & F @ $4m$ from A

Net unbalance vertical shear at D

Consider part (DA) is

$175 \uparrow - 20 X 4 \downarrow$

= 95KN

$N_D = 187.5 cos\theta \swarrow + 95 sin\theta\swarrow$

$N_D = 187.5 cos (25.64) + 95 sin (25.64)$

$\boxed{N_D = 210.14 KN\swarrow}$

$F_D = 95 cos\theta\searrow - 187.5 sin\theta\nwarrow$

$F_D = 95 cos (25.64) - 187.5 sin (25.64)$

$\boxed{F_D = 4.51KN\searrow}$

To find $\theta$

$y = \frac{4h}{l^2} (lx - x^2)$

$y = \frac{4 X 4}{20^2} (20x - x^2)$

Diff w.r.t to x

$\frac{dy}{dx} = \frac{1}{25}[20-2x]$

$\frac{dy}{dx} = \frac{1}{25} (20 - 2(4)) AT x = 4m$

$tan\theta = 0.48$

$\theta = tan^{-1}(0.48)$

$\boxed{\theta = 25.64^\circ}$

4. Max +ve and -ve B.M

Take section x - x distance x from A.

Consider part [CA]:

$B.M_x = 175 \times x - 187.5 \times y - 20 x \frac{x}{2}$ ------------- (1)

We know that

$y = \frac{4h}{l^2}(lx - x^2)$

$y = \frac{4 X 4}{20^2} (20x - x^2)$

$y = \frac{1}{25}(20x - x^2)$

Put in equation (1) we get

$B.M_x = 175 \times x - 187.5 \times \frac{1}{25} (20x - x^2) - 10x^2$

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

$175 - \frac{187.5}{25}[20 - 2x] - 20x = 0$

$\boxed{x = 5m}$

Put x = 5 min $M_x$ equation

$M_x = 175 \times 5 - 187.5 \times \frac{1}{25} (20 \times 5 - 5^2) - 10 \times 5^2$

$\boxed{B.M_x = M_{max} = 62.5 KN.m}$

Max positive BM = 62.5 KN.m occuring at 5m from A span CA.

Consider part [CB]

$B.M_x = 75 \times x - 187.5 \times y$

$B.M_x = 75 \times x - 187.5 \times \frac{1}{25} (20x - x^2)$ ---------- (1)

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

$75 - \frac{187.5}{25} (20 - 2x) = 0$

$x = 5m$

Put x= 5m in equation (1)

$75 \times 5 - \frac{187.5}{25} (20 \times 5 - 5^2)$

$M_x = -187.5 KN.m$

Max negative $BM_x = -187.5 KN.m$ occuring at 5m from span CD.

5. B.M.D