Gantry Girder:

Wheel load on Gantry Girder, W1 Wr*(B-g)/2B

Where, Wr Weight of Trolley

B Distance b/w the gantry girders

g Dist. b/w the CG of the trolley and gantry

Weight of Crane girder Wc

Wheel load due to Crane girder weight, W2 Wc/4

Lateral Load (electrically operated) 10 % of wt of trolley & lifting load

Lateral Load (hand operated) 5 % of wt of trolley & lifting load

Longitudinal Load, Fg (Drag force) 5 % of wheel load

Reaction due to wheel load Fg*e/L

Moment due to wheel load (Fg*e/L)*bMoment due to wheel load (Fg*e/L)*b

Load Analysis

Longitudinal spacing of column, L 5.65 m

Centre to centre distance of gantry girder, B = 7.5 m

Wheel spacing, a 1.5 m

Edge distance, g 0.75 m

Weight of crane girder, Wc 1.8 ton

18 KN

Weight of trolley car, Wr 10 KN

Weight to be lifted, Wk 30 KN

Wheel Load due to crane girder, W1 18/4 KN

4.5 KN

Wheel load from car & lifting load, W2 (10+30)X(7.5-0.75)/2X7.5

18 KN

(10+30)X(0.75)/2X7.5

2 KN

Impact Load, W3 25% for mechanically operated

0.25X((30+10)X(7.5-0.75)/(2X7.5)+4.5)

Wheel load from car & lifting load, W2 (at

the other end)

5.625 KN

Impact Load, W3 (at other end) 25% for mechanically operated

0.25X((30+10)X(0.75)/(2X7.5)+4.5)

1.625 KN

Total wheel load, W W1+W2+W3

4.5+18+5.625

28.125 KN

Say 28.5 KN

Total wheel load, W (at other end) W1+W2+W3

4.5+2+1.625

8.125 KN

Say 8.5 KN

Lateral Load, Fl 10% of Wt of trolley & Lifting Load

0.1X(10+30)

4 KN

Drag Force, Fg 5 % of wheel load

(5/100)X28.5 (5/100)X28.5

1.425 KN

Drag Force, Fg (at other end) 5 % of wheel load

(5/100)X8.5

0.425 KN

Design for Bending Moment

The maxm BM on the beam occurs when one of the loads is at a dist,

X L/2-a/4

5.65/2-1.5/4 m

2.45 m

Maxm BM due to vertical load, M1 2W/L*(L/2-a/4)^2

(2X28.5/5.65)*(5.65/2-1.5/4)^2

60.5561947 KN-m

Reaction due to drag force, Ra Fg*e/L

1.425X(0.35+0.15)/5.65

0.126 KN

Bending Moment due to drag, M2 Ra*(L/2-a/4)

0.126X(5.65/2-1.5/4)

0.30896018 KN-m

Assuming selfweight + rail load , wg 1.5 KN/m

Bending moment due to self wt & rail load, M3= wg*L*x/2-wg*x^2/2

(1.5X5.65X2.45/2)-(1.5X2.45^2/2)

5.88 KN-m

Total design Bending moment, Mx M1+M2+M3

60.56+0.309+5.88

66.7451549 KN-m

Say 65 KN-m

Assume the section to be IS MB 400 whose properties are

Zx 1020 cm3

Ix 20458.4 cm4

Depth, D 0.4 m

web, tw 8.9 mm

flange, tf 16 mm

rmin 28.2 mmrmin 28.2 mm

Slenderness ratio L/rmin

5650/28.2

200.35461

tf/tw 16/8.9

1.79775281 < 2

d1/tw (400-2X16)/8.9

41.35 < 85

D/tf 400/16

25

using the above data, value of stress allowed, bc = 79 N/mm2 (IS : 800 - 1984)

Z(reqd) Mx/bc

65X10^6/79

822784.81 mm3

822.78481 cm3 < 1020 cm3

OK, SAFE

Check for Shear

Maxm Shear force occurs when one of the loads is at a distance D from the support.

V W X (L-D)/L + W X (L-D-a)/L

2 X W X(L-D-a/2)/L

2 X 28.5 X (5.65 - 0.4 - 1.5/2)/5.65

45.40 KN

Shear Stress, v V/D X t

45398/(400X8.9)

12.8 N/mm2

Per. Sheer Stress 0.4*fy

0.4X250

100 N/mm2

OK, SAFE

Check for deflection

The deflection is computed using 'conjugate beam method', by

placing loads summetric with the mid span.

a

L

W(L-a)/2EI

Bending Moment under the load W*(L-a)/2

28.5 X (5.65-1.5)/2

59.1375 KN-m

Reaction of conjugate beam, Ra 1/2(Area of (M/EI) diagram)

W*(L-a)/2EI*(1/2)*[((L-a)/2)+a]

W/8EI*(L^2-a^2)

deflection, (mid span) Ra*(L/2)-(W(L-a)/2EI)*1/2*((L-a)/2)*(L/2-2/3*(L-a)/2)-W((L-a)/2)*a/2*a/4

(W(L-a)/48EI)*(2L^2+2aL-a^2)

(28500X(5650-1500)/48X200000X204584000)*(2X5.65^2+2X1.5X5.65-1.5^2)10^6

4.73008126 mm

placing loads summetric with the mid span.

Permissible deflection L/750 (IS : 800 - 1984)

5650/750

7.53333333 OK, SAFE

Live Load due to crane on bracket:

1.5

2.85 t 2.85 t 0.65 KN/m

Ra 5.65 Rb

Reaction at one end due to this load

Ra (28.5X(5.65+(5.65-1.5))/5.65)+0.65X5.65/2

51.3 KN 51.3 KN

Say 55 KN

Moment 55X0.385+(25X0.55^2/2X0.3X(0.6+0.3)/2)

21.7 KN-m

Lateral Load 4.00 KN

Drag Force 1.425 KN

0.65 KN/m

1.5

Ra 5.65 Rb

Reaction at other end, Rb (8.5X(5.65+(5.65-1.5))/5.65)+0.65X5.65/2

16.6 KN

Say 18 KN

Moment 18X0.385+25X0.3X(0.6+0.3)/2X0.55^2/2

7.44 KN-m

0.85 t 0.85 t

Lateral Load 4.00 KN

Drag Force 1.425 KN

Dead Load due to crane on bracket (near end) It can be found out by putting the lifting load = 0Wheel load at near end 11.5 KN

Reaction on bracket (11.5X(5.65+(5.65-1.5))/5.65)+0.65X5.65/2

21.8 KN

Moment 21.78X0.385+25X0.55X0.3X(0.6+0.3)/2

10.24 KN-m

say 10.5 KN-m

(Same as above, to

be on higher side)