Cantilever Depression Analysis by Dr. S. D. More

Topic Expression for depression of cantilever Presented by Dr. S. D. More Assistant Professor & Head Department of Physics Deogiri College, Aurangabad

Case:II Expression for depression of cantilever by considering externally attatched load and weight of the cantilever itself Consider a cantilever AB fixed at end A. Let l be the length of cantilever and W1 be the weight of the cantilever. Let initially the cantilever be in the horizontal position.

Consider a small element of cantilever of length dx at a distance of x from fixed end (i.e. at point P) A load W is attatched to the free end B of the cantilever. The small element at point P is depressed through a small distance y due to load W . In the equilibrium position two moment of forces are at acting at point P. 1) External depression moment 2) Bending moment of beam 1) External depression moment at point P is due to weight of cantilever and externally attatched load W. The weight of cantilever affecting at point P will the weight of cantilever for length (l-x)

The weight of cantilever affecting at point P will the weight of cantilever for length (l-x) Total weight of cantilever = W1 Length of cantilever = l Weight per unit length of cantilever W = 1 l W l ( ) = Weight of cantilever of length ( - x) - x ...(1) 1 l l ( l ) is The weight of cantilever for section - x supposed l - x centre its at act to distance at i.e. of form point P. 2

The moment of force point at l weight to due P of cantilever = of length ( - x) weight of cantilever l distance of point of length ( - x) from P centre its W l - x l ( ) = - x 1 l 2 ( 2 ) 2 W - l x l = ...(2) 1 The moment of force of weight W about ( ) = P W - x ...(3) l

= Total depression moment about P Moment of force due to Moment + of force due to W weight of cantilever ( ) 2 W - x) l ( ) = + W - x) ...(4) 1 l 2 l Y = Bending moment of cantilever I ...(5) g R = Where = R Radius of curvature at P Ig Geometrica M. l I. of cantilever equilibriu At m External = depression moment bending moment

( ) 2 Y W - x) l ( ) = + I W - x) ...(6) 1 l g R 2 l ( Y ) ( ) 2 1 W - x) W - x) l l = + ...(7) 1 R 2 I Y I l g g However, 2 1 d y 2 = R dx 2 d y 2 W W ( ) ( ) 2 = + - x) - x) ...(8) 1 l l dx 2 Y I Y I l g g Integratin above g equation (8) w. r. t. x 2 d y 2 W W ( ) ( ) 2 = + dx - x) - x) dx 1 l l dx 2 Y I Y I l g g

dy W W ( ) ( ) 2 = + - x) dx - x) dx 1 l l dx 2 Y I Y I l g g dy W W ( ) ( ) 2 = + - x) dx - x) dx 1 l l dx 2 Y I Y I l g g dy W W ( ) = + + 2 2 ( 2x x ) dx - x) dx 1 l l l dx 2 Y I Y I l g g dy W W = + + 2 2 dx 2x dx x dx dx x dx 1 l l l dx 2 Y I Y I l g g dy W W = + + 2 2 dx 2 x dx x dx dx x dx 1 l l l dx 2 Y I Y I l g g 2 3 2 dy W x W x x = + + + 2 x - 2 x K 1 l l l 1 dx 2 Y I 2 3 Y I 2 l g g

3 2 dy W W x x = + + + 2 2 x - x x K ...(9) 1 l l l 1 dx 2 Y I 3 Y I 2 l g g = = Applying boundry condition find to out K at x and 0 y 0 1 dy = 0 dx = + + 0 0 = 0 K 1 K 0 1 3 2 dy W W x x = + + 2 2 x - x x ...(10) 1 l l l dx 2 Y I 3 Y I 2 l g g Integratin above g = equation w r. . t. to x 3 2 dy W W x x + + 2 2 dx x - x x dx 1 l l l dx 2 Y I 3 Y I 2 l g g

3 2 W W x x = + + 2 2 y x - x dx x dx 1 l l l 2 Y I 3 Y I 2 l g g 3 2 W W x x = + + 2 2 1 y x - x dx x dx l l l 2 Y I 3 Y I 2 l g g 3 W x = dx + 2 2 y x dx x dx 1 l l 2 Y I 3 l g 2 W x + dx x dx l Y I 2 g 2 2 3 4 2 3 W x x x W x x l l l = + + + y K 1 2 2 Y I 2 3 12 Y I 2 6 l g g Applying boundry condition find to K 2

= = K i. e. x 0, y 0 = + 0 + 0 0 0 2 = K 2 2 2 3 4 2 3 W x x x W x x l l l = + + y 1 2 Y I 2 3 12 Y I 2 6 l g g To calculate yB depression i.e. depression at end B, Substitute x=l 2 3 2 2 2 3 4 2 3 W W l l l l l l l l = + + y 1 l B 2 Y I 12 Y I 6 l g g 4 4 4 3 3 W W l l l l = + + y 1 l B 2 Y I 2 3 12 Y I 2 6 l g g + 4 4 4 3 3 W 6 4 W 3 l l l l = + y 1 B 2 Y I 12 Y I 6 l g g

4 3 W 3 W 2 l l = + y 1 l B 2 Y I 12 Y I 6 l g g 4 3 W W l = + y 1 B 2 Y I 4 Y I 3 l g g 4 3 W W l l = + y 1 B 2 Y I 4 Y I 3 l g g 3 3 W W l l = + y 1 B 8 Y I 3 Y I g g 3 3 W l = + y W 1 B 3 Y I 8 g

Depression of a beam supported at its end and loaded at center. Let a bar of unifrom cross-section be supported horizontally on two knife edges at A and B. Let AB = l and C be the center of AB. Let a load W be attatched to the end at its center, So that CC = y = Maximum depression of rod.

A normal reaction equal to W/2 is acting at point A and B in upward direction. Tangent drawn at C to arc is horizontal. C is considered as fixed point and C A can be considered as inverted cantilever of length l/2 loaded at end A with weight W/2. Thus the depression of cantilever C A = depression CC . The expreesion for depression of a cantilever loaded at its free end is given by, 3 Load (length) = y ...(1) I Y 3 g

W l = = In this case, load and length 2 2 3 W l 2 2 = y I Y 3 g 3 W l = y I 2Y 8 3 g 3 W l = y ...(2) 48 I Y g This is the expreesion for depression of a bar loaded at center.

Special Cases Case:I If the bar is rectangular having breadth b and depth d , then d b I g = 3 12 Equation of modified gets y as, 3 3 W 12 W l l = = y 3 48 Y d b 3 d b l 48 Y 12 3 W = y ...(3) 3 4 Y d b

Case:II If the bar is circular having crosection radius r, then 4 r = I g 4 3 3 W W 4 l l = = Depression , y 4 48 Y r 4 r 48 Y 4 3 W l = y ...(4) 4 12 Y r