2. Supports which resist forces in two directions. Fig Hinge. Rough Surface. Fig Rocker. Roller. Frictionless Surface

Transcription

1 4. Structural Equilibrium 4.1 ntroduction n statics, it becomes convenient to ignore the small deformation and displacement. We pretend that the materials used are rigid, having the propert or infinite stiffness. The necessar and sufficient condition for a bod ling in a plane to be in structural equilibrium is b satisfing the static equilibrium equations as follows F = F = M = 4. Supports Bodies are usuall supported b different tpes of supports. There are three tpes of supports: 1. Supports which resist forces in one direction onl e.g. Roller, Rocker and Frictionless support Roller Rocker Frictionless Surface Fig Supports which resist forces in two directions Hinge Fig. 4. Rough Surface

2 . Support which resist forces in all directions e.g. fied or rigid, or built-in-support F Fig. 4. M F 4. Free Bod Diagram n order to analse a rigid bod in equilibrium, it is necessar to idealize the two or three dimensional bod to a line diagram showing the forces acting on it. This diagram is referred to as Free Bod Diagram. The first major step in solving a problem in static equilibrium is to draw an accurate free bod diagram of the bod together with the forces and reactions acting on it. The following guideline should be followed in drawing the free bod diagram: Separate the bod to be analsed Draw the schematic diagram of the bod b showing its eternal boundar. ndicate all the eternal forces acting on the bod. ndicate all the reactions acting on the bod. ndicate all the dimensions and choice of coordinates aes (where necessar) on the diagram. 4.4 First Moment of rea First moment or area of an element about an ais in the plane of the area is given b the product of the area of the nd element and the perpendicular distance between the element and the ais. Centroid Centroids: The center of mass or centre of gravit ma visualized as the location of the resultant of a set of parallel forces. First Moment of rea about -ais (Statical Moment) is mathematicall given as: z C.G d Fig. 4.4

3 Q = d Total rea = d The position for the centroid for the total area is given as : = d Q = ; d = d Q = d Since for most structural applications the areas involved are regular geometric shapes, the integral becomes i i = ( ) i i = ( ) i i Eample4.1 Determine the centroid location of the T shape in the figure below. ll dimensions are in mm. 1 1 =7 6 = 15 Fig. E4.1(a) Solution: The shape consists of two rectangles with their centroids as shown in Fig. E4.1(a) Fig. E4.1 B smmetr, is = + = ( )( 75) + ( 15)( 6) = ( 75) + 15( 1) = 55m m 1 The centroidal location is shown in Fig. E4.1(b)

4 Centroidal aes =55 Fig. E4.1(b) Eample 4. rectangular plane section 7mm 9mm is having a square hole mm mm in it as shown in Fig. E4.. Locate the centroid of the plane area. 1 hole 4 Fig. E4.(a) Solution: The centroid along the and aes are located b subtracting the square hole from the rectangular section. 1 1 = 1 5( 7 )( 9 ) 5( )( ) = 7 ( 9 ) ( ) = 7 mm 4 5( 9 )( 7 ) 5 5( )( ) = 9 ( 7 ) ( ) = 4 mm The centroidal location is shown in Fig. E4.(b)

5 1 =7 hole C.G =4 Fig. E4.(b) 4. Second Moment of rea Moment of nertia is a mathematical concept that is used to quantif the resistance of various sections to bending or buckling. beam section with large Moment of nertia, value, will have smaller stress and deflections under a given load than one with a lesser value. long, slender column will not buckle laterall if of its cross-section is sufficient. Moment of inertia of rea of an element shown in Fig. 4.5 about an ais is defined as the product of the area of the element and the square of the perpendicular distance between the element and the ais. = = d d d n value has units of length Fig. 4.5 to the 4 th power. Hence, it is sometimes called Second Moment of rea. t ma help to understand the concept of moment of inertia if we draws an analog based upon real inertia due to motion and mass with equal area. magine the two shapes in the Fig.4.6 below to be cut out of heav sheet material and placed on an ale () so the could spin about it. Fig. 4.6(a) Fig. 4.6(b)

6 t would be much harder to start the first shape Fig. 4.6(a) and once moving much harder to stop than the second shape in Fig. 4.6(b) The same principle is involved when a skater spins on ice. With arms held close in, the skater will rotate rapidl, and with arms outstretched (creating more moment of inertia, ), the skater slows down. Eample 4. Find the Moment of nertia,. of a rectangle Solution: The rectangle is shown in Fig. E4.. Dividing the rectangle into strips parallel to the - ais, we obtain h/ h/ d = = h h d bd b b h h = = bh = 1 h h b Fig. E4. Eample 4.4 Find the moment of inertia about the a and aes for the shape shown in Fig. E4.4. ll dimensions are in mm.

7 Solution: The shapes could be divided into three rectangles to determine the. To compute the, a rectangle could be used for the entire area and then subtract two rectangular spaces from it as follows. = va lu es o f th e rec tan g les = rec tan g le rec tan g ular sp aces = + Web Flange bh bh = Web = + (8) mm 16( 1) = = gross sp ace Flange 6 4 = 1( 5) mm 46( ) = Parallel is Theorem Fig. E4.4 This provides a simple wa to compute moment of inertia about an ais parallel to centroid ais. Consider the moment of inertia of a small elemental area d with respect to ais as shown in Fig. 4.7 which could be written as = ' d

8 ' d C.G ' d Fig. 4.7 ' Since = + d, this could be epressed as ' = ( ' + d ) d = ' d + d ' d + d d The second integral represents the first moment of the area with respect to its own centroidal ais which must be zero. Hence, = + ' d This epression is known as the parallel ais theorem. Parallel ais theorem :The parallel ais theorem for the second moment states that the second moment of area about an ais is equal to the second moment area about a parallel ais through the centroid of the area plus the product of the area and the square of the perpendicular distance between the two aes. Eample 4.5 Determine the moment of inertia of the T-shape in Eample 4.1 with respect to its horizontal centroidal ais as shown in Fig. E4.5(a). = Fig. E4.5(a) Solution The T-shape could be divided into two rectangles with their centroids as shown in Fig

9 E.4.5(b) 1 d1=15 5 Fig. E4.5(b) Since neither rectangle has it centroid coincident in the centroid of the entire section, as shown in Fig. E4.5(b) we have [ ] [ ] = + d + + d b1h1 b h = + d d ( ) = ( 6 ) 7 5( )( 1 5) 1 5( 6 )( 5) 1 1 = 1. 1 mm 6 4 Eample 4.6 Determining the moment of inertia, for the retaining wall shown in Fig. E4.6. 1m 9m 1m 1m 4m Solution The retaining wall could be divided into a rectangle and two triangles. The moment of inertia with respect to the aes, is the computed as follows: [ ] [ ] = + + d = m rect tria ng le 9 1 = ( ) (. ) ( )(. )( ) 4

10 Eample 4.7 The section of a steel is as shown in Fig. E4.7. The section is smmetrical about -ais. Determine the moment of inertia of the cross-section with respect to -ais. 7.5mm 1mm 7.5mm 1mm 8mm 8mm Fig. E4.7 Solution Moment of inertia of (i) Outer Bo, = 16 = ( 16 )/1 = 686mm 4 (ii) nner Bo, = /1 = 51mm 4 ( iii) Outer Half Circle, = R 4 Π Π 4 R Π + = mm 4 4 Π Π 4 r ( iv) nner Half Circle, = r Π + = mm 4 ( v) Small hatched rectangle, = bh /1 + d

11 = 171mm 4 Moment of nertia about, = (686-51) + ( ) +(171) = mm 4