SLOPE AND DEFLECTION FOR A CANTILEVER BEAM SUBJECTED TO EXTERNAL MOMENT Fig 1 Fig 2 Consider a cantilever beam PQ of length L subjected to external moment M on the free end side of the beam. Fig 2 show the M/EI diagram for the beam subjected to external moment. The effect of moment remains same throughout the beam; hence there is invariability in M/EI diagram.        Slope at the free end = Area of M/EI diagram ϴQ = (L) (-M/ EI) ϴQ = (-ML/ EI) Deflection at Q = (Area of M/EI diagram) (Centroidal distance from Q to O) YQ = (L) (-M/ EI) (1/2L) YQ= -ML2 ..read more

CANTILEVER BEAM SUBJECTED TO UNIFORMLY DISTRIBUTED LOAD THROUGHOUT THE ENTIRE SPAN Fig 1 Fig 2: M/EI diagram Consider a cantilever beam subjected PQ (shown in fig 1) of span L, subjected to uniformly distributed load of w/m throughout the entire span. Fig 2 shows bending moment diagram of the cantilever beam with uniformly distributed load throughout the span. Slope at the free end = Area of M/EI diagram (As per 1stmoment area theorem) Area of parabolic diagram = (1/3) (base) (height) ϴQ = (1/3) (L) (-wL2/ 2EI) Therefore, ϴQ = -Wl3/ 6EI ϴQ = Wl3/ 6EI rad (clockwise with tangent from P ..read more

CANTILEVER BEAM SUBJECTED TO POINT LOAD AT FREE END Fig 1 Fig 2: Deflected shape of beam Fig 3: M/EI diagram of a beam Consider a cantilever beam PQ (fig 1) of span L subjected to point load of magnitude W KN at free end. Fig 2 shows the deflected shape of the beam.Fig 3 shows bending moment diagram of the cantilever beam with concentrated load. Let ϴ be the slope and y is the deflection for the deflected beam. Slope at the free end = Area of M/EI diagram (As per 1stmoment area theorem) ϴB = ½ (L) (WL/EI) Therefore, ϴB = (WL2/2EI) Consider the M/EI diagram in which O is the centroi ..read more

SIGN CONVENTION FOLLOWED IN MOMENT AREA METHOD FOR DETERMINING SLOPE AND DEFLECTION. 1.     For Right hand side of the support ,anticlockwise moments are taken as positive ,clockwise moments are taken as negative and vise versa in case of left hand side support. 2.   Slope from Right hand side is taken as positive when it makes anticlockwise angle with left hand side tangent and vise versa with slope of left hand side tangent. 3.     Deflection is taken as positive if the right hand side tangent is above the left hand side tangent and vise ..read more

2ndMOMENT AREA THEOREM Consider the simply supported beam AB subjected to the load W. Let C and D be the two points between the supports A and B in order to determine the deflection for elemental length. Let ∆ be the deflection between the two points C and D. Let X be the distance from D to the meeting point of tangent. Let ϴCD be the angle between the tangents drawn from points C and D. From property of circles, Referring to the figure ∆ = x (ϴCD) From 1st moment area theorem W k t ϴCD = C∫D   (M/ E I) (dx) ∆ = C∫D   (M/ E I) (x) (dx) Therefore 2nd  theorem of mome ..read more

1ST THEOREM OF MOMENT AREA METHOD Consider a simply supported beam of span L with supports at A and B, subjected to point load of magnitude W. Fig 1     Fig 2    Fig 3 Consider figure 2, which indicates the deflected shape of the simply supported beam subjected to point load. Let C and D be the two points between the supports A and B in order to determine the slope for elemental length. Let dx be the elemental length between CD to determine the slope value which resembles shape of an arc and projected to meet at point O making an angle dϴ. Let R be the radi ..read more

MOMENT AREA METHOD Introduction 1.     Whenever a structure subjected to external load, due to action of external load on the structure, beyond elastic limit the structure will deform with an eccentric distance with reference to its initial position. 2.     The deformation values are most important to be known in order to design a structure. 3.     The deformation in a structure should be within the range and if its values are large then it causes crack and damage to the structure. 4.     The most important facto ..read more

LENGTH OF CABLE SUBJECTED TO UNIFORMLY DISTRIBUTED LOAD Consider the figure in which cable of span l is subjected to uniformly distributed load of w/m throughout the entire span. Since cable is a flexible structure, it deflects in a parabolic way when subjected to udl. Let h is the central dip of the cable. The equation of cable is considered by taking point C as origin. We know that the equation for dip for a parabola is given by Y = 4hx (l -x)/ (l)2 ..........(1)       Considering a section X- X of span x, for which the deflected length of the cable has t ..read more

CABLES SUBJECTED TO UNIFORMLY DISTRIBUTED LOAD Consider a cable of span L, subjected to uniformly distributed load of w/m throughout the entire span. Let A and B are the two pinned supports which have the vertical reactions Va and Vb, horizontal reaction H at both the ends. Let h is the central dip (vertical distance) of the cable. Due to symmetry, the reactions Va and Vb are equal Therefore, Va = Vb = wL/2 Taking moment about point C to determine the horizontal thrust Va (L/2) – H(h) -w(L/2) (L/4) = 0 (Note: The value of the moment is taken as zero, since the cable structure will alwa ..read more

NUMERICAL ON THE DETERMINATION OF TENSILE FORCES WHEN CABLE SUBJECTED TO THE CONCENTRATED LOAD A cable supported at its ends with span 40m apart carries a load of 20KN,10KN and 12KN at the distances of 10m, 20m and 30m respectively from left support. If the vertical distance of the point where the 10KN load is carried is 13m below the level of end supports. Determine  1. Support reactions  2. Tension forces at different parts of the cable 3. Total length of cable Step 1: Determination of the support reactions V a + V b = 20 + 10 + 12= 42KN…. (1) Taking moment about support ..read more