Beam and Frame Analysis: Force Method, Part I by S. T. Mau 115 For the thrust diagram, we designate tension force as positive and compression force as nagative. For shear and moment diagrams, we use the same sign convention for both beams and columns. For the vertically orientated columns, it is customary to equate the “in-side” of a column to the “down-side” of a beam and draw the positive and nagetive shear and moment diagrams accordingly. Thrust, shear and moment diagrams of the example problem. Example 10. Analyze the loaded frame shown and draw the thrust, shear and moment diagrams. 3 m 3 m 4 m 4 m 3 m 3 m 4 m 4 m 3 m 3 m 4 m 4 m T V M 3.75 kN -3.75 kN -5 kN -3.75 kN 5 kN 5 kN 15 kN-m -15 kN-m 15 kN-m -15 kN-m Beam and Frame Analysis: Force Method, Part I by S. T. Mau 116 Statically determinate frame example problem. Solution. The inclined member requires a special treatment in finding its shear diagram. (1) Find reactions and draw FBD of the whole structure. FBD for finding reactions. ∑M c =0, 60(2)− R(8)=0, R=15 kN (2) Draw the thrust, shear and moment diagrams. Before drawing the thrust, shear, and moment diagrams, we need to find the nodal forces that are in the direction of the axial force and shear force. This means we need to decompose all forces not perpendicular to or parallel to the member axes to those that are. The upper part of the figure below reflects that step. Once the nodal forces are properly oriented, the drawing of the thrust, shear, and moment diagrams is achieved effortlessly. 5 m 3 m 4 m 15 kN/m 5 m 3 m 4 m 60 kN R =15 kN 15 kN 60 kN c Beam and Frame Analysis: Force Method, Part I by S. T. Mau 117 Thrust, shear, and moment diagrams of the example problem. 60 kN 48 kN 36 kN 15 kN 12 kN 9 kN 7.2 kN/m 9.6 kN/m 12 kN 9 kN 39 kN 48 kN 60 kN 15 kN 15 kN 60 kN T V -12 kN -48 kN -60 kN 75 kN-m 9 kN -39 kN 15 kN M -75 kN-m 75 kN-m 48 kN/5m=9.6 kN/m 36 kN/5m=7.2 kN/m 0.94 m 5m (9)/(39+9) =0.94 m 4.22 kN-m M (x=0.94)=(9)(0.94)-(9.6)(0.94) 2 /2=4.22 kN-m 0.94 m 5 m Beam and Frame Analysis: Force Method, Part I by S. T. Mau 118 Problem 2. Analyze the beams and frames shown and draw the thrust (for frames only), shear and moment diagrams. (1) (2) (3) (4) (5) (6) (7) (8) (9) (10) Problem 2. Beam Problems. 3 m5 m 10 kN 3 m 5 m 10 kN 3 m5 m 3 kN/m 3 m 5 m 3 kN/m 3 m5 m 10 kN-m 3 m 5 m 10 kN-m 6 m 10 kN-m 6 m4 m 2 kN/m 5@2m=10m 10 kN 5@2m=10m 2 kN/m 4 m Beam and Frame Analysis: Force Method, Part I by S. T. Mau 119 (11) (12) (13) (14) (15) (16) Problem 2. Frame problems. 5 m 5 m 10 kN 5 m 5 m 10 kN-m 5 m 5 m 10 kN 5 m 5 m 10 kN-m 5 m 5 m 10 kN 5 m 5 m 10 kN-m Beam and Frame Analysis: Force Method, Part I by S. T. Mau 120 121 Beam and Frame Analysis: Force Method, Part II 5. Deflection of Beams and Frames Deflection of beams and frames is the deviation of the configuration of beams and frames from their un-displaced state to the displaced state, measured from the neutral axis of a beam or a frame member. It is the cumulative effect of deformation of the infinitesimal elements of a beam or frame member. As shown in the figure below, an infinitesimal element of width dx can be subjected to all three actions, thrust, T, shear, V, and moment, M. Each of these actions has a different effect on the deformation of the element. Effect of thrust, shear, and moment on the deformation of an element. The effect of axial deformation on a member is axial elongation or shortening, which is calculated in the same way as a truss member’s. The effect of shear deformation is the distortion of the shape of an element that results in the transverse deflections of a member. The effect of flexural deformation is the bending of the element that results in transverse deflection and axial shortening. These effects are illustrated in the following figure. dx x T V M dx dx dx Beam and Frame Analysis: Force Method, Part II by S. T. Mau 122 Effects of axial, shear, and flexural deformations on a member. While both the axial and flexural deformations result in axial elongation or shortening, the effect of flexural deformation on axial elongation is considered negligible for practical applications. Thus bending induced shortening, ∆ , will not create axial tension in the figure below, even when the axial displacement is constrained by two hinges as in the right part of the figure, because the axial shortening is too small to be of any significance. As a result, axial and transverse deflections can be considered separately and independently. Bending induced shortening is negligible. We shall be concerned with only transverse deflection henceforth. The shear deformation effect on transverse deflection, however, is also negligible if the length-to- depth ratio of a member is greater than ten, as a rule of thumb. Consequently, the only effect to be included in the analysis of beam and frame deflection is that of the flexural deformation caused by bending moments. As such, there is no need to distinguish frames from beams. We shall now introduce the applicable theory for the transverse deflection of beams. Linear Flexural Beam theory—Classical Beam Theory. The classical beam theory is based on the following assumptions: (1) Shear deformation effect is negligible. (2) Transverse deflection is small (<< depth of beam). Consequently: ∆ Beam and Frame Analysis: Force Method, Part II by S. T. Mau 123 (1) The normal to a transverse section remains normal after deformation. (2) The arc length of a deformed beam element is equal to the length of the beam element before deformation. Beam element deformation and the resulting curvature of the neutral axis(n.a). From the above figure, it is clear that the rotation of a section is equal to the rotation of the neutral axis. The rate of change of angle of the neutral axis is defined as the curvature. The reciprocal of the curvature is called the radius of curvature, denoted by ρ . ds dè = rate of change of angle =curvature ds dè = ρ 1 (6) For a beam made of linearly elastic materials, the curvature of an element, represented by the curvature of its neutral axis , is proportional to the bending moment acting on the element. The proportional constant, as derived in textbooks on Strength of Materials or Mechanics of Deformable bodies, is the product of the Young’s modulus, E, and the moment of inertia of the cross-section, I. Collectively, EI is called the sectional flexural rigidity. M(x)= k )( 1 x ρ , where k = EI. n.a ds dx d θ d θ ρ ds Beam and Frame Analysis: Force Method, Part II by S. T. Mau 124 Re-arrange the above equation, we obtain the following moment-curvature formula. EI M = ρ 1 (7) Eq. 7 is applicable to all beams made of linearly elastic materials and is independent of any coordinate system. In order to compute any beam deflection, measured by the deflection of its neutral axis, however, we need to define a coordinate system as shown below. Henceforth, it is understood that the line or curve shown for a beam represents that of the neutral axis of the beam. Deflection curve and the coordinate system. In the above figure, u and v are the displacements of a point of the neutral axis in the x- and y-direction, respectively. As explained earlier, the axial displacement, u, is separately considered and we shall concentrate on the transverse displacement, v. At a typical location, x, the arc length, ds and its relation with its x- and y-components is depicted in the figure below. Arc length, its x- and y-component, and the angle of rotation. The small deflection assumption of the classical beam theory allows us to write Tan θ ~ θ = dx dv = v’ and ds = dx (8) where we have replaced the differential operator dx d by the simpler symbol, prime ’. A direct substitution of the above formulas into Eq. 6 leads to y , v x , u v(x) x , u y , v dv dx ds θ [...]... of diagrams below Readers are encouraged to verify all numerical results P Pa Reactions 2a a a P/2 P/2 Pa/2 Moment Diagram −Pa Pa/2EI Pa/EI Conjugate Beam 2a a a Pa/2EI Pa/EI Reactions 2a a a 5Pa2 /12 EI 11 Pa2 /12 EI a/6 Shear(Rotation)Diagram ( Unit: Pa2 /EI ) 1/ 6 1/ 12 1 Moment(Deflection) Diagram ( Unit: Pa3 /EI ) 1/ 3=−0.33 − 27 01/ 77 76=−0.35 Solution process for rotation and deflection diagrams 13 1... process for rotation and deflection diagrams 13 1 5 /12 Beam and Frame Analysis: Force Method, Part II by S T Mau Problem 3 Draw the rotation and deflection diagrams of the loaded beams shown EI is constant in all cases (1) 1 kN-m L/2 L/2 (2) 1 kN L/2 L/2 (3) Mo= 1 kN-m L/2 L/2 (4) 1 kN L/2 L/2 (5) 2Pa 2a a Problem 3 13 2 a Beam and Frame Analysis: Force Method, Part II by S T Mau Energy Method – Unit Load... and Eq 2, which are reproduced below in equivalent forms V= ∫ − qdx (1) M= ∫∫ − qdxdx (2) Clearly the operations in Eq 11 and Eq 12 are parallel to those in Eq 11 and Eq 12 If we define 12 6 Beam and Frame Analysis: Force Method, Part II by S T Mau − M as “elastic load” in parallel to q as the real load, EI then the two processes of finding shear and moment diagrams and rotation and deflection diagrams...Beam and Frame Analysis: Force Method, Part II by S T Mau 1 dè dè = = = θ’ = v” ds ρ dx (9) which in turn leads to, from Eq 7, M EI = 1 ρ = v” (10 ) This last equation, Eq 10 , is the basis for the solution of the deflection curve, represented by v(x) We can solve for v’ and v from Eq 9 and Eq 10 by direct integration Direct Integration If we express M as a function of x from the moment diagram,... Reactions 7a2Mo/6EI 2aMo/3EI AMo/3EI 5aMo/3EI Solution process to find tip rotation and deflection At the right end (tip of the real beam): Shear = 5aMo/3EI θ = 5aMo/3EI Moment = 7a2Mo/6EI v = 7a2Mo/6EI Example 14 Draw the rotation and deflection diagrams of the loaded beam shown EI is constant P 2a a a Beam example on rotation and deflection diagrams 13 0 Beam and Frame Analysis: Force Method, Part II... because the effect of shear and axial forces on deflection is negligible We introduce a new entity, strain energy, U, which is defined as the work done by internal forces Then Wint = U = Σ ∫ 1 1 M 2 dx Mdθ = Σ ∫ 2 2 EI 13 3 (15 ) Beam and Frame Analysis: Force Method, Part II by S T Mau where the summation is over the number of frame members, and the integration is over the length of each member For... expression is derived as follows The angle of rotation of an infinitesimal element induced by a pair of internal moments is illustrated in the figure below x dx dθ M M dx Change of angle induced by internal moments The change of angle is related to the internal moment, according to Eq 7 and Eq 9, by dθ = Mdx EI which leads to Eq 15 Example 15 Find the rotation at the tip of the beam shown EI is constant, and... deflection diagram ( the moment diagram of the conjugate beam) MoL2 /2EI Moment(Deflection) diagram indicating upward deflection Example 13 Find the rotation and deflection at the tip of the loaded beam shown EI is constant Mo 2a a Find the tip rotation and deflection Solution The solution is presented in a series of diagrams below 12 9 Beam and Frame Analysis: Force Method, Part II by S T Mau Mo Reactions... versa 12 7 Beam and Frame Analysis: Force Method, Part II by S T Mau V=0 M=0 V≠0 M≠0 V=0 M=0 θ=0 v=0 θ≠0 v≠0 θ≠0 v=0 θ≠0 v=0 V=0 M=0 V≠0 M≠0 V≠0 M=0 V≠0 M=0 V≠0 M≠0 V≠0 M=0 V≠0 M=0 θ=0 v=0 θ≠0 v≠0 V=0 M=0 V≠0 M≠0 Real Conjugate Conversion from a real beam to a conjuagte beam We can now summarize the process of constructing of the conjugate beam and drawing the rotation and deflection diagrams: (1) Construct... integration Direct Integration If we express M as a function of x from the moment diagram, then we can integrate Eq 10 once to obtain the rotation θ = v’= ∫ M (11 ) dx EI Integrate again to obtain the deflection v= ∫∫ M (12 ) dxdx EI We shall now illustrate the solution process by two examples Example 11 The beam shown has a constant EI and a length L, find the rotation and deflection formulas Mo x A cantilever . kN 5@2m =10 m 2 kN/m 4 m Beam and Frame Analysis: Force Method, Part I by S. T. Mau 11 9 (11 ) (12 ) (13 ) (14 ) (15 ) (16 ) Problem 2. Frame problems. 5 m 5 m 10 kN 5 m 5 m 10 kN-m 5 m 5 m 10 kN 5 m 5 m 10 . Beam P a/2EI P a/EI 2aaa R eactions P a/2EI P a/EI 11 Pa 2 /12 EI 5Pa 2 /12 EI Shear(Rotation)Diagram ( Unit: Pa 2 /EI ) 1 1/ 12 5 /12 1/ 6 M oment(Deflection) Diagram ( Unit: Pa 3 /EI ) a/6 − 27 01/ 77 76=−0.35 1/ 3=−0.33 P /2 P /2 Beam. diagrams of the example problem. 60 kN 48 kN 36 kN 15 kN 12 kN 9 kN 7. 2 kN/m 9.6 kN/m 12 kN 9 kN 39 kN 48 kN 60 kN 15 kN 15 kN 60 kN T V -12 kN -48 kN -60 kN 75 kN-m 9 kN -39 kN 15 kN M -75 kN-m 75