Beams Session Subject: S1014 / MECHANICS of MATERIALS Year: 2008
Bina Nusantara Beams
Bina Nusantara What is Bending Stresses ?
Bina Nusantara What is Bending Stresses ?
Bina Nusantara Normal Stress A normal stress is a stress that occurs when a member is loaded by an axial force. The value of the normal force for any prismatic section is simply the force divided by the cross sectional area.
Bina Nusantara Normal Stress
Bina Nusantara What is Bending Stresses ?
Bina Nusantara What is Bending Stresses ? When a member is being loaded similar to that in figure 1 bending stress (or flexure stress) will result. Bending stress is a more specific type of normal stress.
Bina Nusantara What is Bending Stresses ? When a beam experiences load like that shown in figure 1 the top fibers of the beam undergo a normal compressive stress.
Bina Nusantara What is Bending Stresses ? The stress at the horizontal plane of the neutral is zero. The bottom fibers of the beam undergo a normal tensile stress.
Bina Nusantara What is Bending Stresses ? It can be concluded therefore that the value of the bending stress will vary linearly with distance from the neutral axis.
Bina Nusantara What is Bending Stresses ?
Bina Nusantara Shear Stress Normal stress is a result of load applied perpendicular to a member. Shear stress however results when a load is applied parallel to an area.
Bina Nusantara Shear Stress Like in bending stress, shear stress will vary across the cross sectional area.
Bina Nusantara Shear Stress
Bina Nusantara ELASTIC CURVE The deflection diagram of the longitudinal axis that passes through the centroid of each cross-sectional area of the beam is called the elastic curve, which is characterized by the deflection and slope along the curve. E.g.
Bina Nusantara ELASTIC CURVE Moment-curvature relationship: Sign convention:
Bina Nusantara ELASTIC CURVE Moment-curvature relationship: Sign convention:
Bina Nusantara ELASTIC CURVE
Bina Nusantara ELASTIC CURVE
Bina Nusantara ELASTIC CURVE Consider a segment of width dx, the strain in are ds, located at a position y from the neutral axis is ε = (ds’ – ds)/ds. However, ds = dx = ρdθ and ds’ = (ρ-y) dθ, and so ε = [(ρ – y) d θ – ρd θ ] / ( ρdθ), or 1 ρ = – ε y
Bina Nusantara ELASTIC CURVE Comparing with the Hooke’s Law ε = σ / E and the flexure formula σ = - My/I We have 1 ρ = M EI or 1 ρ = – σ Ey
Bina Nusantara SLOPE AND DISPLACEMENT BY INTEGATION (CONT.) Boundary Conditions: The integration constants can be determined by imposing the boundary conditions, or Continuity condition at specific locations
Bina Nusantara SLOPE AND DISPLACEMENT BY INTEGATION (CONT.) Boundary Conditions: The integration constants can be determined by imposing the boundary conditions, or Continuity condition at specific locations
Bina Nusantara SLOPE AND DISPLACEMENT BY INTEGATION (CONT.) Boundary Conditions: The integration constants can be determined by imposing the boundary conditions, or Continuity condition at specific locations
Bina Nusantara SLOPE AND DISPLACEMENT BY INTEGATION (CONT.) Boundary Conditions: The integration constants can be determined by imposing the boundary conditions, or Continuity condition at specific locations
Bina Nusantara This assumes that the system is linear-elastic, and therefore the deflection is a linear function of F. , Deflection at B Load, F
Bina Nusantara The total strain energy stored in the system is the sum of the individual strain energies in each of the truss members numbered i=1 to 7.
Bina Nusantara Beam Elements: y x y dx dA z y F(x) A beam that is symmetrical in x-section about the z-axis, is subjected to bending. Consider a infinitesimal volume element of length dx and area dA as shown. This element is subjected to a normal stress: s x =My/I
Bina Nusantara Beam Elements: y x y dx dA z y F(x) The Strain Energy Density on this element is: For linear elastic material
Bina Nusantara Substituting, and multiplying by the Volume of the element Hence, the Strain Energy for a slice of the beam, of width dx, is y x dx
Bina Nusantara Strain Energy in Entire Beam Consider the cantilever beam as shown I F L x y
Bina Nusantara Deflection I L x y F External Work, Strain Energy Linear-elastic, F Classical Solution
Bina Nusantara y x P L/2 Shear Force PL/4 P/2 -P/2 Moment Determine Elastic Strain Energy due to bending for simply supported 3-point bending member of constant X-section. For 0 x L/2: M=Px/2 Note by symmetry we can find the total strain energy by doubling the strain energy of the LHS.
Bina Nusantara y L/2 P BB Determine B…….
Bina Nusantara Elastic Strain Energy due to Transverse Shear Stress y x xy a xy xy
Bina Nusantara Shear Strain Energy y dA z y x dx F(x) f is called a form factor: Circle f=1.11 Rectangle f=1.2 Tube f=2.00 I section f=A/A web
Bina Nusantara Applications: Castigliano’s 2 nd theorem can be used to determine the deflections in structures (eg, trusses, beams, frames, shells) and we are not limited to applications in which only 1 external force or moment acts. Furthermore, we can determine the deflection or rotation at any point, even where no force or moment is applied externally.
Bina Nusantara Design Considerations Stress – Yield Failure or Code Compliance Deflection Strain Stiffness Stability – Important in compressive members Stress and strain relationships can be studied with Mohr’s circle Often the controlling factor for functionality
Bina Nusantara Deflection [Everything’s a Spring] When loads are applied, we have deflection Depends on – Type of loading Tension Compression Bending Torsion – Cross-section of member – Comparable to pushing on a spring We can calculate the amount of beam deflection by various methods
Bina Nusantara Superposition Determine effects of individual loads separately and add the results May be applied if – Each effect is linearly related to the load that produces it – A load does not create a condition that affects the result of another load – Deformations resulting from any specific load are not large enough to appreciably alter the geometric relations of the parts of the structural system
Bina Nusantara Deflection --- Energy Method There are situations where the tables are insufficient We can use energy-methods in these circumstances Define strain energy
Bina Nusantara Deflection --- Energy Method Define strain energy density** V – volume
Bina Nusantara Deflection --- Energy Method Put in terms of ,
Bina Nusantara Example – beam in bending
Bina Nusantara Example – beam in bending
Bina Nusantara Castigliano’s Theorem Deflection at any point along a beam subjected to n loads may be expressed as the partial derivative of the strain energy of the structure WRT the load at that point
Bina Nusantara Castigliano’s Theorem We can derive the strain energy equations as we did for bending Then we take the partial derivative to determine the deflection equation AND if we don’t have a force at the desired point: – If there is no load acting at the point of interest, add a dummy load Q, work out equations, then set Q = 0
Bina Nusantara Stability Up until now, 2 primary concerns – Strength of a structure It’s ability to support a specified load without experiencing excessive stress – Ability of a structure to support a specified load without undergoing unacceptable deformations Now, look at STABILITY of the structure – It’s ability to support a load without undergoing a sudden change in configuration Material failure
Bina Nusantara Unit Load Method Deflection at C ??? A B C L q
Bina Nusantara Unit Load Method Procedure 1 : Determine M o A B C L q
Bina Nusantara Unit Load Method Procedure 1 : Determine M o q VAVA HAHA VBVB M A =0 w. ½ L- V B.L = 0 qL ½ L- V B.L= 0 ½ qL 2 – VB.L = 0 V B = ½ qL
Bina Nusantara Unit Load Method Procedure 1 : Determine M o q VAVA HAHA VBVB V A = ½ qL M B =0 - w. ½ L+ V A.L = 0 - qL ½ L+ V A.L= 0 - ½ qL 2 + V A.L = 0
Bina Nusantara Unit Load Method Procedure 1 : Determine M o q VAVA HAHA VBVB H=0 H A = 0 H A = 0 V=0 V A + V B = w ½ qL + ½ qL = qL OK!!!
Bina Nusantara Unit Load Method Procedure 1 : Determine M o ( cont’ ) for A-C 0<x 1 < 1/2 L HAHA VAVA q A w = qx1 x1 M x =0 w.( ½.x 1 ) –V A. x 1 + M x1 = 0 ( qx 1.x 1 ).( ½.x 1 ) – qL.x1 + M x1 = 0 M x1 = qLx 1 – ½ x 1 2 M x1 = qLx 1 – ½ x 1 2 Mx1
Bina Nusantara Unit Load Method Procedure 1 : Determine M o ( cont’ ) for B-C 0<x 2 < 1/2 L M x =0 w.( ½.x 2 ) –V B. x 2 + M x2 = 0 ( qx 1.x 2 ).( ½.x 2 ) – qL.x 2 + M x2 = 0 M x2 = qLx 2 – ½ x 2 2 M x2 = qLx 2 – ½ x 2 2 VBVB B x2 q w = qx2 Mx2
Bina Nusantara Unit Load Method Procedure 2 : Put P = 1 unit on C ( without external loads ) VAVA HAHA VBVB M A =0 P. ½ L- V B.L = 0 1 ½ L- V B.L= 0 ½ L – VB.L = 0 V B = ½ P = 1 unit
Bina Nusantara Unit Load Method Procedure 2 : Put P = 1 unit on C ( without external loads ) VAVA HAHA VBVB V A = ½ M B =0 - P. ½ L+ V A.L = ½ L+ V A.L= 0 - ½ L + V A.L = 0 P = 1 unit
Bina Nusantara Unit Load Method Procedure 2 : Put P = 1 unit on C ( without external loads ) VAVA HAHA VBVB H=0 H A = 0 H A = 0 V=0 V A + V B = w ½ + ½ = 1 OK!!! P = 1 unit
Bina Nusantara Unit Load Method Procedure 3 : Determine m for A-C 0<x 1 < 1/2 L HAHA VAVA A x1 M x =0 –V A. x 1 + m x1 = 0 – ½.x1 + m x1 = 0 m x1 = ½ x 1 m x1 = ½ x 1 mx1
Bina Nusantara Unit Load Method Procedure 3 : Determine m ( cont’ ) for B-C 0<x 2 < 1/2 L M x =0 –V B. x 2 + M x2 = 0 – ½.x 2 + M x2 = 0 M x2 = ½ x 2 M x2 = ½ x 2 VBVB B x2 q Mx2
Bina Nusantara Unit Load Method Procedure 4 : Determine deflection at C M x1 = qLx 1 – ½ x 1 2 M x2 = qLx 2 – ½ x 2 2 m x1 = ½ x 1 M x2 = ½ x 2 A-C 0<x 1 < ½ L B-C 0<x 2 < ½ L
Bina Nusantara Unit Load Method Procedure 4 : Determine deflection at C
Bina Nusantara Unit Load Method Procedure 4 : Determine deflection at C
Bina Nusantara Unit Load Method Deflection at C = A B C L q
Bina Nusantara Unit Load Method Slope Deflection C ??? A B C L q
Bina Nusantara Unit Load Method Procedure 1 : Determine M o A B C L q
Bina Nusantara Unit Load Method Procedure 1 : Determine M o q VAVA HAHA VBVB M A =0 w. ½ L- V B.L = 0 qL ½ L- V B.L= 0 ½ qL 2 – VB.L = 0 V B = ½ qL
Bina Nusantara Unit Load Method Procedure 1 : Determine M o q VAVA HAHA VBVB V A = ½ qL M B =0 - w. ½ L+ V A.L = 0 - qL ½ L+ V A.L= 0 - ½ qL 2 + V A.L = 0
Bina Nusantara Unit Load Method Procedure 1 : Determine M o q VAVA HAHA VBVB H=0 H A = 0 H A = 0 V=0 V A + V B = w ½ qL + ½ qL = qL OK!!!
Bina Nusantara Unit Load Method Procedure 1 : Determine M o ( cont’ ) for A-C 0<x 1 < 1/2 L HAHA VAVA q A w = qx1 x1 M x =0 w.( ½.x 1 ) –V A. x 1 + M x1 = 0 ( qx 1.x 1 ).( ½.x 1 ) – qL.x1 + M x1 = 0 M x1 = qLx 1 – ½ x 1 2 M x1 = qLx 1 – ½ x 1 2 Mx1
Bina Nusantara Unit Load Method Procedure 1 : Determine M o ( cont’ ) for B-C 0<x 2 < 1/2 L M x =0 w.( ½.x 2 ) –V B. x 2 + M x2 = 0 ( qx 1.x 2 ).( ½.x 2 ) – qL.x 2 + M x2 = 0 M x2 = qLx 2 – ½ x 2 2 M x2 = qLx 2 – ½ x 2 2 VBVB B x2 q w = qx2 Mx2
Bina Nusantara Unit Load Method Procedure 2 : Put M = 1 unit on C ( without external loads ) M A =0 M - V B.L = V B.L= 0 V B = 1/L VAVA HAHA VBVB M = 1 unit
Bina Nusantara Unit Load Method Procedure 2 : Put M = 1 unit on C ( without external loads ) M B =0 M + V A.L = V A.L= 0 V A = -1/L VAVA HAHA VBVB M = 1 unit
Bina Nusantara Unit Load Method Procedure 2 : Put M = 1 unit on C ( without external loads ) VAVA HAHA VBVB M = 1 unit H=0 H A = 0 H A = 0 V=0 -V A + V B = 0 - ½L + ½L = 0 OK!!!
Bina Nusantara Unit Load Method Procedure 3 : Determine m for A-C 0<x 1 < 1/2 L HAHA VAVA A x1 M x =0 –V A. x 1 + m x1 = 0 – -1/L.x 1 + m x1 = 0 m x1 = -1/L x 1 m x1 = -1/L x 1 mx1
Bina Nusantara Unit Load Method Procedure 3 : Determine m ( cont’ ) for B-C 0<x 2 < 1/2 L M x =0 –V B. x 2 + M x2 = 0 – 1/L.x 2 + M x2 = 0 M x2 = 1/Lx 2 M x2 = 1/L x 2 VBVB B x2 q Mx2
Bina Nusantara Unit Load Method Procedure 4 : Determine deflection at C M x1 = qLx 1 – ½ x 1 2 M x2 = qLx 2 – ½ x 2 2 m x1 = -1/Lx 1 M x2 = 1/Lx 2 A-C 0<x 1 < ½ L B-C 0<x 2 < ½ L
Bina Nusantara Unit Load Method Procedure 4 : Determine deflection at C
Bina Nusantara Unit Load Method Procedure 4 : Determine deflection at C
Bina Nusantara Unit Load Method Slope Deflection C = 0 A B C L q