CEG2301

1. Tie rod
   1. Tensile forces
   2. Slender
   3. Common cross sections >>
2. Beam
   1. Straight horizontal members
      1. <<Support types
   2. Vertical loading
   3. ^^Most effcient^^ cross section
3. Colums
   1. Vertical members
   2. Resist axial compression

1. ^^Truss^^
2. Cable^^
3. ^^Arch^^
4. Frame>>
5. << Surface structure

1. Dead loads

   Anmerkungen:

   • Wieght of the structure and premenent objects
2. Live loads

   Anmerkungen:

   • natural forces , moving loads and tempoary objects
3. Wind loads
4. Accidental loads

   Anmerkungen:

   • Earthquakes and other desasters

1. Pinned>>
2. Torsion>>
   1. K= spring constant
3. Fixed>>
4. Reactions>>

Anmerkungen:

• Summing all forces and find the eqivelent force required for the same effects
• material must behave in a liner elastic manner and the geometry of the structure must not change significantly

1. The total displacement or internal loadings (stress) at a point in a structure subjected to several external loadings can be determined by adding together the displacements or internal loadings (stress) caused by each of the external loads acting separately.
   1. Diagram of idea

Anmerkungen:

• 6 total 3 moments in x y and z and 3 forces in x y and z

1. (sum)Fo=0
   1. Forces in each direction much be 0
2. (sum)Mo=0
   1. Moments in each direction must be 0

1. r=3n
2. n=parts r=reactions

1. r<3n
   1. Unstable
2. r>3n
   1. Unstable if reactions are concurent or parellell or components form a collapseble mechanisium

1. ^^Compound truss^^
2. ^^Simple truss^^

1. <<Roofing
2. Bridges>>

1. The members are joined by smooth pins and members are conccurent at a point
2. All loading as applied at the joints
3. All members act with axial force
   1. ^^Compression ^^
   2. ^^Tension ^^

1. (b+r=2j)=determenent
2. (b+r>2j)=indeterminate
3. Degree of indeterminacy = (b+r)=2j
   1. j=joints r=external reactions b=number of bars

1. (b+r)<2j = internally unstable
   1. Will collapse as there are not enough bars or reactions to constrain the joints
2. Externally unstable
   1. ^^Parallel external forces^^
   2. ^^Conccurent external forces^^

1. Methord
   1. separate each joint out of the frame and draw in arows to show the forces that use trig to resolve forces to keep the point in eqilibrium repeat for each joint until all required forces are found keep sign convention consistant for all forces and signs will work out
2. Other notes
   1. Assume truss is in equilibrium
   2. Assume all unknown forces to be in tension or compression for signs to work out
   3. Some members can be zero and carry no force but are required for stability

1. Metord
   1. Cut the truss through the required members and treat each part as a separate objects in equilibrium treating the cut member forces as external forces apply the equilibrium equations to find the forces required
   2. <<Example diagram to draw
2. Assume forces and stick to the sign convention

1. Equations
   1. b+r<3j = Unstable truss
   2. b+r=3j = Statistically determinate - Check stability
   3. b+r>3j = Statically indeterminate - Check stability
   4. ^^Force Components^^
      1. l=Sqrt(x^2+y^2+z^2)
      2. Fx=F(X/l)
      3. Fy=F(y/l)
      4. Fz=F(z/l)
      5. F=Sqrt(Fx^2+Fy^2+Fz^2)

1. This can be done at any point using the method of sections

1. Loads represent stress over a cross sectional area>>>>
2. Sign conventions
   1. Normal force , N
      1. + Force = elongation of segment
   2. Shear force, v
      1. + Shear = rotates the segments clock wise
   3. Bending moment, M
      1. + Bending = bends the segments upwards (concave shape)

1. (dV/dx) = W(x)
   1. Slope of the shear diagram is equal to the intensity of the distributed load
2. (dM/dx) = V
   1. Slope of moment diagram is equal to the intensity of the shear
3. (delta)V = Inter( W(x) dx)
   1. Change in shear over a length is equal to the area under the loading diagram
4. (delta)M = Inter( W(x) dx)
   1. Change in bending moment over a length is equal to the area under the shear diagram

1. Loads
2. Tempreture
3. Fabrication errors
4. Settlement

1. Beam deflection mainly from internal bending
2. Truss deflections mainly from internal axial forces

1. Positive moment = Upward bend Negative moment = Downward bend
2. If the shape of the moment diagram is known an elastic curve can be constructed and visa versa
   1. << Example diagrams

1. Force
   1. dUe = F dx
      1. Force applied gradually
         1. Ue = 0.5 P (delta)
      2. Force already applyed
         1. Ue' = P (delta)'
2. Moment
   1. dUe = M d(theta)
      1. Force applied gradually
         1. Ue + 0.5 M (theta)
      2. Moments already applied
         1. Ue' = M (theta)'

1. Axial force
   1. Ui = (N^2L)/(2AE)
      1. Work done gradually will be converted to strain energy and stored in the bar
2. Bending
   1. Ui = Inter(o->L) ((M^2dx)/(2EI))
      1. Load applied gradually causes a moment leading to a rotation

1. External virtual work
   1. 1.(delta)
2. Internal virtual work
   1. u.dL
3. First apply virtual load to find real displacement by proportions
   1. Works for loads and moments, beams and trusses.
      1. Loads
         1. 1.(delta)=Sum(u.dL)
      2. Moments
         1. 1.(theta)=Sum(u(theta).dL)
      3. Trusses
         1. 1.(delta)=Sum((nNL)/(AE))
            1. Use a table with columns for: member, n, N, L, nNL. For each member work out the values sum the nNL column and devide by the A and E values whith should be given
      4. Beams
         1. Forces
            1. 1.(delta)=Inter(0->L)(mM/EI)dx
         2. Moments
            1. 1.(theta)=Inter(0->L)(m(theta)M/EI)dx
         3. Tables can be used for Inter(0->L)(m(x)m'(x)dx) when the shapes are known

1. The external work is a function of external loads and if the force is increased by a differential amount the new strain also increases and the increase is not effected by the order of the applied loads
   1. (delta)j = (lDelta)Ui/(ldelta)Pj)
      1. So for a truss
         1. (delta) = Sum(N((ldelta)N/(ldelta)P)(L/AE))
            1. Use a table with columns: member, N, (ldelta)N/(ldelta)P, N(P=0), L and N((ldelta)N/(ldelta)P)L. Work out the values for each column for each member and then sum the final column and devide by the given A and E values

1. Maximum stress and deflection is genrally smallered than statically determinate structures
2. Tendency to redistribute loads to redundant supports when there is fault or overloading

1. Redundant supports can cause differential displacements that introduce internal stress to the structure

1. Factors to satisfy
   1. Equilibrium
      1. Reactive forces hold the structure at rest
   2. Compatibility
      1. Segments in the structure fit together with out breaks or overlaps
   3. Force-displacement
      1. depends on material response (linear elastic)
2. Force method
   1. Writing equations that satisfy the compatibility and force-displacement requirements in order to determine the redundant forces
3. Displacement method
   1. Analysis based on first writeing force-displacement relations then satisfying the equilibrium requirements
   2. Displacements unknown
      1. reactions can be determended from the compatibility equations and force-displacements equations
4. Force method Details
   1. Compatibility equations
      1. Moments
         1. 0=(theta)A+ MA(alpha)AA
            1. MA=-(theta)A/(alpha)AA
               1. (alpha)AA= angular flexibility coeffcient
                  1. the angular displacement at A caused by a unit couple moment at A
      2. Forces
         1. 0= -(delta)B +(delta)'BB
            1. (delta)'BB=ByFBB
               1. By=(delta)B/FBB
                  1. FBB= Linear flexibility coefficient
                     1. The displacement at B caused by a unit load acting in the direction of By
      3. << Visual representation of equations and methord

1. The displacement and rotation at point B on a structure due to a unit load acting on it at point A is the same as the displacement and rotation of point A when the unit load is applied to point B
   1. ^^FAB = FBA^^