1. 1 /19 M.Chrzanowski: Strength of Materials SM1-05: Statics 4: statically determined bar structures STATICALLY DETERMINED PLANE BAR STURCTURES (FRAMES, ARCHES)

2. 2 /19 M.Chrzanowski: Strength of Materials SM1-05: Statics 4: statically determined bar structures Formal definition: A frame is a plane (2D) set of beams connected at stiff and/or hinged joints (corners) Joints have to be in the equilibrium!  X = 0  Y = 0  M K = 0 ? ? !?! Frames STIFF JOINT HINGED JOINT

3. 3 /19 M.Chrzanowski: Strength of Materials SM1-05: Statics 4: statically determined bar structures What is the difference between beams and frames? Why do we need to make frames? BeamFrame Beam or frame? Hey, you! Frames

4. 4 /19 M.Chrzanowski: Strength of Materials SM1-05: Statics 4: statically determined bar structures Equilibrium equations  X = 0  Y = 0  M K = 0 M A2 = 0 M A1 = 0 M An = 0 ………… + For n hinged joints (if any!) at A 1, A 2 …A n points + kinematic stability of a structure (c.f. Theoretical Mechanics) Centre of instability Examples of unstable structures HYPER-STIFF UNSTABLE Frames

5. 5 /19 M.Chrzanowski: Strength of Materials SM1-05: Statics 4: statically determined bar structures After we determine reactions and check stability we can deal with a frame as a set of individual beams, applying all techniques which have been demonstrated for beams. But, besides of diagrams of M and Q we have to make diagrams of N, too.  Some problems can be encountered with sloping members x y W q’=q·Δx/Δs=q ·cos  W = q·Δx W ΔsΔs ΔxΔx  x s y q’ q s =q’·cos  = q·cos 2  Δx/Δs = cos  q qsqs s n s =q’·sin  = q·sin  ·cos  = q’·Δs qsqs  q’ nsns Frames nsns

6. 6 /19 M.Chrzanowski: Strength of Materials SM1-05: Statics 4: statically determined bar structures Example: Diagrams of M, Q, N for a simple frame 1,5 kN/m 2 kN 2 m 1,5 m 1 m 2 kN 1 kN 2 kN  sin  =0,6 cos  =0,8 + - - + Q [kN] M [kNm] 0,8 3 1 3 0,33 1 2 1,6 2 1,2 0,6 2 2 0,5 m + - - 2 N [kN] 2 Frames Q N o n 3kN

7. 7 /19 M.Chrzanowski: Strength of Materials SM1-05: Statics 4: statically determined bar structures 2 Checking the equilibrium at a joint M [kNm] 3 1 3 0,25 1 + - 2 1,2 0,6 - 2 N [kN] - + Q [kN] 0,8 1,6 2 2 2 0,5 m + - 2 3 3 3 3 2 3 3 2 2 2 Q N o n Frames

8. 8 /19 M.Chrzanowski: Strength of Materials SM1-05: Statics 4: statically determined bar structures FRAME ARCH M N + - - Stones and other brittle materials do not sustain an extension Arches Formal definition: An arch is a plane (2D) set of curved beams connected at stiff and/or hinged joints

9. 9 /19 M.Chrzanowski: Strength of Materials SM1-05: Statics 4: statically determined bar structures  X = 0  Y = 0  M K = 0 M c = 0 C Arches

10. 10 /19 M.Chrzanowski: Strength of Materials SM1-05: Statics 4: statically determined bar structures x y 2l  xCxC x yCyC C C y Parabolic arch Semi-circular arch To determine reactions we only need to know position and magnitude of loads and position of the hinge and supports But to determine the cross-sectional forces we do need the equation describing shape of the arch: including coordinates of any point and its tangent. y = a + bx + cx 2 x = r·cos  r CC rCrC x C, y C r,  C h a,b,c from: for x = 0 y = 0 for x = l y = h for x = 2 l y = 0 (Symetric arch) n  n   = arctg d y /d x Parabolic arch Semi-circular arch (in polar coordinates) Arches

11. 11 /19 M.Chrzanowski: Strength of Materials SM1-05: Statics 4: statically determined bar structures Example: parabolic arch under concentrated force x y ll h C AB M c = 0 RARA RBRB VBVB HBHB  X = 0  M A = 0 P· l - V B ·2 l =0 VAVA HAHA V B = P/2 H A /V A = H B /V B = l/h Higher the ratio l/h ( i.e. lower the ratio h / l) – higher the value of horizontal reaction H H A = H B = V A · l/h P  Y = 0 = V A Arches

12. 12 /19 M.Chrzanowski: Strength of Materials SM1-05: Statics 4: statically determined bar structures Arches RBRB x y ll h C AB RARA VBVB V A = P/2 H A = (P/2) ·(l/h) P Symmetry axis RARA RBRB M RBRB Δ M = R A ·Δ ΔRAΔRA N Q NANA QAQA N C =H A Q A = V A

13. 13 /19 M.Chrzanowski: Strength of Materials SM1-05: Statics 4: statically determined bar structures    V A =P/2 H A =(P/2)( l/h ) QVQV QHQH NVNV NHNH Bar axis   Q N n n N Q A C At C:  =0 At A:  ǂ 0 For (l/h)>>1 – shallow arch For (l/h)<<1 – steep arch

14. 14 /19 M.Chrzanowski: Strength of Materials SM1-05: Statics 4: statically determined bar structures RBRB N Q |NA|>P/2 N C =H A =- (P/2) (l/h) + - - - + Q A <P/2 Q A = V A =P/2 Anti-symmetric axis P/2 Symmetry axis Q N o n

15. 15 /19 M.Chrzanowski: Strength of Materials SM1-05: Statics 4: statically determined bar structures P r r r P/2 M  0,3 r  0,2P r P/2 Q + - + - 0,7P N Q N M o n tension + - + Example: semi-circular arch under horizontal force

16. 16 /19 M.Chrzanowski: Strength of Materials SM1-05: Statics 4: statically determined bar structures r P r r r P r r r P r r r r P r r r Quantitative comparison of frame, quasi-arch and arch 

17. 17 /19 M.Chrzanowski: Strength of Materials SM1-05: Statics 4: statically determined bar structures r r P P r P/2 M QN + - + PrPr P r /2 P/2 P P Q N M o n „Frame”

18. 18 /19 M.Chrzanowski: Strength of Materials SM1-05: Statics 4: statically determined bar structures P P/2 r r P r M Q tension P r /2 0,55 P r P/2 P + - N P - + ~1,1P Quasi-arch

19. 19 /19 M.Chrzanowski: Strength of Materials SM1-05: Statics 4: statically determined bar structures Comparison M QN 0,2Pr + P P/2 Pr/2 0,55Pr Pr/2 Pr Pr/2 - P/2 P + + - P - + P 1,1P P/2 + - + + - 0,7P P/2

20. 20 /19 M.Chrzanowski: Strength of Materials SM1-05: Statics 4: statically determined bar structures  stop