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SM1-05: Statics 4: statically determined bar structures STATICALLY DETERMINED PLANE BAR STURCTURES (FRAMES, ARCHES) M.Chrzanowski: Strength of Materials 1/19 SM1-05: Statics 4: statically determined bar structures Frames Formal definition: A frame is a plane (2D) set of beams connected at stiff and/or hinged joints (corners) HINGED JOINT STIFF JOINT Joints have to be in the equilibrium! ? X = 0 Y = 0 MK = 0 ? !?! !?! M.Chrzanowski: Strength of Materials 2/19 SM1-05: Statics 4: statically determined bar structures Frames What is the difference between beams and frames? Why do we need to make frames? Hey, you! Beam M.Chrzanowski: Strength of Materials Beam or frame? Frame 3/19 SM1-05: Statics 4: statically determined bar structures Frames Equilibrium equations X = 0 Y = 0 MK = 0 + + MA1 = 0 MA2 = 0 ………… MAn = 0 For n hinged joints (if any!) at A1, A2 …An points kinematic stability of a structure (c.f. Theoretical Mechanics) Examples of unstable structures Centre of instability M.Chrzanowski: Strength of Materials 4/19 SM1-05: Statics 4: statically determined bar structures Frames 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 W q W = q·Δx = q’·Δs qs =q’·cos= q·cos2 Δx/Δs = cos ns =q’·sin= q·sin ·cos y s y W q’ q’ q’=q·Δx/Δs=q ·cos x s qs ns x Δx M.Chrzanowski: Strength of Materials 5/19 SM1-05: Statics 4: statically determined bar structures Frames Example: Diagrams of M, Q, N for a simple frame 3 3 2 kN 1 sin=0,6 cos =0,8 1,5 kN/m 3kN 1,5 m 1 kN 2 kN 1m 2m n 2 kN M [kNm] N o 2 2 Q [kN] - 2 2 Q - + - + + N [kN] 2 0,5 m 2 M.Chrzanowski: Strength of Materials 6/19 SM1-05: Statics 4: statically determined bar structures n Frames N o Checking the equilibrium at a joint Q 3 3 1 3 2 3 M [kNm] Q [kN] 2 2 - 2 2 3 - + + M.Chrzanowski: Strength of Materials + - 2m 0,5 3 2 - 3 2 2 N [kN] 3 2 2 7/19 SM1-05: Statics 4: statically determined bar structures Arches Formal definition: An arch is a plane (2D) set of curved beams connected at stiff and/or hinged joints + - - N M FRAME ARCH Stones and other brittle materials do not sustain an extension M.Chrzanowski: Strength of Materials 8/19 SM1-05: Statics 4: statically determined bar structures Arches C X = 0 Y = 0 MK = 0 Mc = 0 M.Chrzanowski: Strength of Materials 9/19 SM1-05: Statics 4: statically determined bar structures Parabolic y arch Arches y C n rC xC n C C yC h Semi-circular arch r x 2l x To determine reactions we only need to know position and magnitude of loads and position of the hinge and supports r, C xC , yC 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 + cx2 a,b,c from: for x = 0 for x = 2l (Symetric arch) for x = l y=0 y=0 y=h x = r·cos (in polar coordinates) = arctg dy/dx M.Chrzanowski: Strength of Materials 10/19 SM1-05: Statics 4: statically determined bar structures Arches Example: parabolic arch under concentrated force P C y h A HA l RA MA = 0 Mc = 0 B VA P·l - VB ·2l =0 HA/VA = HB/VB = l/h x l VB HB RB VB = P/2 = VA X = 0 Y = 0 HA = HB = VA· l/h Higher the ratio l/h (i.e. lower the ratio h / l) – higher the value of horizontal reaction H M.Chrzanowski: Strength of Materials 11/19 SM1-05: Statics 4: statically determined bar structures Arches P C Symmetry axis y Symmetry axis Δ M = RA·Δ h HA= (P/2) ·(l/h) A l RA VB N N A M.Chrzanowski: Strength of Materials RB x l VA = P/2 NC=HA M RA B ΔRA RB QA = VA QA Q 12/19 SM1-05: Statics 4: statically determined bar structures Q QV QH VA cos H A sin Q P 2cos l hsin HA=(P/2)(l/h) N NV N H VA sin H A cos NH Bar axis Q QH C QV At C: =0 QC P 2 NC P 2l h At A: ǂ0 For (l/h)<<1 – steep arch NV A VA=P/2 N N P 2sin l hcos 2 sin 1, cos 0 QA P 20 l h P 20 0 0 N A P 21 0 P 2 For (l/h)>>1 – shallow arch nn tg sin h l 0 l h cos 1 N QA P 21 l hsin P 21 1 tg sin P 21 cos 0 Q M.Chrzanowski: Strength of Materials N A P 20 l h N A 13/19 SM1-05: Statics 4: statically determined bar structures n N o NC=HA =- (P/2) (l/h) Q QA = VA=P/2 Anti-symmetric axis Symmetry axis QA<P/2 |NA|>P/2 + - N M.Chrzanowski: Strength of Materials P/2 P/2 + - Q 14/19 SM1-05: Statics 4: statically determined bar structures tension P Example: semi-circular arch under horizontal force r P/2 P/2 Q P/2 P/2 P/2 0,7P n + P/2 N N P/2 P/2 0,2Pr o r Q M r P/2 M P/2 P/2 0,3r P/2 + P/2 P/2 - M.Chrzanowski: Strength of Materials P/2 - + P/2 P/2 - P/2 + P/2 15/19 SM1-05: Statics 4: statically determined bar structures Quantitative comparison of frame, quasi-arch and arch P r r r P r P r r P r r r M.Chrzanowski: Strength of Materials r r r 16/19 SM1-05: Statics 4: statically determined bar structures „Frame” P M r Pr/2 P Pr Pr/2 r N r o n Q P/2 M P/2 N + Q + - P P/2 P M.Chrzanowski: Strength of Materials 17/19 SM1-05: Statics 4: statically determined bar structures Quasi-arch tension P r M P 0,55 Pr P/2 P/2 r Pr/2 r N Q P/2 + P ~1,1P M.Chrzanowski: Strength of Materials P/2 P + P/2 18/19 SM1-05: Statics 4: statically determined bar structures Comparison Pr/2 M 0,55Pr 0,2Pr Pr/2 Pr Pr/2 P + Q - P/2 P P/2 P/2 + + - 1,1P P/2 + P M.Chrzanowski: Strength of Materials + P P/2 + 0,7P P/2 N - P/2 + P/2 P/2 P/2 - P/2 19/19 SM1-05: Statics 4: statically determined bar structures stop M.Chrzanowski: Strength of Materials 20/19