Lecture 5: Pure bending of beams

Transcription

1 Lecture 5: Pure bending of beams Each end of an initiall straight, uniform beam is subjected to equal and opposite couples. The moment of the couples is denoted b (units of F-L. t is shown in solid mechanics that the beam deforms into a plane circular arc. L undeformed beam deformed beam Lec. 5: 1 of 17

2 Neutral ais & cross sections remain plane Some line elements originall parallel to the - ais, loosel called fibers, are elongated and some are shortened in the deformed beam. Hence, one fiber does not change length. t is called the neutral ais, and we take the z-ais in the undeformed beam to be the neutral ais. The fiber on the neutral ais bends but does not change length. Plane cross sections perpendicular to the - ais in the undeformed beam remain plane and perpendicular to the -ais in the deformed beam Lec. 5: 2 of 17

3 Normal strain is linear in the -coordinate Show that: ε - ρ PQ P * Q * RPQ R * P * Q * π - 2 R P 0 d undeformed S Q ρ R * P * Q * dθ S * ε R * S * RS RS RS PQ P * Q * ε R * S * P * Q * P * Q * ρ deformed radius of curvature of the neutral ais ε ( ρdθ ρdθ ρdθ therefore: ε - ρ Lec. 5: 3 of 17

4 Hooke s law σ Eε Static equivalence N da σda σda A A A σda differential normal force N normal force and moment Lec. 5: 4 of 17

5 Hooke s law (continued Substitute ε - and σ Eε into the N and ρ integrals, recognizing E and ρ are independent of integration over the cross section, to get N E --- da ρ E --- ρ 2 da A A But the net aial force N 0 Therefore, we must have in pure bending. d A 0; i.e., the origin of the cross-sectional coordinates is at the centroid of the area. A Lec. 5: 5 of 17

6 Hooke s law (continued The second area moment about z-ais through the centroidal coordinates in the cross section is denoted b, where A The dimensional units of are L 4 ; e.g.,in 4 or m 4. 2 da z da centroid A Lec. 5: 6 of 17

7 Hooke s law (concluded Hence, for centroidal coordinates in the cross section we find E or ρ 1 - ρ E which is Hooke s law for the beam. That is, Hooke s law for the beam relates the reciprocal of the radius of curvature of the - ais in the deformed beam to the bending moment divided b the bending stiffness E of the cross section.beams with a large value of the bending stiffness have less curvature, or equivalentl, a larger radius of curvature. Lec. 5: 7 of 17

8 Fleure formula Substitute Hooke s law to get σ into σ Eε E E ρ - This is the fleure formula. t shows that the normal stress due to bending is proportional to the bending moment, proportional to the - coordinate in the cross section, and inversel proportional to the second area moment. At 0 the bending normal stress is zero, and in the section the ma magnitude σ ma ma. 1 - ρ Lec. 5: 8 of 17

9 Bending normal stresses σ σ For > 0 the upper fibers ( > 0 are in tension and the lower fibers ( < 0 are in compression. f < 0, then the upper fibers are in compression and the lower fibers are in tension. Bending is a combination of tension and compression. Lec. 5: 9 of 17

10 Normal stresses in wing bending p airload intensit in lb/ft p L π b b b/2 Stead level flight with L total lift (both wings, and b/2 the semi-span. How do we size the spar? Lec. 5: 10 of 17

11 Shear force and bending moment p( ξdξ p( ξ 2L π ( b 2 1 2ξ b V FBD & Equilibrium ξ ξ dξ b 2 b 2 V ( p( ξ dξ b 2 ( ( ξ p( ξ dξ Lec. 5: 11 of 17

12 Shear force and bending moment distribution VêL êhlbê2l êhbê2l êhbê2l -0.2 shear force distribution bending moment distribution V ( 0 L 2 ( 0 ( bl ( 3π For L 20, 000 lb and b 65 ft, at the wing root V 10, 000 lb and 137, 934 lb-ft Lec. 5: 12 of 17

13 Wing spar: thickness t and spar cap area A s The magnitude of the maimum shear force and bending moment are at the wing root. So we wish to size the spar to carr these loads. dealized spar. h 2 z h 2 t spar cap, area A s spar web, thickness spar cap, area A s h Usuall dimension is given b the thickness to chord ratio for the airfoil. t V Lec. 5: 13 of 17

14 Spar sizing Requirement Select area A s and thickness t such that magnitude of the normal stresses in the spar are less that the ield stress of material. We need formulas for the normal stresses due to bending and the shear stresses due to bending. The formula for the normal stress due to bending is called the fleure formula, and we alread derived this important formula as σ Lec. 5: 14 of 17

15 Second area moments for some sections t f h h t h 12 t w t f h 2 bt f 2 h 3 t w 12 t b thin-walled -section r t πr 3 t h 2 h 2 A s t h 2 A s 2 + h t 12 thin-walled tube A s spar Lec. 5: 15 of 17

16 Spar cap areas Sa the thickness to chord ratio is 0.14 of the airfoil at the wing root with a chord of 8 ft. So h 1.12 ft. for the spar. The material ield strength is 42 ksi (2024-T4, and assume a factor of safet of 1.5 against ielding in the stead level flight condition. Then the allowable stress is 28 ksi. n the fleure formula, the maimum bending normal stress occurs at ± h/2 at the wing root where -137,934 lb-ft. For a thin-walled web we neglect the contribution of the web to the second area moment and take h 2 A s /2. σ h ( 2 ma ( h 2 A s ha s Lec. 5: 16 of 17

17 Spar cap areas (concluded So 28000lb/in 2 ( , 934 lb-ft, or ( 1.12 fta s A s ( 137, 934 lb-ft ( 1.12 ft lb in 2 A s 4.40 in 2 Lec. 5: 17 of 17