F = 0 can't be satisfied.

Transcription

1 11/9/1 Equilibrium of eams 1. asics 1.1. Reactions: Draw D ount number of reactions (R) and number of internal hinges (H). If R < H 3 > unstable beam staticall determinate staticall indeterminate (a) (b) (c) If (b), write 3 eq. of equilibrium for each section separated b a hinge and solve. Note that the above is a necessar condition onl! Eample (n unstable, staticall indeterminate beam) Here, can't be satisfied. There are 3 reactions, but onl two equations to solve for them. (, ) 1

2 E 6 - eams - Equilibrium 11/9/1 S.P. 1: R3, H, (no internal hinges) R - H staticall determinate Z 1 5' 1kips - 9' 5kips 1 95 ft - kips 7.9 kips 7.9kips -1kips - 5kips 7.8 kips

3 E 6 - eams - Equilibrium 11/9/1 1.. oments of Distributed loads w()force per unit length dincremental length w()dincremental force d Z O Define Z O [ w( ) d] w( ) d Z O w( ) d w( ) d as the first moment of the distributed load about O. The centroidal distance is given b w( ) d w( ) d (first moment of force) / force 3

4 E 6 - eams - Equilibrium 11/9/1 4 Often it is eas to compute d w ) ( if and d w ) ( are known, e.g. ) ( ) ( ma 1 ma 1 1 w d w d w (the area under the triangular load.) 1 ma 1 3 (tabulated) 3 w Z O

5 E 6 - eams - Equilibrium 11/9/1 S.P. : R - H - 3 oth horiontal and vertical reaction components are present The three reactions are not concurrent (staticall determinate) Z ( w / ) ( / 4) w 8 w / 5w 8 5

6 E 6 - eams - Equilibrium 11/9/ nalsis at Hinges n internal hinge can transmit horiontal and vertical force components, but no moment. D Idealiation: Two analsis approaches 1. reak structure apart at hinge and insert the two reaction forces shown above. Then write equations of equilibrium for both halves.. Since moment is ero at the hinge, sum moments of all forces to one side of the hinge to ero about the hinge. This provides an etra "equation of condition". We'll illustrate these approaches below. 6

7 E 6 - eams - Equilibrium 11/9/1 7 S.P. 3: Solution b "breaking" the structure at the hinge. R 4, H 1, R - H o.k. (Reactions are not concurrent or parallel) / / / / / / Pa a a P P P P P P Z Y X X Z

8 E 6 - eams - Equilibrium 11/9/1 8 S.P. 4: lternate solution to S.P. 3 Equation of condition: or the left side of the beam, / / P P left or the entire beam, ) ( ) ( ) ( / Pa a P a a P a P P P Z in agreement with S.P. 3.

9 E 6 - eams - Equilibrium 11/9/ Stress Resultants Internal stress resultants (ISR) are necessar to achieve equilibrium To find internal stress resultants at a point Draw D of section to left, or right of point. Write eq. of equilibrium Solve for stress-resultants or planar loading, 3 stress resultants are needed: moment shear V aial resultant N Sign convention (deformation) (Note: these sign conventions are widel used, but not universal.) 9

10 E 6 - eams - Equilibrium 11/9/1 S.P. 5: or the beam shown, assume we've computed the reactions. Determine the ISR as a function of (The reactions are 3w /8, 17w / 8) (i) < / : ake a cut to the left of midspan N 3w 3w w V V w 8 8 3w ( w) 8 Z 3w 8 w 1

11 E 6 - eams - Equilibrium 11/9/1 11 (ii) : / < ake a cut between midspan and. N (b observation) w w V V w w w 8 7 ) ( w w w w w w Z X (iii) >: (Details left to ou.) One possible D is 3, 3 w w V

12 E 6 - eams - Equilibrium 11/9/1. Equilibrium of Differential Elements.1. undamental Differential Relationship beam is transversel loaded with w() as shown. t a distance from the end, an small section of the beam is isolated as a D. ssume that w() is approimatel constant over d. or the differential length element, V ( V dv ) w( ) d dv w( ) d dv w() d (a) Z d ( V d d d ) V d w( ) w( ) d / d V (b) (The load term is higher order in d and is ignored.) d 1

13 E 6 - eams - Equilibrium 11/9/1 Substituting (b) into (a), gives a single differential equation of equilibrium... d d w( ) (c).. Initial Values onsider a differential length element at the left end. limv d w( ) d V ( ) Z lim d d Z Z d q( ) Z ( d) Therefore, the shear approaches the vertical reaction, and the moment approaches the applied moment at the left end. 13

14 E 6 - eams - Equilibrium 11/9/1.3. Stress-Stress Resultant Relationship rom the notes on stress and strain, σ d where the moments of the stresses are summed about the centroid of the section. lso V τ d Here, a sign change has been introduced to reconcile the V and τ sign conventions. Note: These equations are true regardless of the actual stress distribution. 14

15 E 6 - eams - Equilibrium 11/9/1.4. Stress Resultants b integration and summation (i) Integrate relationship (a) from section.1. dv w() V ( ) V ( 1) d w( ) d The change in shear between two points equals the "area" under the w() curve between the two points. (ii) Integrate relationship (b) from section.1. d V d ( ) ( 1) 1 1 V ( ) d The change in moment between two points equals the "area" under the () curve between the two points. V 15

16 E 6 - eams - Equilibrium 11/9/1 (iii) What if a concentrated load is applied at? Then V Vd ( V ( d P V ) P V P ) Vd P d discrete jump in V equal in magnitude, and corresponding in sign to an applied concentrated load occurs. The change in moment across the concentrated load is infinitesimal. (The latter result is because the moment arm of the concentrated load is infinitesimal.) 16

17 E 6 - eams - Equilibrium 11/9/1 (iv) What about a concentrated moment? V ( V V ) V (No change in shear at a concentrated moment.) ( ) appl V d appl (The shear couple's moment is infinitesimal and can be ignored because the moment arm is infinitesimal) t a concentrated applied moment, the shear does not change, and the moment jumps b an amount equal and opposite to the applied moment. In applications, we can often perform the integration graphicall. 17

18 E 6 - eams - Equilibrium 11/9/1 S.P. 6: Determine the shear and moment for the beam shown. 18

19 E 6 - eams - Equilibrium 11/9/1 S.P. 7: Determine the shear and moment for the beam shown. 19

20 E 6 - eams - Equilibrium 11/9/1 S.P. 8: Determine the distribution of shear and moment for the beam shown.

21 E 6 - eams - Equilibrium 11/9/1 S.P. 9: 1

22 E 6 - eams - Equilibrium 11/9/1 S.P. 1:

23 E 6 - eams - Equilibrium 11/9/1 S.P. 11: Note: the concentrated moment causes no local change in the shear, but a jump in the moment. 3