V. DEMENKO MECHANICS OF MATERIALS LECTURE 6 Plane Bending Deformation. Diagrams of Internal Forces (Continued)


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1 V. DEMENKO MECHNCS OF MTERLS LECTURE 6 Plne ending Deformtion. Digrms of nternl Forces (Continued) 1 Construction of ending Moment nd Shering Force Digrms for Two Supported ems n this mode of loding, first of ll, we should determine the support rections. Let us consider the method in following emple. Emple 1 nternl forces induced by concentrted force Given: F,, b. R.D.: Q( ), My( ). Let us find the support rections R nd R from the equtions of equilibrium of the bem: M = 0 R + b F = 0, (1) R = F + b, () M = 0 R ( + b) Fb = 0, (3) Fig. 1 R = Fb + b. (4) Let us check the support rections The result is true. Fb F F = 0, F + = 0. (5) + b + b Let us divide the bem into portions nd write the equtions of shering forces nd bending moments for n rbitrry crosssections of two portions:  nd . From the conditions of equilibrium of the left () or right () portion of the mentlly cut rod it follows tht for portion t distnce from support internl forces re described by the equtions.
2 V. DEMENKO MECHNCS OF MTERLS 015 Fb b = =, M ( ) R + 0 Q R Fb Fb = = = 0 = + b = = + b. (6) y Portion. n rbitrry section is situted t distnce within the limits F 0 b: Q ( ) = R = + b, (7) F Fb My ( ) = R= = 0 = + b = 0 = b + b. (8) Thus, in ech portion of the bem, the force Q is constnt in vlue nd is positive for portion nd negtive for portion. The moment depends linerly on ; it increses in portion from 0 to Fb ( + b) nd increses from ero to this vlue in portion. The digrms of Q ( ) nd M ( ) cn now be creted using results of these clcultions. y t should be noted tht the digrm of shering forces hs n brupt in the point of ppliction of specified eternl force, which is equl to the mgnitude of the force. R, F, For instnce, the Q ( ) digrm hs three brupts corresponding to the forces R. There re no ny brupt in the digrm of bending moments. Fig. Emple nternl forces induced by concentrted moment Given: M,, b. R.D.: Q (), M y (). From the equtions of equilibrium, the rections re since R = M R = + b, (9) M = 0; R + b M = 0, R = M, R = R. (10) + b Equtions of internl forces: Portion : 0 M Q ( ) = R = + b, (11)
3 Portion : 0 b V. DEMENKO MECHNCS OF MTERLS M M M ( ) = R = + b = = + b. (1) y 0 = = 0 M Q ( ) =+ R =+ + b, (13) M Mb M ( ) = R = + b = = + b y 0 = b = 0 (14) Note: point of n eternl moment ppliction, the bending moment digrm hs jump (brupt). t is equl to the mgnitude of the ctive moment pplied in this point. Emple 3 nternl forces induced by uniformly distributed lod Given: q, l. R.D.: Q( ), My( ). Let us determine the rections of supports nd from conditions of equilibrium: R Equtions of internl forces: n portion (0 l) : ql = R =. (15) Fig. 3 ql ql ql Q ( ) = q = = =, (16) 0 = l ql q ql My( ) = = 0 = = 0. (17) 8 = l = 0 = l
4 4 V. DEMENKO MECHNCS OF MTERLS 015 The shering force decreses linerly from ql to ql nd the bending moment vries long the bem length nonlinerly nd ttins the mimum vlue in the bem midsection. Emple 4 nternl forces induced by linerly distributed lod Given: q m, l. R.D.: Q( ), My( ). Clcultion of the support rections: M = 0, ql 1 Rl l = 0, (18) 3 ql R =, (19) 6 M = 0, ql Rl l = 0, (0) 3 ql R =, (1) 3 Checking: Fig. 4 ql ql ql F = 0, + = 0. () 6 3 The equtions of Q( ) nd My( ) re s follows: q ql q ql ql Q ( ) = R + = 6 + l = l 6 = = l 3 ; (3) Etreme vlue of bending moment is clculted substituting into its eqution corresponding coordinte of the point. The lst one is determined equting to ero the derivtive of bending moment function which is described by sher force (check if you disgree): = 0 ql q e e 0 e l Q = + = =. (4) 6 l 3
5 V. DEMENKO MECHNCS OF MTERLS q ql q ql y 0 0 l l = l = e = 0 M = R = = = =. (5) Reltionships etween ending Moment M y, Shering Force Q nd Distributed Lod ntensity q() Let rod be fied in rbitrry mnner nd subjected to distributed lod of intensity q q( ) = in generl cse. There re other lods pplied to the bem. The direction dopted for q is considered positive (see Fig. 5). moments solte n element of length d from the rod nd pply to both its boundries the M y nd My dm y + nd lso shering forces Q nd Q + dq. Fig. 5 The lod q my be considered to be uniformly distributed within the infinitesimlly smll segment d (see Fig. 6). Equte to ero the sum of the projections of ll forces on verticl is : 0 Q + q d Q dq =. (6) Mking simplifictions, we obtin dq d ( ) = q. (7) Fig. 6 Equte to ero the sum of the moments with respect trnsverse is y (point ) d My + Q d+ q( ) d My dm y = 0. (8)
6 6 V. DEMENKO MECHNCS OF MTERLS 015 Mking simplifictions nd rejecting the smll quntity of higher order, we obtin dm y d ( ) ( ) = Q. (9) Thus the shering force in fct represents the derivtive of the bending moment with respect to the length of the rod. This conclusion ws proved in the emple mentioned bove. The derivtive of the shering force equls to the intensity of eternl distributed lod q. Emple 5 Generl cse of cntilever bem loding Given: = m, b = 1m, c = 3m, F = 50kN, q = 40 knm, M = 60kNm. R.D.: Q( ), My( )? Equtions of internl forces in n rbitrry crosssections of the portions: 0 < < 0 50 Q =+ F q = = = = = 30kN, My ( ) =+ F q = 0 = = 0 = 0 knm. = Fig. 7 dm d e M y m? dm Q F q; d F = F qe = 0 e = = 1.5m. q = = Q ( ) = 0. e
7 V. DEMENKO MECHNCS OF MTERLS q e M = M 50( 1.5) ym y e = F e = = = knm. 0 < < b 30 Q =+ F q = kn, My ( ) = F( + ) q + = 0= 50 ( + 0) 40 1 () = 0 = 1= 10 knm. 0 < < b 30 Q = F q = kn, My ( ) = F( + b+ ) q + b+ M = 0= 10 = 1= 70kNm. Emple 6 Generl cse of two supported bem loding Fig. 8 Given: q = 0 knm, F = 50kN, M = 130 knm, = 3m, b= c= d = 1m. R.D.: Q( ), My( )? 1) Clcultion of rections R, M = 0 =+ M + F + b + + q R ( + b+ c+ d), R =+ 70 kn. R : The plus sign mens coincidence of R ctul direction with originlly selected.
8 8 V. DEMENKO MECHNCS OF MTERLS 015 M 0 M F c d q = =+ + + b+ c+ d + R + b+ c+ d, R =+ 40 kn. Checking: F = 0= R R + F + q = , i.e. the rections re determined correctly. ) Equtions of internl forces in n rbitrry crosssections of corresponding portions: 0< < 1 70 Q = R = kn, 0 1 My =+ R = = 0 = = 70 knm. 0< < 1 70 Q = R = kn, 0 1 My =+ R d + M = 60 = = 10kNm. 0< < Q = R q = = 40 = = 0 knm, M y m? = = 0 e e Q R q e R 40 e = = = m, q 0 M ( e) = Re q =+ 40kNm. y V V V 0< < 1 0 Q = R q = kn, V My ( ) =+ R( + ) q + = 0= 30 = 1= 10 knm.
9 V. DEMENKO MECHNCS OF MTERLS Emple 7 Generl cse of two supported bem loding Given: q = 0 kn/m, F = 100kN, M = 60kNm, = m, b = 1m, c = 3m, d = 4 m. R.D.: Q( ), My( )? 1) Clcultion of rections R, R : orig M = 0 = R b+ c + F b+ c+ d c+ d qc ( + d) + b M + M q = = R (), 4= 190, R Fig. 9 R orig ct 190 = R = = kn. 4 The plus sign mens coincidence of direction with originlly selected. (b) M = 0=+ Fd + M M + R ctul c d orig + qc qd + R b+ c q + b+ c = 9 = 100( 4) orig orig 0 + R ( 4) 40( ) = R ( 4) 190, + R = 47.5 kn. The minus sign mens noncoincidence of Thus, originl direction of R ctul direction with originlly selected. R must be chnged on opposite. ct orig (c) Checking: F = 0 =+ R R + F + q qc ( + d) = 0, i.e. the rections re determined correctly.
10 10 V. DEMENKO MECHNCS OF MTERLS 015 ) Equtions of internl forces in n rbitrry crosssections of corresponding portions: 0 < < = 0= 0 = 40 kn, My( ) q = 0 = Q q = = 0 < < b 87.5 Q = q R = kn, ctul = = 0 = 40 knm. My ( ) = q + Rctul+ M = 0=+ 80 = 1= 7.5 knm. 0 < < d = 0 = 4 Q =+ F q =+ 100kN =+ 0kN, My ( ) = F+ q = 0= 0 = 4 = 40kNm. V V 0 < < c = 0 = 3 V Q =+ F q d + R = 7.5kN = 87.5kN, V My ( ) = F( d + ) + R+ q( d + ) d + = 0= 40kN = 3 = 67.5kNm. 3 Construction of nternl Force Digrms for Stticlly Determinted Frmes y br system is ment ny structure consisting of rodshped elements. f the elements of structure ct primrily in tension or compression, the br system is clled truss. f the elements of br system re primrily in bending or torsion, the system is clled frme. We will consider plne frmes. y stticlly determinte system is ment system for which ll the rections of the supports cn be determined by mens of equtions of equilibrium nd the internl forces t ny cross section cn lso be found by the method of sections.
11 V. DEMENKO MECHNCS OF MTERLS Given: F,, b. R.D.: N( ), Q( ), My( ) functions nd their grphs. Solution Equtions of internl forces in two portions re: 0 < < N ( ) = 0, 0 Q =, Fig y = = M = F = = F, 0 < < b =, Q ( ) = 0, N F Corresponding grphs re: M = F. y Fig. 11
12 1 V. DEMENKO MECHNCS OF MTERLS 015 Emple 9 Clcultion of internl forces in plne frme Given: q = 0 kn/m, l = 3m. orig v R = 15kN (ctul direction of R.D.: N( ), Q( ), My( ). t first let us determine the rections in supports nd : ) M = 0, R = 0, R =+ 15kN; b) F = 0, h R = qb =+ 60kN: c) M = 0, orig v R = 0, ct R v is opposite to originl). Let us write the equtions of the internl forces: N ( ) = 0 ; 15 N =+ R =+ kn; 60 N =+ R =+ kn; h 15 Q = R = kn; 0 Q = q 0 60 = = = = kn; l ct 15 Q =+ R = kn; v 0 My = R 0 45 = = = = knm; l q y = 3 = 0 M = R l =+ 45 = 45 knm; Fig. 1 ct 0 3 My = Rv 0 45 = = = = knm.
13 V. DEMENKO MECHNCS OF MTERLS Using this equtions the bending moment digrm nd the digrms of norml nd shering forces my be constructed: N ( ), kn Q ( ), kn My, knm Fig. 15 Checking the results, i.e. the equilibrium of the frme ngles. Fig. 16 Emple 10 Clcultion of internl forces in plne frme Given: q = 40 kn/m, F = 50kN, M = 40kNm, = m, b = 4m, c = 3m. R.D.: N( ), Q( ), My( )? 1) Clcultion of support rections: b orig orig () M = 0= M M + F q + qb + R 65 v b Rv = kn. The minus sign mens noncoincidence of R v ctul direction with originlly selected. Thus,
14 14 V. DEMENKO MECHNCS OF MTERLS 015 originl direction of R v must be chnged on opposite. orig ct (b) F = 0= F + R R + q + b R = 15 kn, i.e. ctul direction of R is upwrds. v (c) F = 0= R F R =+ 50 kn (right directed). h h ) Equtions of internl forces in n rbitrry crosssections of the portions (see Fig. 17): 0 < < 0 N =, 0 Q =+ F q = = e F 50 = = = 1.5 m. q 40 Fig. 17 = 50 = 30 kn, = My( ) = F q = 0 = = 0 = =+ 0 knm, = 0 e e Q = F q =, 1 My( e) = Mym = Fe qe = knm. 0 < < c = = 15kN, Q ( ) F 50 N R 0 3 My = F = = 0 = = 150kNm. 0 < < c =+ =+ kn, = = 65kN, Q ( ) R 50 N R v 0 3 = = kn, My =+ Rh+ M = = 40 = =+ 190kNm. h
15 V V V 0 < < c V. DEMENKO MECHNCS OF MTERLS V =+ =+ 50 kn, 0 4 N R h Q = Rv + q = = 65 = = 95kN, V My ( ) = q + Rv+ M + Rhc = 0= 190 = 4 = 130kNm. V ct R 65 Q 0 v e = Rv + qe = e = = = 1.65 m, q 40 ct V V e ct ct y ( e) ym v e h ( 1.65) M = M = q + R + M + R c= = 8.4 knm. Fig. 18 Fig. 19 Fig. 0 3) Checking the results i.e. n equilibrium of the rods connection res (see Figs 1 ).
16 16 V. DEMENKO MECHNCS OF MTERLS 015 Fig. 1 Fig. 4 Construction of the Q () nd M y () Digrms for Curviliner ems Given: R, F. R.D: N( ), Q( ), My( ). The force F cn be resolved long the nd es into the components F sinα. F cosα nd Fig. 3 F = Fsinα, F = Fcosα, 0 N α = F = Fsinα 0 F 0, α = = α = π = α = π = 0 Q α =+ F = Fcos α F 0 F, α = = α = = π α = = π 0 My α = Fh = FRsinα 0 FR 0 α = = α = π = α = π =. Fig. 4
17 V. DEMENKO MECHNCS OF MTERLS Emple 11 Clcultion of internl forces in curviliner frme Given: M, F, R. R.D.: N( ), Q( ), My( ). Solution Equtions of internl forces re: 0 < α < π / N ( α ) = 0; 0 Q α = ; Fig. 5 My α = M. 0 < α < π / ( α) sinα 0 α = 0 α = π / N = P = = P, ( α) α α = 0 α = π / Q = Pcos = P = 0, y ( α) sinα α = 0 α = π / M = M + PR = M = M + PR. The grphs of internl forces re: Fig. 6
18 18 elements V. DEMENKO MECHNCS OF MTERLS 015 Emple 1 Clcultion of internl forces in plne frmes with curviliner Given: M = 40kHm, q = 10kN/m, = 1m. R.D.: N, Q, M y functions nd their grphs. Solution 1) Clcultion of support rections: F = 0 q R = 0 = 0 kn, R H H M F = 0 R M q = 0 = 30 kn, R rbitrry crosssections of corresponding portions: 0 = = 30kN, Q ( ) = 0, 0 N R 0 α π 0 N = R cos = 30 = 0, α α = α = π α α = 0 α = π y V V F = 0 R R = 0 R = 30 kn. ) Equtions of internl forces in n M =. Q = R sin = 0 = 30 kn, ( α) y α= 0 α= π M = R 1 cos = 0 = R = 30 knm. 0 α π α α = 0 α = π N = R sin = 0 = 30 kn, y Fig. 7 α α = 0 α = π Q = R cos = 30 = 0,
19 V. DEMENKO MECHNCS OF MTERLS ( α) = 0 = M y = R 1+ sin M = 10 = 0 knm. α α π V V V 0 α V = = 30 kn, Qy ( ) RH q = 0 = N R V V q y АH = = 0 M = R = 0 = 0 knm. = = 0 = 0, Fig. 8 Glossry Stticlly determinte system (рус. статически опpеделимая система, укр. статично визначувана система) the system, rections nd internl forces of which could be determined solely from freebody digrms nd equtions of equilibrium. Truss (рус. ферма, укр. ферма) frme, generlly of steel, timber, concrete, or light lloy, built from members in tension nd compression.
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