bai tap lon so 2 full pro
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GVHD:TS.DNG C HNG
LP XDCTN-K55
Trng i hc M a Cht MN HC
B mn Sc bn vt liu C HC KT CU 2
Bi tp ln s 2
TNH KHUNG SIU TNH BNG PHNG PHP LC
5-2
1)S liu nh sau : STT Kch thc hnh hc Ti trng
L1 L2 q(kN/m) P (kN)M(kN/m)
210840140120
2)S tnh :S 5
yu cu v Th T thc hin
1.Tnh h siu tnh do ti trng tc dng
1.1,v cc biu ni lc:mmen un M lc ct Qlc dc N trn h siu tnh cho.bit F=10J/L(m)
1) Xc nh bc siu tnh v chn h c bn
2)Thnh lp cc phng trnh dng chnh tc dng tng qut
3)Xc nh cc h v s hng t do ca phng trnh chnh tc,kim tra cc kt qu tnh c.
4)Gii h phng trnh chnh tc
5)V biu mmen trn h siu tnh cho do ti trng tc dng M.kim tra cn bng cc nt v iu kin chuyn v.
6)V biu lc ct Q v lc dc N trn h siu tnh cho
1.2.Xc nh chuyn v ngang ca 1 im hoc gc xoay ca tit din K bit
E=2.10kN/m.J=10L(m)
2.Tnh h siu tnh tc dng c 3 nguyn nhn(ti trng,nhit thay i v gi ta di ch).
2.1 Vit h phng trnh chnh tc dng s
2.2 Trnh by 1) V biu M do 3 nguyn nhn ng thi tc dng ln h siu tnh cho v cch kim tra
2)Tnh chuyn v nu nh trn mc 1.2 Bit:
-Nhit thay i trong thanh xin : th trn T=+45,th di l T=+30
-Tnh thanh xin c chiu cao tit din h=0.12(m)
-H s gin n di v nhit =10
-Chuyn v gi ta:
Gi D dch chuyn sang phi mt on =0,001 L (m)
Gi H b ln xung mt on =0,001 L (m)BI LM1.Tnh h siu tnh do ti trng tc dng
1.1,V cc biu ni lc:mmen un M lc ct Qlc dc N trn h siu tnh cho.bit F=10J/L(m):1. Xc nh bc siu tnh v chn h c bna)Xc inh bc siu tnh:
+iu kin cn:
n=T+2K+C+3H-3D (*)vi (K=1,H=5,C=4,D=6,T=0) thay vo cng thc (*) ta c n=0+2.1+4+3.5-3.6=3
+iu kin
Coi (II) l tri t v coi n l 1 ming cng c ni vi ming cng 1 bng 1 khp
m v thanh nf h bt bin hnhVy h cho l h siu tnh bc 3 b)Chn h c bn
2)Thnh lp cc phng trnh chnh tc dng tng qut
(11X1 + (12X2 + (13X3 + (1P = 0
(21X1 + (22X2 + (23X3 + (2P = 0
(31X1 + (32X2 + (33X3 + (3P = 0
3. Xc nh cc h s v s hng t do ca h phng trnh chnh tc,kim tra cc kt qu tnh ton Cc biu mmen un ln lt do X1 = 1; X2 = 1; X3 = 1 v ti trng gy ra trong h c bn nh trn:
Kim tra cc kt qu tnh ton
-Kim tra hang i:
+Hng 1
(NG)
+Hng 2
(NG)+Hng 3
(NG) Kim tra cc h s ca n :
(NG) Kim tra cc s hng t do ca phng trnh chnh tc
(NG)4) Gii h phng trnh chnh tc
H tnh nh tng ng :
5)V biu momen Mp trn h siu tnh cho do tai trng tc dng.Kim
tra cn bng cc nt v kim tra iu kin chuyn v.
Kim tra cn bng cc nt-Kim tra nt K:
-Kim tra nt A -Kim tra nt C: Nt C=0-Kim tra nt C:
Kim tra bng iu kin chuyn v
EMBED Equation.DSMT4
Ta thy cc chuyn v cc gi ta rt nh v ph hp vi yu cu tnh ton c chuyn chuyn v nh trn l do sai s khi tnh ton.6)V biu lc ct QP v lc dc NP trn h siu tnh cho.
1,2)Xc nh chuyn v ngang ti mt ct I(trng tm)hoc gc xoay ca mt
ct ti K.Bit E=2.108 kN/m2 ,J=10-6 L41 (m4)
Vy mt ct K xoay cng chiu kim ng h 1 gc
2.Tnh h siu tnh chu tc dng ca c 3 nguyn nhn (Ti trng ,nhit
Thay i v chuyn v gi ta )
2.1 Vit h phng trnh chnh tc.
2.2 Th t thc hin1.V biu momen un M do c 3 nguyn nhn ng thi tc dng trn h siu tnh cho v kim tra kt qu :
-Cc h s v s hng t do bit:
Tnh cc h s do tc ng ca thay i nhit
Biu lc dc Ni
(NG) Tnh cc h s do chuyn v cng bc gy ra
-i vi lc X1=1
i vi lc X2=1
i vi lc X3=1
Thay tt c cc h s v s hng t do vo phng trnh chnh tcTa c :
Gii h ta c:
H tnh nh tng ng do tc dng ca 3 nguyn nhn gy ra(ti trng,nhit thay i v chuyn v cng bc)
Biu mmen un M do c 3 nguyn nhn gy ra
Tch nt kim tra:
Nt B mmen =0
Kim tra iu kin chuyn v
Tnh chuyn v gc xoay
Vy mt ct K xoay ngc chiu vi chiu kim ng h 1 gc :
SV:Nguyn Vn Qunh MSSV:1021050081
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