受力分析 ∑F=0,FBc9a-1=0小A B FBA YOB ∑ F=0 Fpc sin a- Ip Fnc=56.6KN NBe FMn=40KN NAB
ফߚᵤ ∑ = ( α − = ( ∑ = α − = = % $ = $
■强度条件 AB杆的截面直径 AB 4F NAB 4x{40×103 d≥ =178×10-3m zll1Vz×160×0 d=18×10°m BC杆的截面边长 Bc =a> NBO 566×103 68,7×10°m 12×10 a=70×103m
ᔎᑺᴵӊ [ ] σ π = ≥ [ ] ( ) ( ) &% − = × × × × × ≥ = π σ π & − = × [ ] σ = ≥ [ ] & % % − = × ×× ≥ = σ − = × ᴚⱘ䴶Ⳉᕘ ᴚⱘ䴶䖍䭓
■例7-3已知BC和EF为圆截面杆,直径均为 d=30mm,{d=160MPa,试求该结构所能 承受的许可荷载。 30075016003200
՟ ˉ Ꮖⶹ Ў䴶ᴚˈⳈᕘഛЎ " ˈ σ ˙ˈ䆩∖䆹㒧ᵘ᠔㛑 ˈ䆩∖䆹㒧ᵘ᠔㛑 ᡓফⱘ䆌ৃ㥋䕑DŽ
受力分析 ∑M4=0 3 F=0. 8F B NI 3.75 300017506003200 ∑ F,×3.8 3.2×0.5 3200 0.8×3.8 E F=1.9F 3.2×0.5 FM=F
ফߚᵤ ∑ = &% % = = ∑ = % % % &% &% % % &% = × × = × × =
■强度条件 受力大的EF杆为危险杆 19F、×4 22S nd 314×010160 =5952×103N 1.9×4 5932KN
ᔎᑺᴵӊ ফⱘᴚЎॅ䰽ᴚ [ ] σ π ≤ [ ] σ π ≤ × % ( ) ( ) % % % = × × × × × × ≤ − [ ]= %$