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求解材料力学题目

求解材料力学题目

.一、取整体为受力分析对象,取sin45≈0.7:

. ΣMe =0,即:Fax.3-Fay.6+10(3)+10(3x0.7)+90 =0,

. 简化得银拆:Fax =2Fay -47, ...①,

.二、取ABC为受力分析对象,C铰接,不传递90KNm力偶,

. ΣMc =0,即:Fax.6-Fay.3+10(3x0.7) =0,

. 简化得:2Fax -Fay +7=0 ,...②,

. 将①代入②得:Fay =29 (KN) ( ↑ ), 

. 将计得的Fay=29KN代入①得:Fax =11(KN) (→),

. ΣFx =0, 即:Fax -10 +10(0.7)+Fcx =0,

. 将计得的Fax=11KN值代入上式得:Fcx = -8(KN) (←),

. ΣFy =0, 即:Fay -10(0.7) +Fcy =0, 

. 将计得的Fay=29KN代入上式得:Fcy = -22(KN) ( ↓ ),

.三、再碰搏激取整体为受力分析对象:

. ΣFx =0, 即:Fax -10 +10(0.7)+Fex =0,

. 将计得的Fax=11KN 值代笑袜入上式得:Fex = -8KN (←),

. ΣFy =0, 即:Fay -10(0.7) +Fey =0,

. 将计得的Fay=29KN代入上式得:Fey = -22KN ( ↓ ),