工程力学习题答案-范钦珊-蔡新着-工程静力学与材料力学-第二版

5-1

习题4-2图

图 a

图 b

图 c

13

图 d

5-2 1 b 5-3

5-4

解:(a)

A截面: FQ =b/(a+b)FP,M=0

C截面: FQ =b/(a+b) FP,M=ab/(a+b) FP D截面: FQ =-a/(a+b) FP,M=ab/(a+b) FP B截面: FQ =-a/(a+b) FP,M=0

14

(b)

A截面: FQ =M0/(a+b),M=0

C截面: FQ =M0/(a+b),M=a/(a+b)M0

D截面: FQ =- M0/(a+b),M=b/(a+b) M0 B截面: FQ =- M0/(a+b),M=0 (c)

A截面: FQ =5/3qa,M=0

C截面: FQ =5/3qa,M=7/6qa2 B截面: FQ =-1/3qa,M=0 (d)

A截面: FQ =1/2ql,M=-3/8qa2 C截面: FQ =1/2ql,M=-1/8qa2 D截面: FQ =1/2ql,M=-1/8qa2 B截面: FQ =0,M=0 (e)

A截面: FQ =-2 FP,M=FPl C截面: FQ =-2 FP,M=0 B截面: FQ =FP,M=0 (f)

A截面: FQ =0,M= FP l/2 C截面: FQ =0,M= FP l/2 D截面: FQ =- FP,M= FP l/2 B截面: FQ =-FP,M=0

5-5 (a)

FQ ( x ) =-M/2 l, M( x) =-M/2 l x ( 0 ≤ x ≤ l) FQ ( x ) =-M/2 l，M( x) =-Mx/2 l + M ( l ≤ x ≤ 2 l) FQ ( x ) = -M/2 l, M( x) = -Mx/2 l + 3M ( 2 l ≤ x ≤ 3 l) FQ ( x ) = -M2 l, M( x) = -Mx/2 l + 2M ( 3 l ≤ x ≤ 4 l)

( b)

FQ ( x ) = -(1/4)ql-qx , M( x) = ql2-(1/4)ql x –(1/2)qx2 ( 0 ≤ x ≤ l) FQ ( x ) = -(1/4)ql, M( x) =(1/4)ql(2l- x) ( l ≤ x ≤ 2 l)

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( c)

FQ ( x ) = ql-qx , M( x) = ql x + ql2-(1/2)qx2 ( 0 ≤ x ≤ 2 l) FQ ( x ) = 0 , M( x) = ql2 ( 2 l ≤ x ≤ 3 l)

(d)

FQ ( x) =（5/4）ql-qx, M( x) =（5/4）qlx-(1/2)qx2 (0≤x≤2l) FQ ( x) =-ql + q(3 l-x) , M( x) = ql(3l-x) –(1/2)q( 3l-x)2 (2 l≤x≤3 l)

(e)

FQ ( x) = qx , M( x) =(1/2)qx2 (0 ≤ x ≤ l)

FQ ( x) = ql-q( x-l) , M( x) = ql(x -1/2)-(1/2)q( x-l)2 ( l ≤ x ≤ 2 l)

(f)

FQ ( x) = -ql/2+ qx , M( x) = -(1/2)qlx +(1/2)qx2 ( 0 ≤ x ≤ l) FQ ( x) =-ql/2+ q(2l-x) , M( x) = (ql/2)(2 l-x)-(1/2)q(2l-x)2( l≤x≤2l)

5-6画出5-5图示各梁的剪力图和弯矩图，并确定 |FQ|max、Mmax。

解：（a）?MA?0，FRB? ?Fy?0，FRA?M（↑） 2l?M（↓） 2lFQFQM |FQ|max?， |M|max?2M

2l

（b）?MA?0

(ql)CAM2lBA14B1454l?ql?ql??ql?l?FRB?2l?0

21 FRB?ql（↑）

42 (a-1) (b-1) ?Fy?0，FRA?MC?FRB?1ql（↓） 411?l?ql?l?ql2（＋） 44AM2CDEM2BA1 C14B MA?ql

2

M32M2MMql2 |FQ|max?

5ql, |M|max?ql2 4 (a-2) (b-2) 16