sm7 67

________________________________________________________________________
PROBLEM (7.67) A hollow box- beam supports the loads shown in Fig. P7. 67. Compute the
maximum value of P for all
σ
= 5 MPa and all
τ
= 1.2 MPa.
SOLUTION
0)4()2()5(40:0 =+−−=
∑BA RPM
50 0: 10
22
ByA
PP
RFR=+ = =−
∑
y
z
200 mm
50 mm
50 mm
200 mm
C
40 kN
B
2 m 1 m
R
B=P/2+50
C
A
R
A=P/2-10
2 m
P
V, kN
P
/2-10
-
P
/2-10
40
x
M
, kN⋅m
40
2P-20
x
33 6
1[200(300) 100(200) ] 383.33 10
12 4
I
mm=−=×
33
max 1017505010010075150200 mmQ ×=××−××=
We have
36
663
( 10)10 (1750 10 )
2
;1.210 383.33 10 (2 50 10 )
all
P
VQ
Ib
τ
−
−−
−×
=×=
×××
or
1.2(38,333)
10 , 72.57
2 1750
PPk−= = N
33
66
(2 20)10 (150 10 )
;510 383.33 10
all Mc P
I
σ
−
−
−×
=×= ×
or
2kNPP 39.16,78.1220
=
=−
Thus,
kNPall 39.16=
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