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Unformatted text preview: Hw it: .a a .. ' .2 Suppose that the length of the beam in Example 8 2' Is L: 8 ﬂ
and it is made of ASTM A36 structural steel The maximum
anticipated magnitude of the distributed load is WI): 2400 lblft.
Determine the dimension n so that the beam has a factor of safety 8= 3.
FTQQ 39d}: Diagram: ’ ,.. "l 2,400llo/Ft
/,#«’ I
A B
A y B
OLUTION: Y Summing moments about point B to ﬁnd the reaction at point A: ' 1 lb 3
2M, = o = [5[2400E](8 ML 19]— A118 fr) Ay=3200b T Summing vertical forces to ﬁnd 3,:
2Fsf = 0 =  1/: (2400 Ibi‘ft) (8ft) 4 3200 lb + By =6400Ibt Draw the F80 for the lefthand portion of the beam and ﬁnd the location where the shear force is zero (M is
max): ft F. , M
4619ft X_.1Tj Summing moments about a cut through the beam at x = 4. 619 ft:
M = 0 =(32001b)(4.619ﬁ}[—[300%)(4 691ﬁX4 ﬁlgﬁ)][4.6;9ﬂ] + M
M = 9853 ftlb y 1 w (300xxb/Ft
as; = o = 32001.5 — E[300—400 (x) Using the factor of safety (given) to ﬁnd the maximum allowable normal stress: 36, oooib
crm=%=———— 3/ =12,ooo!/ The moment of inertia for the square crosssection would be: I _ Mk 3 # h‘
12 12
Using equation (812) to determine the dimensions of the square (and remembering to convert ftlb to inlb):
(9853 f: mung— 11—12:]
1b = _..__ —> h3 = 59.12 in3
12,000 All *7
12
ANS h = as in 41? ..13.?.— 1.__1.‘ .'"'.;:_'_‘_'.:'. ‘F’F; ._._ .._. : \
l
..
a'j
Ii.
1:

I The maximum anticipated load on the beam is shown.
Choose a material from Appendix B and design a cross
section for the beam so that it has a factor of safety 5‘ = 2. Free Body Diagram: / «1 EU’UUOWN ﬂ !‘}~20kNim
A ’ " / I t .I B I . '
lI—3nI—‘'f'——Jrn . 3m——4—3m B
sgcynon: y i
From Appendix B for 70T5T6 aluminum. o, = 480 MPa. A A crosssection is chosen such that its height is twice its
width. The moment of inertia for the crosssection is:
3
I = ﬂ = ‘0'"): = 0.667x‘m‘
12 12
Ex
Summing moments about point B to determine the reaction at point A: l
1
2MB 1 o 1 [3(2000001 f mksm)](4m) — Mam) l—x—l A, = 20.000 N t The bending moment is maximum at the location where the shear force is zero, or the sum of vertical forces
is zero. Summing vertical forces on the FBD: 2F), = 0 = 20,000N —[~1—(<5,i567mm2 xm)](xm) 1 x = 2.45 m
2 Summing moments on the FBI): W = 0 = (20,000NX245m)  [%(5,66?N im)(2.45mX2.45m)12':5m] + M
M=32.659 Nm
The maximum allowable normal stress due to bending is found using the factor of safety. a 10W; 2
a...” = :3: = “if—m = 240::10‘th2 Now using Equation (8:12) to determine the dimension. x: 240x10iiwm2 = W 1 x3 = 0.000204 m3 0.667x‘m“ ANS x = 9.0589 m = 5&3 mm 421 ...
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 Spring '06
 Keil

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