bending problems
Post on 11-Jun-2015
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Prob. 4.1: Knowing that the couple shown acts in a vertical plane, determine the stress at (a) point A. (b) point B.
Solution:
I=bh3
12=[ (80 ) (120 )3
12−
(40 ) (80 )3
12 ]×10−12=9.8133×10−6m4
σ=− yMI
At A:
σ=−(0.04 )(15000)9.8133×10−6 =61.14MPa (Compressive stress)
Bending Problems
Part1
ϵ=− yρ→σ=− yE
ρ,M= EI
ρ→σ=− yM
I
At B:
σ=−(−0.06 )(15000)9.8133×10−6 =91.7MPa (Tensile stress)
Prob. 4.2: Knowing that the couple shown acts in a vertical plane, determine the stress at (a) point A. (b) point B.
Solution:
From the symmetry of the cross section, we can see obviously that the centroidal axis is z-axis.
I z=b h3
12−2[ 14 π r4 ]
I z=(120)(60)3
12−2[ 14 π (19)4]=1.12922519∗10−8m4
σ=− yMI
At point A:
σ=− (0.03 )(2000)1.12922519∗10−8=−5.31GPa
At point B:
σ=− (−0.019 )(2000)1.12922519∗10−8=3.365GPa
Prob. 4.3: A beam of the cross section shown is extruded from an aluminum alloy for which σ y=250MPa∧σU=450MPa.Using a factor of safety of 3.00, determine the largest couple that can be appliedto the beam when it is bent about the z-axis.
Solution:
σ all=σUF . S
=4503
=150MPa
Since σ all<σ ythe beam remains in the elastic range and we can apply our relations mentioned above.
I=bh3
12=[ (16 ) (80 )3
12−
(16 ) (32 )3
12 ]×10−12=1.409×10−6m4
σ all=− yMI M largest=
−σ all∗Iymax
∴M largest=−150∗106∗1.409∗10−6
−0.04=5.283KN .m
Prob. 4.5: The steel beam shown is made of a grade of steel for which σ y=250MPa∧σU=400MPa. Using a factor of safety of 2.50, determine the largest couple that can be appliedto the beam when it is
bent about the x-axis.
Solution:
σ all=σUF . S
=4002.5
=160MPa
Since σ all<σ ythe beam remains in the elastic range and we can apply our relations mentioned above.
I=[ (10 ) (228 )3
12+2∗[ (200 ) (16 )3
12+ (200 ) (16 ) (122 )2] ]×10−12
¿1.0527109×10−4m4
M largest=−σ all∗Iymax
=−160∗106∗1.0527109×10−4
−0.13
¿129.564KN .m
Prob. 4.10: Two equal and opposite couples of magnitude M=25KN .m are applied to the channel-shaped beam AB. Observing that the couples cause the beam to bend in a horizontal plane,
determine the stress at (a) point C, (b) point D. (c) point E.
Solution:
I y Z`i A Z`i AreaPart#
¿3.443∗10−6m4
¿5.213∗10−6m4
¿5.213∗10−6m4
0.12(0.036)3
12+0.00432(0.02625)2
0.03¿¿
0.03¿¿
77760
216000
216000
18 60
60
4320
3600
3600
1
2
3
z=∑ Z i A
∑ A=50976011520
=44.25mm
I y=I y1+ I y2+ I y3=1.386936∗10−5m4
At point (C):
σ=−zMI
=−−0.04425∗250001.386936∗10−5
=79.76MPa (Tensile)
At point (D):
σ=−zMI
=−0.07575∗250001.386936∗10−5
=−136.54 MPa(compressive)
At point (E):
σ=−zMI
=−−0.00825∗250001.386936∗10−5
=14.87MPa (Tensile)
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