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)