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User Generated
Subject
Engineering
School
Ferris State University
Type
Homework
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Question 1
Consider the beam with rectangular cross section shown. The beam has internal resultant
loadings on the cross section at point C with directions shown in the free body diagram of the left
side section. The magnitudes of these internal loadings are N
C
=312 N, V
C
= 180 N, and M
C
=360
N-m.
a.) Determine the state of stress located at point D on the beam’s cross section.
Use the formula to calculate the bending stress:
D
N My
AI
=
where:
N= load
A= area
My= the internal bending moment about the neutral axis
y- the perpendicular distance from the neutral axis to the point on the section.
I= moment of inertia
Since y= 0
0
My
I
=
312
0 62.4
(0.1)(0.3)
DD
kPa

= + =
(I didn’t have the chance to calculate the bending stress for letter a, I only solved for the
shear stress.)
Use the formula to calculate the shear stress:
D
VQ
It
=
Where:
V= value of the shear force at the section
Q= first moment of the area
I= moment of inertia
T= width of the cross-section

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2
3
(180 )(0.025 )(0.05 )
54
1
(0.05 )(0.1 ) (0.05 )
12
DD
N m m
kPa
m m m

= =
(For the shear stress I used the alternative formula
max
3
2
V
A
=
)
b.) Determine the state of stress located at point E on the beam’s cross section
Use the formula to calculate the bending stress:
E
N My
AI
=
where:
N= load
A= area
My= the internal bending moment about the neutral axis
y- the perpendicular distance from the neutral axis to the point on the section.
I= moment of inertia
3
312 (360 )(0.05 )
4.38
1
(0.1 )(0.05 )
(0.05 )(0.1 )
12
EE
N N m m
kPa
mm
mm

= + =
(For the bending moment I didn’t consider the normal stress so I end up getting 4.32 kPa
as my answer)
Use the formula to calculate the shear stress:
E
VQ
It
=
Where:
V= value of the shear force at the section
Q= first moment of the area
I= moment of inertia
T= width of the cross-section
Since Q=0
0
EE
VQ
It

= =
c.) Draw a 2-D element representing the state of stress for part (b) at point E.

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Question 1 Consider the beam with rectangular cross section shown. The beam has internal resultant loadings on the cross section at point C with directions shown in the free body diagram of the left side section. The magnitudes of these internal loadings are NC=312 N, VC= 180 N, and MC=360 N-m. a.) Determine the state of stress located at point D on the beam’s cross section. Use the formula to calculate the bending stress: N My D =   A I where: N= load A= area My= the internal bending moment about the neutral axis y- the perpendicular distance from the neutral axis to the point on the section. I= moment of inertia Since y= 0 My =0 I 312 D = + 0 →  D = 62.4kPa (0.1)(0.3) (I didn’t have the chance to calculate the bending stress for letter a, I only solved for the shear stress.) Use the formula to calculate the shear stress: VQ D = It Where: V= value of the shear force at the section Q= first moment of the area I= moment of inertia T= width of the cross-section D = (180 N )(0.025m)(0.05m) 2 →  D = 54kPa 1 3 (0.05m)(0.1m) (0.05m) 12 (For the shear stress I used the alternative formula  max = 3V ) 2A 3(180) →  max = 54kPa 2(0.05 x0.1) b.) Determine the state of stress located at point E on the beam’s cross section  max = Use the formula to calculate the bending stress: E =  N My  A I where: N= load A= area My= the internal bending moment about the neutral axis y- the perpendicular distance from the neutral axis to ...
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