Structural Mechanics
2.080 Recitation 4
Semester Yr
Recitation 4
4.1
Find the max. P that this beam can support without
yielding.
Cross-section
h
h
FBD:
Rx = P cos ✓
(4.1)
Rz = P sin ✓
MR = P sin ✓L
M (x) =
4-1
P sin ✓L(1
x
)
l
(4.2)
Structural Mechanics
In general,
xx
=
N
A
|{z}
2.080 Recitation 4
+
axial force
In this case:
Semester Yr
Mz
(eqn. 4.47).
I}
|{z}
|{z
bending
N = P cos ✓
A = h2
h4
12
M (x) =
(4.3)
I=
x
)
L
P sin ✓L(1
Axial component is constant along length and height, =
P cos ✓
(tension).
h2
Bending component is maximum tension at x = 0, z =
h
:
2
=
b
)
First yield occurs when
h4
12
=
xx,max
xx,max
=
h
2)
P sin ✓L(
y:
P =
=
6P sin ✓L
h3
P cos ✓ 6P sin ✓L
+
h3
h2
L
P
(cos ✓ + 6 sin ✓ ) =
2
h
h
yh
(4.4)
(4.5)
y.
2
cos ✓ + 6 sin ✓(L/h)
(4.6)
* What is the ratio of the 2 stress components?
axial
bending
Example: if ✓ = 45 , L = 20h !
P cos ✓
1 h cos ✓
h2
= ( )( )(
=
)
6P sin ✓L
6 L sin ✓
h3
axial
bending
=
1
120
4-2
(4.7)
Structural Mechanics
4.2
2.080 Recitation 4
Semester Yr
Compare the max. shear stress in this beam to the max.
tensile stress.
Recall that shear stress xz is a parabolic
function of z (eqn. 4.55)

z2
3V
1
(4.8)
(z)
=
xz
2A
(h/2)2
In our case, V = P sin ✓
xz
is max at z = 0, constant W.R.T. x !
P
L
(cos ✓ + 6 sin ✓ )
2
h
h
3 P sin ✓
3 sin ✓
2 h2
=
=
P
L
2(cos ✓ + 6 sin ✓L/h)
(cos ✓ + 6 sin ✓ )
h2
h
xx,max
xz
xx
Same example:
xz,max
=
3 P sin ✓
2 h2
=
xz
xx
=
3
' 0.012
2(1 + 120)
4-3
(4.9)
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2.080J / 1.573J Structural Mechanics
Fall 2013
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