MEM
MEM202 Engineering Mechanics - Statics
Shear Forces and Bending Moments in Beams
A beam is a structural member or machine component that is designed to
support primarily forces acting perpendicular to the axis of the member.
Types of Beams
Simply
Supported
Propped
(One pin, one roller)
(One fixed end
and one roller)
Overhanging
Continuous
(One pin, one roller)
Cantilever
(One fixed end)
Statically Determinate
(Several pins and
rollers)
Built-in
(Both ends fixed)
Statically Indeterminate
1
MEM202 Engineering Mechanics - Statics
MEM
Shear Forces and Bending Moments in Beams
Types of Supports
Roller:
one unknown
Pin:
two unknowns
Fixed:
three unknowns
Types of Loads
Concentrated
loads
Distributed
loads
Concentrated
moments
2
MEM202 Engineering Mechanics - Statics
MEM
Shear Forces and Bending Moments in Beams
Procedure for determining shear forces and bending moments
1. Determine the reactions, if necessary,
using the free-body diagram of the
overall beam.
2. Cut the beam at cross section where
the shear force and bending moment
are to be determined. Draw the freebody diagram.
3. Set up equilibrium equations for the
free-body diagram and use them to
determine the shear force and
bending moment at the cross section.
4. Repeat steps 2 through 4 for as many
cross sections as needed.
∑F
y
∑M
= 0 ⇒ Vr = R − wx − P1
a −a
wx 2
= 0 ⇒ M r = Rx −
− P1 ( x − a )
2
3
MEM202 Engineering Mechanics - Statics
MEM
Shear Forces and Bending Moments in Beams
P
B
A
a
Reactions
⎛a⎞
(
)
(
)
=
−
=
⇒
=
M
B
L
P
a
0
B
⎜ ⎟P
∑ A y
y
⎝ L⎠
⎛ a⎞
=
+
−
=
⇒
=
F
A
B
P
0
A
⎜1 − ⎟ P
∑ y y y
y
⎝ L⎠
∑ Fx = Ax = 0
P
A
Ax
B
a
L
Ay
For 0 ≤ x ≤ a
By
Mx
A
x
A
Ay
∑F
y
Vx
Ay
For a ≤ x ≤ L
L
∑M = −A
y
P
Mx
a
x
⎛ a⎞
= Ay − Vx = 0 ⇒ Vx = Ay = ⎜1 − ⎟ P
⎝ L⎠
Vx
∑F
y
⎛ a⎞
⋅ x + M x = 0 ⇒ M x = Ay ⋅ x = ⎜1 − ⎟ Px
⎝ L⎠
⎛a⎞
= Ay − P − Vx = 0 ⇒ Vx = −⎜ ⎟ P
⎝ L⎠
∑ M = −A
y
⎛ x⎞
⋅ x + P( x − a) + M x = 0 ⇒ M x = ⎜1 − ⎟ Pa
⎝ L⎠
4
MEM
MEM202 Engineering Mechanics - Statics
Shear Forces and Bending Moments in Beams
q
B
A
L
Reactions
∑ M A = By (L) − q(L)(L 2) = 0 ⇒ By =
q
B
A
Ax
L
Ay
By
Mx
A
x
Ay
Vx
∑ Fy = Ay + By − qL = 0 ⇒ Ay =
∑F
x
qL
2
qL
2
= Ax = 0
⎛L
⎞
=
−
−
=
⇒
=
0
F
A
qx
V
V
q
⎜ − x⎟
∑ y y
x
x
⎝2
⎠
qx 2
qx
(L − x )
0
M
=
−
A
⋅
x
+
+
M
=
⇒
M
=
∑
y
x
x
2
2
5
MEM202 Engineering Mechanics - Statics
MEM
Shear Forces and Bending Moments in Beams
Sign Convention
6
MEM202 Engineering Mechanics - Statics
MEM
Shear Forces and Bending Moments in Beams
RB
RA
∑M
B
= − RA (10 ) + 2000(13)
+ 1000(6)(5) + 2000(2 ) = 0
RA = 6000 lb
∑ F = −2000 + 6000 − 1000(x − 2) − V = 0
∑ M = 2000(3 + x ) − 6000 x
y
a −a
2
(
x − 2)
+ 1000
+M =0
2
V = −1000 x + 6000 lb
M = −500 x 2 + 6000 x − 8000 ft - lb
7
MEM202 Engineering Mechanics - Statics
MEM
Shear Forces and Bending Moments in Beams
Relations among w, V, and M
w( x )
B
A
x
dx
L
∑F
y
= V + wdx − (V + dV ) = 0
or dV = wdx
dV
=w
dx
∫
x2
x1
V2
wdx = ∫ dV = V2 − V1
V1
x2
V2 = ∫ wdx + V1
x1
8
MEM
MEM202 Engineering Mechanics - Statics
Shear Forces and Bending Moments in Beams
Relations among w, V, and M
w( x )
2
B
A
x
w(dx )
∑ M O = − M − Vdx − 2
+ (M + dM ) = 0
dx
L
w(dx )
dM = Vdx +
2
2
or
dM
=V
dx
∫
x2
x1
Vdx = ∫
M2
M1
dM = M 2 − M 1
x2
M 2 = ∫ Vdx + M 1
x1
9
MEM
MEM202 Engineering Mechanics - Statics
Shear Forces and Bending Moments in Beams
q
q
B
A
L
x
qL
2
∫
x
0
∫
x
0
L
Mx
A
Vx
qL
2
wdx = ∫ (− q )dx = − qx = Vx − V0
0
Vdx = ∫
0
qL
2
qL
w = − q V0 =
2
x
x
B
A
⎛ Lx x 2 ⎞
⎛L
⎞
− ⎟⎟ = M x − M 0
q⎜ − x ⎟dx = q⎜⎜
2⎠
⎝2
⎠
⎝ 2
M0 = 0
⎞
⎛L
Vx = q ⎜ − x ⎟
⎠
⎝2
Mx =
qx
(L − x )
2
10