MEM202 Engineering Mechanics - Statics
MEM
Shear Forces and Bending Moments in Beams
Sign Convention
1
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⎠
2
MEM
MEM202 Engineering Mechanics - Statics
Shear Forces and Bending Moments in Beams
For a ≤ x ≤ L
For 0 ≤ x ≤ a
⎛a⎞
F
A
P
V
0
V
=
−
−
=
⇒
=
−
⎜ ⎟P
∑ V y
x
x
⎝ L⎠
⎛ a⎞
∑ FV = Ay − Vx = 0 ⇒ Vx = Ay = ⎜⎝1 − L ⎟⎠ P
∑M = −A
y
⎛ a⎞
⋅ x + M x = 0 ⇒ M x = Ay ⋅ x = ⎜1 − ⎟ Px
⎝ L⎠
∑M = −A
y
P
A
⎛ a⎞
⎜1 − ⎟ P
L⎠
⎝
⎛ a⎞
M max = ⎜1 − ⎟ Pa
⎝ L⎠
⎛ a⎞
⎜1 − ⎟ Px
⎝ L⎠
a
x⎞
⎛
⋅ x − Pa + M x = 0 ⇒ M x = ⎜1 − ⎟ Pa
⎝ L⎠
RB
L
B
⎛a⎞
− ⎜ ⎟P
⎝ L⎠
x⎞
⎛
⎜1 − ⎟ Pa
⎝ L⎠
3
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
4
MEM
MEM202 Engineering Mechanics - Statics
Example
Reactions
∑M
about A
= RB (24 ) − P (9 ) − q(30)(30 2 ) = 0
⇒ RB = 9 k;
RA = 11 k
Free-body diagram #1
∑ F = 11 − 0.2(15) − 14 − V = 0 ⇒ V = −6 k
∑ M = −V (15) − 14(9) − 0.2(15)(15 2) + M = 0
V
x
⇒ M = 58.5 ft - k
15 ft
Free-body diagram #2
∑ F = V + 9 − 0.2(30 − 15) = 0 ⇒ V = −6 k
∑ M = − M + 9(9) − 0.2(15)(15 2) = 0 ⇒ M = 58.5 ft - k
V
9 ft
15 ft
5
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
6
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
7
MEM202 Engineering Mechanics - Statics
MEM
Shear-Force and Bending Moment Diagrams
w=
dV
dx
V=
dM
dx
8
MEM202 Engineering Mechanics - Statics
MEM
Shear-Force and Bending Moment Diagrams
One
Concentrated
Load
Several
Concentrated
Loads
9
MEM202 Engineering Mechanics - Statics
MEM
Shear-Force and Bending Moment Diagrams
Uniform
load over
the entire
span
Uniform
load over
part of the
span.
10
MEM202 Engineering Mechanics - Statics
MEM
Shear-Force and Bending Moment Diagrams
Cantilever
beam with
two
concentrated
loads.
Cantilever
beam with
a uniform
load.
11
MEM202 Engineering Mechanics - Statics
MEM
Shear-Force and Bending Moment Diagrams
Beam with an overhang and
a concentrated moment
12
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
13
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
14