8.2 Shear and Bending-Moment Diagrams: Equation Form
8.2 Shear and Bending-Moment Diagrams: Equation Form Example 1, page 1 of 6
1. Express the shear V and bending moment M as functions of x, the distance from
the left end of the beam to an arbitrary point on the beam. Plot V and M vs. x.
6 kip
9 kip
A
B
3 ft
x
Draw a free-body diagram and find the reactions.
6 kip
9 kip
A
3 ft
5 ft
7 ft
+
→
RA
ΣFy = 0: RA − 9 kip − 6 kip + RB = 0
ΣMA = 0: −(9 kip)(3 ft) − (6 kip)(3 ft + 5 ft) + RB(3 ft + 5 ft + 7 ft) = 0
+
1
7 ft
5 ft
Solving gives
RA = 10 kip and RB = 5 kip
RB
8.2 Shear and Bending-Moment Diagrams: Equation Form Example 1, page 2 of 6
2
Pass a section through the beam at a point between the left end and the 9-kip force.
6 kip
9 kip
x
B
A
3 ft
RA = 10 kip
5 ft
7 ft
0 < x < 3 ft
3
Draw a free-body diagram of the portion of the beam to the left
of the section and find V and M at the section.
x
A
M (sign convention: M positive counterclockwise, on right end of section)
ΣFy = 0: 10 kip − V = 0
ΣMx = 0: −(10 kip)x + M = 0
+
RA = 10 kip
+
→
V (sign convention: V positive down, on right end of section)
Solving gives
V = 10 kip and M = 10x kip⋅ft
valid for 0 < x < 3 ft.
RB = 5 kip
8.2 Shear and Bending-Moment Diagrams: Equation Form Example 1, page 3 of 6
4
Pass a section through the beam at a point between the 9-kip force and the 6-kip force.
6 kip
9 kip
x
B
A
RA = 10 kip
3 ft
5 ft
7 ft
RB = 5 kip
3 ft < x < 8 ft
5
Draw a free-body diagram of the portion of the beam to the left
of the section and find V and M at the section.
+
→
9 kip
ΣFy = 0: 10 kip − 9 kip − V = 0
M
A
ΣMx = 0: −(10 kip)x + (9 kip)(x − 3 ft) + M = 0
+
V
Solving gives
RA = 10 kip
(x − 3 ft)
3 ft
x
V = 1 kip
(3)
M = (x + 27) kip⋅ft
(4)
valid for 3 ft < x < 8 ft.
8.2 Shear and Bending-Moment Diagrams: Equation Form Example 1, page 4 of 6
6
Pass a section through the beam at a point between the 6-kip force and the right end of the beam.
6 kip
9 kip
x
B
A
3 ft
RA = 10 kip
5 ft
7 ft
RB = 5 kip
8 ft < x < 15 ft
Draw a free-body diagram of the portion of the beam to the left
of the section and find V and M at the section.
+
→
7
ΣFy = 0: 10 kip − 9 kip − 6 kip − V = 0
x
+
ΣMx = 0: −(10 kip)x + (9 kip)(x − 3 ft)
6 kip
9 kip
M
A
+ (6 kip)(x − 8 ft) + M = 0
Solving gives
3 ft
RA = 10 kip
(x − 8 ft)
5 ft
(x − 3 ft)
V
V = −5 kip
(5)
M = (−5x + 75) kip⋅ft
(6)
valid for 8 ft < x < 15 ft.
8.2 Shear and Bending-Moment Diagrams: Equation Form Example 1, page 5 of 6
8
Collect the results from Eqs. 1-6:
0 < x < 3 ft
V = 10 kip
M = 10x kip⋅ft
3 ft < x < 8 ft
V = 1 kip
M = (x + 27) kip⋅ft
8 ft < x < 15 ft
V = −5 kip
M = (−5x + 75) kip⋅ft
←Ans.
8.2 Shear and Bending-Moment Diagrams: Equation Form Example 1, page 6 of 6
9
Plot V and M versus x.
6 kip
9 kip
A
RA = 10 kip
B
3 ft
5 ft
7 ft
RB = 5 kip
10
V
(kip)
1
x
−5
M
(kip⋅ft)
35
30
x
8.2 Shear and Bending-Moment Diagrams: Equation Form Example 2, page 1 of 3
2. Express the shear V and bending moment M as functions of x, the distance from
the left end of the beam to an arbitrary point on the beam. Plot V and M vs. x.
2 N/m
20 N⋅m
A
14 m
x
1 Draw a free-body diagram and find the reactions.
14 m
=7m
Resultant = (2 N/m)(14 m) = 28 N
2
2 N/m
MA
20 N⋅m
A
14 m
+
→
RA
2
+
ΣFy = 0: RA − 28 N = 0
A couple-moment reaction must always be
included at a built-in end of a beam.
ΣMA = 0: MA − (28 N)(7 m) + 20 N⋅m = 0
Solving gives
RA = 28 N and MA = 176 N⋅m
8.2 Shear and Bending-Moment Diagrams: Equation Form Example 2, page 2 of 3
3
Pass a section through the beam at an arbitrary point (located by x)
x
2 N/m
MA = 176 N⋅m
20 N⋅m
A
14 m
RA = 28 N
ΣFy = 0: 28 N − 2x − V = 0
Draw a free-body diagram of the portion of the beam to the left
of the section and find V and M at the section.
ΣMx = 0: 176 N⋅m − 28x + (
+
4
+
→
0 < x < 14 m
Resultant = (2 N/m)(x)
MA = 176 N⋅m
x
2
Solving gives
M
A
x
RA = 28 N
x
)(2 N/m)(x) + M = 0
2
V
V = (−2x + 28) N
←Ans.
M = (−x2 + 28x − 176) N⋅m
←Ans.
valid for 0 < x < 14 m.
8.2 Shear and Bending-Moment Diagrams: Equation Form Example 2, page 3 of 3
5
Plot V and M versus x.
2 N/m
MA = 176 N⋅m
20 N⋅m
A
14 m
RA = 28 N
28
V
(N)
x
M
(N⋅m)
20
x
−176
8.2 Shear and Bending-Moment Diagrams: Equation Form Example 3, page 1 of 6
3. Express the shear V and bending moment M as functions of x, the distance from
the left end of the beam to an arbitrary point on the beam. Plot V and M vs. x.
10 lb/ft
B
A
7 ft
10 ft
3 ft
x
1
Draw a free-body diagram and find the reactions.
10 ft
= 5 ft
2
Resultant = (10 lb/ft)(10 ft) = 100 lb
B
A
ΣFy = 0: RA − 100 lb + RB = 0
ΣMA = 0: −(100 lb)(7 ft + 5 ft) + RB(7 ft + 10 ft + 3 ft) = 0
+
RA
+
→
7 ft
Solving gives
RA = 40 lb and RB = 60 lb
10 ft
3 ft
RB
8.2 Shear and Bending-Moment Diagrams: Equation Form Example 3, page 2 of 6
2 Pass a section through the beam at a point between the left end
of the beam and the beginning of the distributed load.
10 lb/ft
x
B
A
7 ft
10 ft
RA = 40 lb
0 < x < 7 ft
M
+
→
Draw a free-body diagram and find the reactions.
ΣFy = 0: 40 lb − V = 0
ΣMx = 0: −(40 lb)x + M = 0
+
3
x
V
Solving gives
RA = 40 lb
V = 40 lb
(1)
M = (40x) lb⋅ft
(2)
valid for 0 < x < 7 ft.
3 ft
RB = 60 lb
8.2 Shear and Bending-Moment Diagrams: Equation Form Example 3, page 3 of 6
4
Pass a section through the beam at a point between the
beginning and end of the distributed load.
x
10 lb/ft
B
A
7 ft
10 ft
3 ft
RB = 60 lb
RA = 40 lb
7 ft < x < 17 ft
Resultant = (10 lb/ft)(x − 7 ft)
+
→
Draw a free-body diagram of the portion of the beam to
the left of the section and solve for V and M at the section.
ΣFy = 0: 40 lb − (10 lb/ft)(x − 7 ft) − V = 0
x − 7 ft
2
ΣMx = 0: −(40 lb)x + [(10 lb/ft)(x − 7 ft)]
+
5
× ( x − 7 ft ) + M = 0
2
M
Solving gives
x − 7 ft
7 ft
x
RA = 40 lb
V
V = (−10x + 110) lb
(3)
M= (−5x2 + 110x − 245) lb⋅ft
(4)
valid for 7 ft < x < 17 ft.
8.2 Shear and Bending-Moment Diagrams: Equation Form Example 3, page 4 of 6
6
Pass a section through the beam at a point between the right end
of the distributed load and the right end of the beam.
x
10 lb/ft
B
A
7 ft
10 ft
3 ft
RB = 60 lb
RA = 40 lb
17 ft < x < 20 ft
7
Draw a free-body diagram of the portion of the beam to the left
of the section and solve for V and M at the section.
Resultant = (10 lb/ft)(10 ft) = 100 lb
10 ft
= 5 ft
2
M
7 ft
10 ft
x
RA = 40 lb
x − 17 ft
V
8
+
→
8.2 Shear and Bending-Moment Diagrams: Equation Form Example 3, page 5 of 6
ΣFy = 0: 40 lb − 100 lb − V = 0
+
ΣMA = 0: −(40 lb)x + (100 lb)[(x − 17 ft) + 5 ft] + M = 0
Solving gives
V = −60 lb
(5)
M = (−60x + 1200) lb⋅ft
(6)
valid for 17 ft < x < 20 ft.
9
Collect the results from Eqs. 1-6:
0 < x < 7 ft
V = 40 lb
M = 40x lb⋅ft
7 ft < x < 17 ft
V = (−10x + 110) lb
M=
17 ft < x < 20 ft
(−5x2
+ 110x − 245) lb⋅ft
V = −60 lb
M = (−60x + 1200) lb⋅ft
←Ans.
8.2 Shear and Bending-Moment Diagrams: Equation Form Example 3, page 6 of 6
10 Plot V and M versus x.
10 lb/ft
B
A
7 ft
10 ft
3 ft
RA = 40 lb
RB = 60 lb
V
(lb) 40
40
x
360
M
(lb⋅ft)
−60
−60
280
180
x
8.2 Shear and Bending-Moment Diagrams: Equation Form Example 4, page 1 of 6
4. Express the shear V and bending moment M as functions of x, the distance from
the left end of the beam to an arbitrary point on the beam. Plot V and M vs. x.
A
B
18 kip⋅ft
27 kip⋅ft
3 ft
x
7 ft
Draw a free-body diagram and find the reactions.
A
18 kip⋅ft
27 kip⋅ft
3 ft
5 ft
+
→
RA
ΣFy = 0: RA + RB = 0
ΣMA = 0: RB(3 ft + 5 ft + 7 ft) − 27 kip⋅ft − 18 kip⋅ft = 0
+
1
5 ft
Solving gives
RA = −3 kip = 3 kip ↓
RB = 3 kip
7 ft
RB
8.2 Shear and Bending-Moment Diagrams: Equation Form Example 4, page 2 of 6
2
Pass a section through the beam at a point between the left end and the 27 kip⋅ft
moment couple.
x
B
A
18 kip⋅ft
27 kip⋅ft
3 ft
RA = 3 kip
5 ft
7 ft
0 < x < 3 ft
Draw a free-body diagram of the portion of the beam to the left
of the section and find V and M at the section.
x
M
A
+
→
3
ΣFy = 0: −3 kip − V = 0
+
ΣMx = 0: (3 kip)x + M = 0
V
RA = 3 kip
Solving gives
V = −3 kip
(1)
M = −3x kip⋅ft
(2)
valid for 0 < x < 3 ft.
RB = 3 kip
8.2 Shear and Bending-Moment Diagrams: Equation Form Example 4, page 3 of 6
4
Pass a section through the beam at a point between the
27 kip⋅ft and 18 kip⋅ft moment couples.
x
B
A
18 kip⋅ft
27 kip⋅ft
RA = 3 kip
3 ft
5 ft
7 ft
RB = 3 kip
3 ft < x < 8 ft
Draw a free-body diagram of the portion of the beam to the left
of the section and find V and M at the section.
ΣFy = 0: −3 kip − V = 0
x
+
→
5
27 kip⋅ft
3 ft
RA = 3 kip
x − 3 ft
V
ΣMx = 0: (3 kip)x − 27 kip⋅ft + M = 0
+
M
A
Solving gives
V = −3 kip
(3)
M = (−3x + 27) kip⋅ft
(4)
valid for 3 ft < x < 8 ft.
8.2 Shear and Bending-Moment Diagrams: Equation Form Example 4, page 4 of 6
6
Pass a section through the beam at a point between the 18 kip⋅ft
moment couple and the right end of the beam.
x
B
A
18 kip⋅ft
27 kip⋅ft
3 ft
5 ft
7 ft
RB = 3 kip
RA = 3 kip
3 ft < x < 8 ft
+
→
Draw a free-body diagram of the portion of the beam to the left
of the section and find V and M at the section.
ΣFy = 0: −3 kip − V = 0
ΣMx = 0: (3 kip)x − 27 kip⋅ft − 18 kip⋅ft + M = 0
+
7
x
M
A
18 kip⋅ft
27 kip⋅ft
3 ft
RA = 3 kip
Solving gives
5 ft
V
V = −3 kip
(5)
M = (−3x + 45) kip⋅ft
(6)
valid for 8 ft < x < 15 ft.
8.2 Shear and Bending-Moment Diagrams: Equation Form Example 4, page 5 of 6
8
Collect the results from Eqs. 1-6:
0 < x < 3 ft
V = −3 kip
M = −3x kip⋅ft
3 ft < x < 8 ft
V = −3 kip
M = (−3x + 27) kip⋅ft
8 ft < x < 15 ft
V = −3 kip
M = (−3x + 45) kip⋅ft
←Ans.
8.2 Shear and Bending-Moment Diagrams: Equation Form Example 4, page 6 of 6
9
Plot V and M versus x.
B
A
18 kip⋅ft
27 kip⋅ft
3 ft
RA = 3 kip
5 ft
7 ft
RB = 3 kip
V
(kip)
x
−3
−3
21
M
(kip⋅ft)
18
3
x
−9
8.2 Shear and Bending-Moment Diagrams: Equation Form Example 5, page 1 of 6
5. Express the shear V and bending moment M in the horizontal portion ACDB of
the beam as functions of x, the distance from the left end of the beam to an
arbitrary point on the beam. Plot V and M versus x.
4 kip
4 kip
A
B
C
2 ft
x
4 ft
2 ft
+
→
Draw a free-body diagram and find the reactions.
4 kip
2 ft
ΣFy = 0: RA − 4 kip − 4 kip + RB = 0
4 kip
ΣMA = 0: −(4 kip)(2 ft) − (4 kip)(10 ft)
+
1
2 ft
D
+ RB(12 ft) = 0
A
RA
B
2 ft
2 ft
4 ft
2 ft
2 ft
Solving gives
RA = 4 kip
RB
RB = 4 kip
8.2 Shear and Bending-Moment Diagrams: Equation Form Example 5, page 2 of 6
2
Pass a section through the beam at a point between the left end and the attachment point for the first arm.
4 kip
4 kip
x
A
B
2 ft
2 ft
4 ft
2 ft
2 ft
RB = 4 kip
RA = 4 kip
Draw a free-body diagram of the
portion of the beam to the left of the
section and find V and M at the
section. Note carefully that the 4-kip
force on the left arm does not act on
this free body.
ΣFy = 0: 4 kip − V = 0
ΣMx = 0: −(4 kip)x + M = 0
+
3
+
→
0 < x < 4 ft
Solving gives
M
A
x
RA = 4 kip
V
V = 4 kip
(1)
M = 4x kip⋅ft
(2)
valid for 0 < x < 4 ft.
8.2 Shear and Bending-Moment Diagrams: Equation Form Example 5, page 3 of 6
Pass a section through the beam at a point between the attachment points of the two arms.
4
4 kip
4 kip
x
A
B
2 ft
2 ft
4 ft
2 ft
2 ft
RB = 4 kip
RA = 4 kip
Draw a free-body diagram of the portion of the beam to
the left of the section and find V and M at the section.
4 kip
ΣMx = 0: −(4 kip)x + (4 kip)(x − 4 ft + 2 ft) + M = 0
x
Solving gives
M
A
2 ft
2 ft
ΣFy = 0: 4 kip − 4 kip − V = 0
+
5
+
→
4 ft < x < 8 ft
x − 4 ft
V=0
(3)
M = 8 kip⋅ft
(4)
V
valid for 4 ft < x < 8 ft.
RA = 4 kip
8.2 Shear and Bending-Moment Diagrams: Equation Form Example 5, page 4 of 6
6
Pass a section through the beam at a point between the point
of attachment of the right arm and the right end of the beam.
4 kip
4 kip
x
A
B
2 ft
2 ft
4 ft
2 ft
2 ft
RB = 4 kip
RA = 4 kip
Draw a free-body diagram of the portion of the beam to
the left of the section and find V and M at the section.
4 kip
Note that the 4-kip force on the right arm acts on the free
body.
ΣFy = 0: 4 kip − 4 kip − 4 kip − V = 0
+
→
7
8
ΣMx = 0: −(4 kip)x + (4 kip)(x − 2 ft)
+
8 ft < x < 12 ft
4 kip
+ (4 kip)(x − 10 ft) + M = 0
x
Solving gives
M
A
2 ft
RA = 4 kip
2 ft
4 ft
2 ft
V
x − 10 ft
V = −4 kip
(5)
M = (−4x + 48) kip⋅ft
(6)
valid for 8 ft < x < 12 ft.
8.2 Shear and Bending-Moment Diagrams: Equation Form Example 5, page 5 of 6
9
Collect the results from Eqs. 1-6:
0 < x < 4 ft
V = 4 kip
M = 4x kip⋅ft
4 ft < x < 8 ft
V = 0 kip
M = 8 kip⋅ft
8 ft < x < 12 ft V = −4 kip
M = −4(x − 12) kip⋅ft
←Ans.
8.2 Shear and Bending-Moment Diagrams: Equation Form Example 5, page 6 of 6
10 Plot V and M versus x.
4 kip
4 kip
B
A
2 ft
2 ft
4 ft
2 ft
2 ft
RB = 4 kip
RA = 4 kip
V
(kip)
4
11 Note that the jumps in the diagrams occur at the
attachment points of the arms, not at the points where
the 4-kip external loads act.
x
−4
16
16
M
(kip⋅ft)
8
8
x
8.2 Shear and Bending-Moment Diagrams: Equation Form Example 6, page 1 of 8
6. Express the shear V and bending moment M as functions of x, the distance from
the left end of the beam to an arbitrary point on the beam. Plot V and M vs. x.
4 kN
8 kN
Hinge
C
A
B
2m
2m
2m
2m
x
Draw a free-body diagram and find the reactions.
4 kN
8 kN
Hinge
C
A
B
2m
RA
2m
2m
RC
+
→
RB
ΣFy = 0: RA + RB − 4 kN − 8 kN + RC = 0
2m
(1)
ΣMA = 0: RB(2 m) − (4 kN)(2 m + 2 m)
+
1
2 Two equations but three unknowns.
An additional equation is needed.
− (8 kN)(2 m + 2 m + 2 m)
+ RC(2 m + 2 m + 2 m + 2 m) = 0
(2)
8.2 Shear and Bending-Moment Diagrams: Equation Form Example 6, page 2 of 8
Pass a section through the beam at a point
immediately to the right of the hinge.
3
4 kN
8 kN
Hinge
2m
4
C
B
A
2m
Draw a free-body diagram of the portion of
the beam to the right of the section.
5
V
C
Because the section is next to a hinge,
the moment is known to be zero there
(that's what we mean by a "hinge").
2m
+
2m
RC
Write the equilibrium equation for the sum of moments about the hinge.
ΣMhinge = 0: −(8 kN)(2 m) + Rc(2 m + 2 m) = 0
7
2m
8 kN
M=0
6
2m
(3)
Note that we don't use the equation ΣFy = 0, because this equation
would introduce an additional unknown, the shear V at the hinge.
8.2 Shear and Bending-Moment Diagrams: Equation Form Example 6, page 3 of 8
8
Solving Eqs. 1-3 gives
RA = −8 kN = 8 kN ↓
RB = 16 kN ↑
RC = 4 kN ↑
9
Pass a section through the beam at a point
between the left end and the reaction at B.
4 kN
8 kN
Hinge
x
C
A
B
2m
RB = 16 kN
2m
2m
RC = 4 kN
+
→
2m
RA = 8 kN
10 Draw a free-body diagram of the portion of the beam to
the left of the section and find V and M at the section.
ΣFy = 0: −8 kN − V = 0
ΣMx = 0: (8 kN)x + Μ = 0
+
0<x<2m
Solving gives
x
M
A
V
RA = 8 kN
V = −8 kN
(4)
M = (−8x) kN·m
(5)
valid for 0 < x < 2 m.
8.2 Shear and Bending-Moment Diagrams: Equation Form Example 6, page 4 of 8
11 Pass a section through the beam at a point
between the the reaction at B and the hinge.
8 kN
4 kN
Hinge
x
A
C
B
2m
2m
RB = 16 kN
RA = 8 kN
2m
2m
RC = 4 kN
2m<x<4m
M
A
B
2 ft
RA = 8 kN
V
RB = 16 kN
ΣFy = 0: −8 kN + 16 kN − V = 0
ΣMx = 0: (8 kN)x − (16 kN)(x − 2 m) + M = 0
+
x
+
→
12 Draw a free-body diagram of the portion of the beam to
the left of the section and find V and M at the section.
Solving gives
V = 8 kN
(6)
M = (8x − 32) kN·m
(7)
valid for 2 m < x < 4 m.
8.2 Shear and Bending-Moment Diagrams: Equation Form Example 6, page 5 of 8
13 Pass a section through the beam at a point
between the hinge and the 8-kN force.
x
4 kN
8 kN
Hinge
A
C
B
2m
RA = 8 kN
2m
RB = 16 kN
2m
2m
RC = 4 kN
4m<x<6m
↑Σ Fy = 0: −8 kN + 16 kN − 4 kN − V = 0
14 Draw a free-body diagram of the portion of the beam to
the left of the section and find V and M at the section.
Σ Mx = 0: (8 kN)x − (16 kN)(x − 2 m)
+ (4 kN)(x − 4 m) + M = 0
4 kN
x
Solving gives
Hinge
M
A
B
2m
RA = 8 kN
2m
RB = 16 kN
V
V = 4 kN
(8)
M = (4x − 16) kN·m
(9)
valid for 4 m < x < 6 m.
8.2 Shear and Bending-Moment Diagrams: Equation Form Example 6, page 6 of 8
15 Pass a section through the beam at a point between
the 8-kN force and end C of the beam.
4 kN
8 kN
x
Hinge
A
C
B
2m
RA = 8 kN
2m
RB = 16 kN
2m
2m
RC = 4 kN
6m<x<8m
↑∑ Fy = 0: −8 kN + 16 kN −4 kN
− 8 kN − V = 0
16 Draw a free-body diagram of the portion of the beam to the
left of the section and find V and M at the section.
4 kN
8 kN
x
M
Hinge
Solving gives
A
B
V
2m
RA = 8 kN
2m
RB = 16 kN
∑ Mx = 0: (8 kN)x − (16 kN)(x − 2 m)
+ (4 kN)(x − 4 m)
+ (8 kN)(x − 6 m) + M = 0
2m
V = −4 kN
(10)
M = (−4x + 32) kN·m
(11)
valid for 6 m < x < 8 m.
8.2 Shear and Bending-Moment Diagrams: Equation Form Example 6, page 7 of 8
17 Collect the results from Eqs. 4-11:
0<x<2m
V = −8 kN
M = −8x kN·m
2m<x<4m
V = 8 kN
M = (8x − 32) kN·m
←Ans.
4m<x<6m
V = 4 kN
M = (4x − 16) kN·m
6m<x<8m
V = −4 kN
M = (−4x + 32) kN·m
8.2 Shear and Bending-Moment Diagrams: Equation Form Example 6, page 8 of 8
18 Plot V and M versus x.
4 kN
8 kN
Hinge
x
A
C
B
2m
RA = 8 kN
V
(kN)
2m
RB = 16 kN
2m
2m
RC = 4 kN
8
8
4
4
x
−4
−4
−8
8
M
(kN⋅m)
x
−16
8.2 Shear and Bending-Moment Diagrams: Equation Form Example 7, page 1 of 7
7. Express the shear V and bending moment M as functions of x, the distance from
the left end of the beam to an arbitrary point on the beam. Plot V and M vs. x.
20 kip
4 kip/ft
Hinge
B
A
5 ft
5 ft
8 ft
x
Draw a free-body diagram and find the reactions.
20 kip
5 ft
5 ft
Hinge
A
8 ft
= 4 ft
2
Resultant = (4 kip/ft)(8 ft) = 32 kip
B
MA
2
RA
8 ft
A couple moment must always be
included at a built-in end of a beam.
RB
+
→
ΣFy = 0: RA − 20 kip − 32 kip + RB = 0
(1)
ΣMA = 0: MA − (20 kip)(5 ft) − (32 kip)(5 ft + 5 ft + 4 ft)
3 Two equations but three unknowns.
An additional equation is needed.
+
1
+ RB(5 ft + 5 ft + 8 ft) = 0
(2)
8.2 Shear and Bending-Moment Diagrams: Equation Form Example 7, page 2 of 7
Pass a section through the beam at a point immediately to the right of the hinge.
4
8 ft
= 4 ft
2
20 kip
Resultant = (4 kip/ft)(8 ft) = 32 kip
Hinge
A
MA
B
5 ft
5 ft
8 ft
RA
5
RB
Draw a free-body diagram of the portion of
the beam to the right of the section.
6
Because the section is next to a hinge,
the moment is known to be zero there
(that's what we mean by a "hinge").
4 ft
M=0
B
V
7 Write the equilibrium equation for the sum of moments about the hinge.
+
ΣMhinge = 0: −(32 kip)(4 ft) + RB(8 ft) = 0
8
Resultant = 32 kip
(3)
Note that we don't use the equation ΣFy = 0, because this equation
would introduce an additional unknown, the shear V at the hinge.
8 ft
RB
8.2 Shear and Bending-Moment Diagrams: Equation Form Example 7, page 3 of 7
9
Solving Eqs. 1-3 gives
RA = 36 kip
RB = 16 kip
MA = 260 kip⋅ft
10 Pass a section through the beam at a point between
the left end and the 20-kip force.
20 kip
x
Hinge
MA = 260 kip⋅ft
4 kip/ft
B
A
5 ft
5 ft
8 ft
RA = 36 kip
RB = 16 kip
+
→
0 < x < 5 ft
ΣFy = 0: 36 kip − V = 0
Solving gives
MA = 260 kip⋅ft
M
A
x
RA = 36 kip
ΣMx = 0: 260 kip⋅ft − (36 kip)x + M = 0
+
11 Draw a free-body diagram of the portion of the beam to
the left of the section and find V and M at the section.
V
V = 36 kip
(4)
M = (36x − 260) kip⋅ft
(5)
valid for 0 < x < 5 ft.
8.2 Shear and Bending-Moment Diagrams: Equation Form Example 7, page 4 of 7
12 Pass a section through the beam at a point between
the 20-kip force and the hinge.
20 kip
4 kip/ft
x
MA = 260 kip⋅ft
Hinge
B
A
5 ft
5 ft
8 ft
RA = 36 kip
RB = 16 kip
5 ft < x < 10 ft
13 Draw a free-body diagram of the portion of the beam to
the left of the section and find V and M at the section.
+
→
20 kip
MA = 260 kip⋅ft
ΣFy = 0: 36 kip − 20 kip − V = 0
+
ΣMx = 0: 260 kip⋅ft − (36 kip)x + 20 kip(x − 5 ft) + M = 0
M
Solving gives
A
5 ft
V
x
RA = 36 kip
V = 16 kip
(6)
M = (16x − 160) kip⋅ft
(7)
valid for 5 ft < x < 10 ft.
8.2 Shear and Bending-Moment Diagrams: Equation Form Example 7, page 5 of 7
14 Pass a section through the beam at a point between
the hinge and the right end of the beam.
20 kip
x
4 kip/ft
Hinge
MA = 260 kip⋅ft
B
A
5 ft
5 ft
8 ft
RB = 16 kip
RA = 36 kip
+
→
10 ft < x < 18 ft
ΣFy = 0: 36 kip − 20 kip
15 Draw a free-body diagram of the portion of the beam to
the left of the section and find V and M at the section.
MA = 260 kip⋅ft
+ (4 kip/ft)(x − 10 ft)( x − 10 ft )
2
+M=0
M
Hinge
A
5 ft
5 ft
x
RA = 36 kip
ΣMx = 0: 260 kip⋅ft − (36 kip)x + 20 kip(x − 5 ft)
+
20 kip
− (4 kip/ft)(x − 10 ft) − V = 0
Resultant = (4 kip/ft)(x − 10 ft)
x − 10 ft
2
V
Solving gives
V = −4x + 56 kip
(8)
M = (−2x2 + 56x − 360) kip⋅ft
(9)
valid for 10 ft < x < 18 ft.
8.2 Shear and Bending-Moment Diagrams: Equation Form Example 7, page 6 of 7
16 Collect the results from Eqs. 4-9:
0 < x < 5 ft
V = 36 kip
M = (36x − 260) kip⋅ft
5 ft < x < 10 ft
V = 16 kip
M = (16x − 160) kip⋅ft
10 ft < x < 18 ft V = (−4x + 56) kip
M = (−2x2 + 56x − 360) kip⋅ft
←Ans.
8.2 Shear and Bending-Moment Diagrams: Equation Form Example 7, page 7 of 7
17 Plot V and M versus x.
x
MA = 260 kip⋅ft
20 kip
4 kip/ft
Hinge
B
A
5 ft
5 ft
8 ft
RB = 16 kip
RA = 36 kip
36
36
V
(kN)
16
16
x
32
−16
x
M
(kN⋅m)
−80
−260
8.2 Shear and Bending-Moment Diagrams: Equation Form Example 8, page 1 of 8
8. Express the shear V and bending moment M as functions of x, the distance from
the left end of the beam to an arbitrary point on the beam. Plot V and M vs. x.
4 kN/m
2 kN/m
B
A
x
1
6m
6m
Draw a free-body diagram and find the reactions.
2 kN/m
2 kN/m
B
A
6m
6m
RA
2
RB
Replace the trapezoidal distributed load by the sum of a rectangular and triangular load.
8.2 Shear and Bending-Moment Diagrams: Equation Form Example 8, page 2 of 8
4
Resultant of rectangular load
Resultant of triangular load
1
(12 m)(2 kN/m)
2
= (12 m)(2 kN/m)
=
= 24 kN
= 12 kN
12 m
5
B
1 (6 m + 6 m) = 4 m
3
(acts through centroid of triangle)
A
6m
+
→
ΣFy = 0: RA − 24 kN − 12 kN + RB = 0
ΣMA = 0: −(12 kN)(2 m) + RB(6 m) = 0
RA = 32 kN
RB = 4 kN
6
6m
RA
Solving gives
2 kN/m
12 m
2 kN/m
+
3
RB
6m−4m=2m
8.2 Shear and Bending-Moment Diagrams: Equation Form Example 8, page 3 of 8
7
Pass a section through the beam at a point between the left end
and the support at A.
x
4 kN/m
2 kN/m
B
A
6m
6m
RA = 32 kN
0<x<6m
8
RB = 4 kN
Draw a free-body diagram of the portion of the beam to the left
of the section and solve for V and M.
2 kN/m
w = distributed load (kN/m) at location x
M
x
V
8.2 Shear and Bending-Moment Diagrams: Equation Form Example 8, page 4 of 8
9
Before we can solve for V and M, we have to express w as a function of x. This can be done
by noting that w is a linear function of x and then using the slope-intercept equation for a line.
w
(0, 2 kN/m)
(12 m, 4 kN/m)
(x, w)
Slope
x
Intercept
w = mx + b
4 kN/m − 2 kN/m
x + 2 kN/m
12 m − 0
x
=
+2
(1)
6
=
10 Now the distributed load on the free-body of length x can be
replaced by the resultant of a rectangular and triangular load.
11 Resultant of rectangular load
Free-body diagram
= (2 kN/m)x
w−2
w
2 kN/m
x
2
=
x
3
=
M
M
x
x
V
12 Resultant of triangular load
V
x
(w − 2)
2
8.2 Shear and Bending-Moment Diagrams: Equation Form Example 8, page 5 of 8
+
→
ΣFy = 0: −(2 kN/m)x − x (w − 2) − V = 0
2
ΣMx = 0: (2 kN/m)(x)( x ) + [ x (w − 2)( x )] + M = 0
3
2
2
+
13
Replacing w in these equations by w = (x/6) + 2 from
Eq. 1 and solving gives
2
V = (− x − 2x) kN
12
3
M = (− x − x2) kN⋅m
36
(2)
(3)
valid for 0 < x < 6 m.
14 Pass a section through the beam at a point between
the support at A and the support at B.
x
4 kN/m
2 kN/m
B
A
6m
6m
RA = 32 kN
6 m < x < 12 m
RB = 4 kN
8.2 Shear and Bending-Moment Diagrams: Equation Form Example 8, page 6 of 8
15 Free-body diagram
x
w(x)
2 kN/m
M
A
6m
x−6m
RA = 32 kN
V
6 m < x < 12 m
16 We can save some work if we note that this free-body diagram is identical to the
previous one except that an additional vertical force of 32 kN is present. This
increases the shear in Eq. 2 by 32 kN and the moment in Eq. 3 by (32 kN)(x − 6 m) so
2
V = ( − x − 2x + 32) kN
12
x3
M= (−
− x2 + 32x − 192) kN⋅m
36
valid for 6 m < x < 12 m.
(4)
(5)
8.2 Shear and Bending-Moment Diagrams: Equation Form Example 8, page 7 of 8
17 Collect the results from Eqs. 4-11:
0<x<6m
V = ( − 1 x2 − 2x) kN
12
M = (− 1 x3 − x2) kN⋅m
36
6 m < x < 12 m
V = ( − 1 x2 − 2x + 32) kN
12
M = (− 1 x3 − x2 + 32x − 192) kN⋅m
36
←Ans.
8.2 Shear and Bending-Moment Diagrams: Equation Form Example 8, page 8 of 8
18 Plot V and M versus x.
x
2 kN/m
4 kN/m
A
B
6m
6m
RA = 32 kN
V
(kN)
RA = 4 kN
17
−4
x
−15
2.03
x
M
(kN⋅m)
−42