Hibbeler R.C. Structural Analysis
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140 CHAPTER 4INTERNAL LOADINGS DEVELOPED IN S TRUCTURAL MEMBERS
4
Procedure for Analysis
The following procedure provides a method for determining the
variation of shear and moment in a beam as a function of position x.
Support Reactions
• Determine the support reactions on the beam and resolve all the
external forces into components acting perpendicular and parallel
to the beam’s axis.
Shear and Moment Functions
• Specify separate coordinates x and associated origins, extending into
regions of the beam between concentrated forces and/or couple
moments, or where there is a discontinuity of distributed loading.
• Section the beam perpendicular to its axis at each distance x, and
from the free-body diagram of one of the segments determine the
unknowns V and M at the cut section as functions of x. On the free-
body diagram, V and M should be shown acting in their positive
directions, in accordance with the sign convention given in Fig. 4–1.
• V is obtained from and M is obtained by summing
moments about the point S located at the cut section,
• The results can be checked by noting that and
where w is positive when it acts upward, away from
the beam. These relationships are developed in Sec. 4–3.
dV>dx = w,
dM>dx = V
©M
S
= 0.
©F
y
= 0
The joists, beams, and girders used to support this floor can be
designed once the internal shear and moment are known
throughout their lengths.
4.2 SHEAR AND MOMENT FUNCTIONS 141
4
EXAMPLE 4.4
Determine the shear and moment in the beam shown in Fig. 4–6a as a
function of x.
2 k/ft
x
30 ft
(a)
Fig. 4–6
20 ft
(b)
30 k
600 kft
30 k
x
(c)
600 kft
30 k
x
—
3
M
V
1
x
__
(
__
) x
2 15
x
w
__
k/ft
15
SOLUTION
Support Reactions. For the purpose of computing the support
reactions, the distributed load is replaced by its resultant force of 30 k,
Fig. 4–6b. It is important to remember, however, that this resultant is
not the actual load on the beam.
Shear and Moment Functions. A free-body diagram of the beam
segment of length x is shown in Fig. 4–6c. Note that the intensity of
the triangular load at the section is found by proportion; that is,
or With the load intensity known, the resultant
of the distributed loading is found in the usual manner as shown in the
figure. Thus,
Ans.
Ans.
Note that and which serves as a
check of the results.
dV>dx =-x>15 = w,dM>dx = V
M =-600 + 30x - 0.0111x
3
600 - 30x + c
1
2
a
x
15
bx d
x
3
+ M = 0d+©M
S
= 0;
V = 30 - 0.0333x
2
30 -
1
2
a
x
15
bx - V = 0+
c
©F
y
= 0;
w = x>15.w>x = 2>30
EXAMPLE 4.5
142 CHAPTER 4INTERNAL LOADINGS DEVELOPED IN S TRUCTURAL MEMBERS
4
14 ft
48 k
60 k
100 k ft
6 ft
108 k
1588 k ft
(b)
48 k
6 ft
108 k
1588 k ft
M
V
x
2
6
x
2
(d)
60 k
x
1
x
3
x
4
x
2
12 ft
20 ft
4 k/ft
100 k· ft
(a)
4x
1
1588 k⭈ft
108 k
M
V
x
1
x
1
—
2
(c)
Fig. 4–7
SOLUTION
Support Reactions. The reactions at the fixed support are
and Fig. 4–7b.
Shear and Moment Functions. Since there is a discontinuity of
distributed load at two regions of x must be considered in
order to describe the shear and moment functions for the entire
beam. Here is appropriate for the left 12 ft and can be used for
the remaining segment.
Notice that V and M are shown in the positive
directions, Fig. 4–7c.
Ans.
Ans.
Fig. 4–7d.
Ans.
Ans.
These results can be partially checked by noting that when
then and Also, note that
and dV>dx = w.dM>dx = V
M =-100 k
#
ft.V = 60 kx
2
= 20 ft,
M = 60x
2
- 1300
1588 - 108x
2
+ 481x
2
- 62+ M = 0d +©M
S
= 0;
V = 60108 - 48 - V = 0,+
c
©F
y
= 0;
12 ft … x
2
… 20 ft,
M =-1588 + 108x
1
- 2x
1
2
1588 - 108x
1
+ 4x
1
a
x
1
2
b + M = 0d+©M
S
= 0;
V = 108 - 4x
1
108 - 4x
1
- V = 0,+
c
©F
y
= 0;
0 … x
1
… 12 ft.
x
2
x
1
x = 12 ft,
M = 1588 k
#
ft,
V = 108 k
Determine the shear and moment in the beam shown in Fig. 4–7a as a
function of x.
4.2 SHEAR AND MOMENT FUNCTIONS 143
4
Fig. 4–8
EXAMPLE 4.6
Determine the shear and moment in the beam shown in Fig. 4–8a as a
function of x.
10 kN/m
30 kN/m
9 m
x
(a)
75 kN
20 kN/m
9 m
(b)
90 kN 90 kN
6 m
4.5 m
105 kN
10 kN/m
x
20(
__
) kN/m
9
10 kN/m
75 kN
10x
x
–
2
x
–
2
x
–
3
M
V
(c)
1
x
__
(20)(
__
) x
2 9
SOLUTION
Support Reactions. To determine the support reactions, the
distributed load is divided into a triangular and rectangular loading,
and these loadings are then replaced by their resultant forces. These
reactions have been computed and are shown on the beam’s free-
body diagram, Fig. 4–8b.
Shear and Moment Functions. A free-body diagram of the cut
section is shown in Fig. 4–8c. As above, the trapezoidal loading is
replaced by rectangular and triangular distributions. Note that the
intensity of the triangular load at the cut is found by proportion. The
resultant force of each distributed loading and its location are
indicated. Applying the equilibrium equations, we have
Ans.
Ans.M = 75x - 5x
2
- 0.370x
3
-75x + 110x2a
x
2
b + c
1
2
1202a
x
9
bx d
x
3
+ M = 0d+©M
S
= 0;
V = 75 - 10x - 1.11x
2
75 - 10x - c
1
2
1202a
x
9
bx d - V = 0+
c
©F
y
= 0;
144 CHAPTER 4INTERNAL LOADINGS DEVELOPED IN S TRUCTURAL MEMBERS
4
FUNDAMENTAL PROBLEMS
F4–1. Determine the internal normal force, shear force,
and bending moment acting at point C in the beam.
F4–4. Determine the internal normal force, shear force,
and bending moment acting at point C in the beam.
F4–2. Determine the internal normal force, shear force,
and bending moment acting at point C in the beam.
F4–5. Determine the internal normal force, shear force,
and bending moment acting at point C in the beam.
F4–3. Determine the internal normal force, shear force,
and bending moment acting at point C in the beam.
F4–6. Determine the internal normal force, shear force,
and bending moment acting at point C in the beam.
A
C
B
2 m 1 m 2 m
1 m
10 kN
20 kN⭈m
F4–1
A
B
8 kN/m
4 kN/m
1.5 m
1.5 m
C
F4–2
A
B
C
6 kN/m
1.5 m
3 m
1.5 m
F4–3
A
B
300 lb
/
ft
3 ft
1.5 ft1.5 ft
C
F4–4
1.5 m
3 m
3 m
45⬚
A
B
C
5 kN/m
F4–5
800 lb
600 lb
A
C
B
150 lb/ft
3 ft 3 ft3 ft 6 ft
F4–6
4.2 SHEAR AND MOMENT FUNCTIONS 145
4
F4–7. Determine the internal shear and moment in the
beam as a function of x.
F4–10. Determine the internal shear and moment in the
beam as a function of x throughout the beam.
F4–8. Determine the internal shear and moment in the
beam as a function of x.
F4–11. Determine the internal shear and moment in the
beam as a function of x throughout the beam.
F4–9. Determine the internal shear and moment in the
beam as a function of x throughout the beam.
F4–12. Determine the internal shear and moment in the
beam as a function of x throughout the beam.
B
3 m
6 kN
18 kN/m
A
x
F4–7
A
B
12 kN/m
6 m
x
F4–8
A
B
8 kN/m
4 m
4 m
x
F4–9
5 kN/m
A
B
2 m 2 m
x
15 kN · m
20 kN · m
F4–10
5 kN/m
A
2 m 2 m
15 kN
x
F4–11
A
B
2 k
/
ft
12 ft
12 ft
x
18 k
F4–12
146 CHAPTER 4INTERNAL LOADINGS DEVELOPED IN S TRUCTURAL MEMBERS
4
4–2. Determine the internal normal force, shear force, and
bending moment in the beam at points C and D. Assume
the support at B is a roller. Point D is located just to the
right of the 10-k load.
*4–4. Determine the internal normal force, shear force,
and bending moment at point D. Take w 150 N兾m.
4–5. The beam AB will fail if the maximum internal
moment at D reaches or the normal force in
member BC becomes 1500 N. Determine the largest load w
it can support.
800 N
#
m
=
4–1. Determine the internal normal force, shear force, and
bending moment in the beam at points C and D. Assume
the support at A is a pin and B is a roller.
4–3. The boom DF of the jib crane and the column DE
have a uniform weight of 50 lb兾ft. If the hoist and load weigh
300 lb, determine the internal normal force, shear force, and
bending moment in the crane at points A, B, and C.
PROBLEMS
6 kN
20 kN · m
B
A
CD
1 m 1 m
2 m 2 m
Prob. 4–1
10 ft 10 ft
10 ft
A
CDB
10 k
25 k · ft 25 k · ft
Prob. 4–2
5 ft
7 ft
C
D
F
E
B
A
300 lb
2 ft 8 ft 3 ft
Prob. 4–3
4 m
A
D
B
C
4 m
4 m
3 m
w
Probs. 4–4/4–5
4.2 SHEAR AND MOMENT FUNCTIONS 147
4
4–6. Determine the internal normal force, shear force, and
bending moment in the beam at points C and D. Assume
the support at A is a roller and B is a pin.
4–9. Determine the internal normal force, shear force, and
bending moment in the beam at point C. The support at A is
a roller and B is pinned.
4–7. Determine the internal normal force, shear force, and
bending moment at point C. Assume the reactions at the
supports A and B are vertical.
4–10. Determine the internal normal force, shear force,
and bending moment at point C. Assume the reactions at
the supports A and B are vertical.
A
B
1.5 m1.5 m
4 kN/m
C
D
1.5 m1.5 m
Prob. 4–6
D
AB
0.5 kN/m
1.5 kN/m
3 m
6 m
Prob. 4–8
3 m
C
AB
0.5 kN/m
1.5 kN/m
6 m
Prob. 4–7
5 kN
A
C
B
3 kN/m
1 m 2 m 2 m
Prob. 4–9
6 ft
A
B
300 lb/ft
4.5 ft
400 lb/ft
4.5 ft
ED
14 ft
Prob. 4–11
8 ft
C
A
B
300 lb/ft
400 lb/ft
12 ft
9 ft
Prob. 4–10
*4–8. Determine the internal normal force, shear force,
and bending moment at point D. Assume the reactions at
the supports A and B are vertical.
4–11. Determine the internal normal force, shear force, and
bending moment at points D and E. Assume the reactions at
the supports A and B are vertical.
148 CHAPTER 4INTERNAL LOADINGS DEVELOPED IN S TRUCTURAL MEMBERS
4
BA
x
a
b
L
P
Prob. 4–12
2 m 1 m1 m
A
x
B
4 kN
6 kN
Prob. 4–13
*4–12. Determine the shear and moment throughout the
beam as a function of x.
4–15. Determine the shear and moment throughout the
beam as a function of x.
4–13. Determine the shear and moment in the floor girder
as a function of x. Assume the support at A is a pin and B is
a roller.
*4–16. Determine the shear and moment throughout the
beam as a function of x.
4–14. Determine the shear and moment throughout the
beam as a function of x.
4–17. Determine the shear and moment throughout the
beam as a function of x.
BA
x
ab
M
0
L
Prob. 4–14
BA
4 m
2 m2 m
x
12 kN⭈m
7 kN
Prob. 4–15
A
B
8 kN/m
3 m3 m
x
Prob. 4–16
1 m 1 m 1 m
8 kN8 kN
4 kN
A
x
Prob. 4–17
4.2 SHEAR AND MOMENT FUNCTIONS 149
4
4–18. Determine the shear and moment throughout the
beam as functions of x.
4–21. Determine the shear and moment in the beam as a
function of x.
4–19. Determine the shear and moment throughout the
beam as functions of x.
4–22. Determine the shear and moment throughout the
tapered beam as a function of x.
*4–20. Determine the shear and moment in the beam as
functions of x.
6 ft 4 ft
2 k/ft
8 k
x
10 k
40 k
⭈ft
Prob. 4–18
A
B
x
4 ft 4 ft
150 lb/ft
6 ft
250 lb
250 lb
Prob. 4–19
x
B
A
w
0
L
__
2
L
__
2
Prob. 4–20
AB
200 lb/ft
10 ft
800 lb
1200 lb⭈ ft
x
Prob. 4–21
8 kN/m
x
9 m
A
B
Prob. 4–22