Problem 7.28
troids.
Determine the coordinates of the ceny
20 mm
60 mm
x
30 mm
70 mm
Solution: Let us solve this problem by using symmetry and by
breaking the composite shape into parts.
l1
20 mm
y
A1
20 mm
h1
l1 = 70 mm
h1 = 70 mm
l2 = 70 mm
h2 = 70 mm
A2
h2
60 mm
60 mm
x
l2
h1 = 20 mm
l2 = 30 mm
h2 = 60 mm
A1 = l1 h1 = 1400 mm2
A2 = l2 h2 = 1800 mm2
By symmetry,
y1 = 70 mm
x2 = 0
y2 = 30 mm
For the composite,
x =
0+0
x1 A1 + x2 A2
=
=0
A1 + A2
320 mm2
y =
y1 A1 + y2 A2
A1 + A2
y =
(70)(1400) + (30)(1800)
152000
=
3200
3200
y = 47.5 mm
x =0
30 mm
70 mm
l1 = 70 mm
x1 = 0
x
Problem 7.34
troids.
Determine the coordinates of the cen-
y
18 in
x
6 in
6 in
6 in
Solution: Divide the object into four areas: (1) The rectangle
18 in by 18 in, (2) The triangle of altitude 18 in and base 6 in, and (3)
the semi circle with radius 9 in and (4) The object itself.
The areas and their centroids are determined by inspection:
(1)
(2)
A1 = 182 = 324 in.2 , x1 = 9 in., y1 = 9 in.
A2 = ( 12 )(18)(6) = 54 in.2 , x2 = 9 in., y2 = 6 in.
(3)
A3 = π9
2
21.8 in.
2
= 127.2 in.2 , x3 = 9 in., y3 = 18 +
4(9)
3π
The composite area: A = A1 − A2 + A3 = 397.2 in.2 .
The composite centroid:
x=
A1 x1 −A2 x2 +A3 x3
A
= 9 in.
y=
A1 y1 −A2 y2 +A3 y3
A
= 13.51 in.
y
18 in.
x
6 in. 6 in. 6 in.
=
Problem 7.38 If the cross-sectional area of the beam
shown in Problem 7.37 is 8400 mm2 and the y coordinate
of the centroid of the area is y = 90 mm, what are the
dimensions b and h?
Solution: From the solution to Problem 7.37
A1 = 120 b, A2 = 200 h
and y =
y =
y1 A1 + y2 A2
A1 + A2
(60)(120 b) + 120 +
h
2
(200 h)
120 b + 200 h
where y1 = 60 mm
y = 90 mm
A1 + A2 = 8400 mm2
Also, y2 = 120 + h/2
Solving these equations simultaneously we get
h = 18.2 mm
b = 39.7 mm
y
200 mm
h
120 mm
x
b
200 mm
h
A2
A1
b
120 mm
Problem 7.52 The distributed load is w = 6x +
0.4x2 N/m. Determine the reactions at A and B.
A
B
2m
4m
Solution:
6
F =
The total distributed load is
6
w(x) dx =
(6x + 0.4x3 ) dx
0
F = 6
0
x2
2
+ 0.4
6
x4
4
0
0.4(36)2
= 3.36 +
4
F = 237.6 N
A
Using the area analogy, the point of application of the equivalent concentrated force F is
6
6
xw(x) dx
(6x2 + 0.4x4 ) dx
x = 0 6
= 0
237.6
w(x) dx
x =
0
3
5
6 x3 + 0.4 x5
6
237.6
0
= 4.436 m
Now to determine the support reactions
y
x
F
Ax
x
4m
Ay
By
Fx :
Ax = 0
Fy :
Ay + By − F = 0
MA :
4By − xF = 0
Solving, we get
Ax = 0
Ay = −25.9 N
By = 263.5 N
B
4m
2m
Problem 7.56 Determine the reactions on member
AB at A and B.
600 N/m
C
A
B
1m
1m
Solution: Divide the beams at the pin, B. Find the equivalent
concentrated load on AB and find two equivalent concentrated loads
on BC. Then find the support reactions
600 N/m
Load 1 w1 (x) = 300x N/m
1
1
F1 =
300x dx = 150x2 = 150 N
0
C
A
B
1m
0
1m
Similarly,
F2 = 150 N
2
2
300 dx = 300x = 300 N
F3 =
1
1
w2 (x)
300 N/m
w1 (x)
300 N/m
x
B
w3 (x) B
1m
A
using the area analogy,
F1
2
= 150 N applied at x = m
3
F2 = 150 N applied at x =
F3
5
m
3
3
= 300 N applied at x = m
2
we now need to write the equilibrium equations.
For AB
Fx :
Fy :
MA :
For BC
Fx :
Fy :
MB :
Ax + Bx = 0
Ay + By − 150 = 0
2
(1)By −
(150) + MA = 0
3
−Bx = 0
−By − 300 − 150 + Cy = 0
2
(1)Cy − (0.5)(300) −
(150) = 0
3
Solving, we get, acting on AB
Ax = 0
Ay = 350 N
MA = 300 N-m
By = −200 N
Bx = 0
also Cy = 250 N
300 N/m
x
1m
C
150 N
MA
AX
2m
3
1m
BY
300 N
BX
0.5 m
AY
BX
BY
1m
150 N
1 m
3
CY
Problem 14.2 Determine the internal forces and moment at A, B, and C.
STRATEGY – In this case you don’t need to determine
the reactions at the built-in support. Cut the beam at the
point where you want to determine the internal forces
and moment and draw the free-body diagram of the part
of the beam to the left of your cut. Remember that P ,
V , and M must be in their defined positive directions in
your free-body diagrams.
Solution:
Cut the beam through point A and draw the FBD:
ΣFx = 0 = PA
ANS:
PA = 0
ΣFy = 0 = 2, 000 N − VA
ANS:
VA = 2, 000 N
ΣMA = 0 = −(2, 000 N)(1 m) + MA
ANS:
MA = 2, 000 N−m
Cut the beam through point B and draw the FBD:
ΣFx = 0 = PB
ANS:
PB = 0
ΣFy = 0 = 2, 000 N − VB
ANS:
VB = 2, 000 N
ANS:
MB = 4, 000 N−m
ΣMB = 0 = −(2, 000 N)(2 m) + MB
Cut the beam through point C and draw the FBD:
ΣFx = 0 = PC
ANS:
PC = 0
ΣFy = 0 = 2, 000 N − VC
ANS:
VC = 2, 000 N
ANS:
MC = 6, 000 N−m
ΣMC = 0 = −(2, 000 N)(3 m) + MC
Problem 14.4
ment at A.
Determine the internal forces and mo-
Solution:
Free Body Diagram:
Cut the bar through point A and draw the FBD:
ΣFx = 0 = (1, 000 lb)(cos30◦ ) − PA
ANS:
PA = 866 lb ←
ΣFy = 0 = (1, 000 lb)(sin30◦ ) − VA
ANS:
VA = 500 lb ↓
ΣMA = 0 = (1, 000 lb)(sin30◦ )(6 ft) − MA
ANS:
MA = 3, 000 ft−lb
Problem 14.5
Determine the internal forces and moment (a) at B; (b)
at C.
Solution:
Free Body Diagram:
Draw the FBD of the entire beam and determine the reactions at points
A and D.
ΣMA = 0 = −(80 lb)(4 ft) + Dy (12 ft) → Dy = 26.67 lb ↑
ΣFx = 0 = Ax → Ax = 0
ΣFy = −80 lb+Dy +Ay = −80 lb+26.67 lb+Ay → Ay = 53.33 lb ↑
(a) Cut the beam through point B and draw the FBD:
ΣFy = 0 = 26.67 lb − VB
ANS:
VB = 26.67 lb ↓ PB = 0
ΣMA = 0 :
ANS:
ΣFx = 0 = PB
80(4)+6(VB )−MB = 320+6(26.67)−MB = 0
MB = 480 ft−lb
(b) Cut the beam through point C and draw the FBD:
ΣFy = 0 = 26.67 lb − VC
ANS:
ΣFx = 0 = PC
VC = 26.67 lb ↓ PC = 0
ΣMD = 0 = VC (3 ft) − MC = (26.67 lb)(3 ft) − MC
ANS:
MC = 80 ft−lb
Problem 14.6 Determine the internal forces and moment at B (a) if x = 250 mm;
(b) If x = 750 mm.
Solution:
Draw the FBD for the entire beam and determine the reactions at points
A and C.
ΣMA = 0 = −20 N−m + Cy (1 m) → Cy = 20 N ↑
ΣFy = 0 = −Ay + Cy = −Ay + 20 N → Ay = 20 N ↓
ΣFx = 0 = Ax → Ax = 0
(a) Cut the beam through the point where x = 250 mm and draw the
FBD:
ΣFy = 0 = −Ay + VB = −20 N + VB
ANS:
ΣFx = 0 = Ax
VB = 20 N ↑ Ax = 0
ΣMB = 0 = Ay (0.25 m) − MB = (20 N)(0.25 m) − MB
ANS:
MB = 5 N−m CW
(b) Cut the beam through the point where x = 750 mm and draw the
FBD:
ΣFy = 0 = −Ay + VB = −20 N + VB
ANS:
ΣFx = 0 = Ax
VB = 20 N ↑ Ax = 0
ΣMA = 0 = −20 N−m+VB (0.75 m)−MB = −20 N−m+(20 N)(0.75 m)+MB
ANS:
MB = 5 N−m CCW
Problem 14.8 Determine the internal forces and moment at A for each loading.
Solution:
(a) Draw the FBD and determine the reactions at the left and right ends
of the beam.
ΣMR = 0 = (8, 000 N)(2 m) − Ly (4 m) → Ly = 4, 000 N ↑
ΣFy = 0 = Ly −8, 000 N+Ry = 4, 000 N−8, 000 N+R = 0 → Ry = 4, 000 N ↑
ΣFx = 0 = Lx → Lx = 0
Cut the beam through point A and draw the FBD:
ΣFy = 0 = Ly − VA = 4, 000 N − VA
ANS:
ΣFx = 0 = PA
VA = 4, 000 N ↓ PA = 0
ΣMA = 0 = −Ly (1 m) + MA = −(4, 000 N)(1 m) + MA
ANS:
MA = 4, 000 N−m
(b) Draw the FBD and determine the reactions at the left and right ends
of the beam.
ΣMR = 0 = 8, 000(2) − 4Ly = 0
Ly = 4, 000 N ↑
ΣFy = 0 = Ly −8, 000 N+Ry = 4, 000 N−8, 000 N+R = 0 → Ry = 4, 000 N ↑
ΣFx = 0 = Lx → Lx = 0
Cut the beam through point A and draw the FBD:
ΣMA = 0 = −(4, 000 N)(1 m)+[(2, 000 N/m)(1 m)](0.5 m)+MA
ANS:
MA = 3, 000 N−m
ΣFy = 0 = Ly −(2, 000 N/m)(1 m)−VA = 4, 000 N−2, 000 N−VA
ANS:
VA = 2, 000 N ↓
PA = 0
ΣFx = 0 = PA
Problem 14.24 Consider the beam in Problem 14.6.
Determine the internal forces and moment as functions
of x for 0 < x < 0.5 m.
Solution:
Draw the FBD for the entire beam and determine the reactions at points
A and C.
ΣMA = 0 = −20 N−m + Cy (1 m) → Cy = 20 N ↑
ΣFy = 0 = −Ay + Cy = −Ay + 20 N → Ay = 20 N ↓
ΣFx = 0 = Ax → Ax = 0
(a) Cut the beam through an arbitrary point on the left-hand portion of
the beam and draw the FBD:
ΣFy = 0 = −Ay + V = −20 N + V
ANS:
ΣFx = 0 = Ax
V (x) = 20 N ↑ Ax = 0
ΣM = 0 = −Ay (x m) + M = (−20 N)(x m) + M
ANS:
M (x) = 20x N−m
Problem 14.25 Consider the beam in Problem 14.6.
Determine the internal forces and moment as functions
of x for 0.5 m < x < 1 m.
Solution:
Draw the FBD for the entire beam and determine the reactions at points
A and C.
ΣMA = 0 = −20 N−m + Cy (1 m) → Cy = 20 N ↑
ΣFy = 0 = −Ay + Cy = −Ay + 20 N → Ay = 20 N ↓
ΣFx = 0 = Ax → Ax = 0
(a) Cut the beam through an arbitrary point on the right-hand portion
of the beam and draw the FBD:
Summing vertical and horizontal forces to find V and P :
ΣFy = 0 = −Ay + VB = −20 N + VB
ΣFx = 0 = Ax
VB = 20 N ↑ Ax = 0
Summing moments about the cut to find the bending moment:
ANS:
ΣMB = 0 = −Cy (1 − x) + M = −(20 N)(1 − x) + M
ANS:
MB = 20(1 − x) N−m
Problem 14.29 Model the ladder rung as a simply
supported (pin-connected) beam and assume that the
200-lb load exerted by the person’s shoe is uniformly
distributed. Draw the shear force and bending moment
diagrams for the rung.
Solution:
Draw the FBD of the rung and determine the reactions at points A and
B.
Summing moments about point A:
ΣMA = 0 = −(50 lb/in)(4 in)(10 in) + By (15 in)
By = 133.3 lb ↑
Summing vertical forces on the FBD:
ΣFy = 0 = −(50 lb/in)(4 in)+Ay +By = −200 lb+133.3 lb+Ay
Ay = 66.7 lb
Problem 14.32 The homogeneous beams AB and CD
weigh 600 lband 500 lb, respectively. Draw the shear
force and bending moment diagrams for beam CD. Remember that the beam’s weight is a distributed load.)
Solution:
Draw the FBD of beam CD and find the reactions at points C and D.
ΣMC = 0 = −(500 lb)(2.5 ft) − (800 lb)(5 ft) + By (2 ft)
By = 2625 lb ↑
ΣFy = 0 = −Cy +By −500 lb−800 lb = −Cy +2625 lb−1300 lb
CY = 1325 lb ↓
Problem 14.33 Draw the shear force and bending moment diagrams for beam AB in Problem 14.32, including
the beam’s weight.
Solution:
Draw the FBD of beam CD and find the reactions at points C and D.
ΣMC = 0 = −(500 lb)(2.5 ft) − (800 lb)(5 ft) + By (2 ft)
By = 2625 lb ↑
ΣFy = 0 = −Cy +By −500 lb−800 lb = −Cy +2625 lb−1300 lb
CY = 1325 lb ↓
Now draw the FBD of beam AB and determine the reactions at point
A (remember that the connections at points C and D are 2-force members):
ΣMA = 0 = −(600 lb)(3 ft)−2625 lb)(6 ft)+(1325 lb)(4 ft)+MA
MA = 12, 250 ft−lb
ΣFy = 0 = 1325 lb − 2625 lb − 600 lb + Ay
Ay = 1900 lb ↑