8.1 Internal Forces in Structural Members
8.1 Internal Forces in Structural Members Example 1, page 1 of 4
1. Determine the normal force, shear force, and
moment at sections passing through a) B and b) D.
4 kN
B
D
C
E
A
2m
2m
4m
4m
8.1 Internal Forces in Structural Members Example 1, page 2 of 4
Part a): Internal forces and moment at B
4 kN
1
Pass a section through point B.
B
E
C
A
Free-body diagram of part of beam to left of B (This is a much
better choice of free body than the part of the beam to the right,
since we won't have to calculate the reactions at C and E.)
2
4
4 kN
VB
B
NB
A
5
MB
Solving Eqs. 1, 2, and 3 gives
NB = 0
Ans.
VB = 4 kN
Ans.
MB = 8 kN·m
Ans.
Free-body diagram showing correct senses of
internal forces and moment at B:
2m
4 kN
3
Equations of equilibrium for part of beam to left of B:
+ Fx

VB = 4 kN
= 0: NB = 0
Fy = 0: VB 4 kN = 0
(2)
MB = 0: (4 kN)(2 m) + MB = 0
(3)
+

+
(1)
B
A
NB = 0
MB = 8 kN·m
2m
8.1 Internal Forces in Structural Members Example 1, page 3 of 4
Part b): Internal forces and moment at D
6
Pass a section through point D.
4 kN
A
7
D
C
E
Free-body diagram of part of beam to right of D
VD
MD
D
E
Ex
ND
Ey
4m
Equations of equilibrium for part of beam to right of D:
8
+

Fx = 0: ND + Ex = 0
Fy = 0: VD Ey = 0
(5)
MD = 0: MD + Ey (4 m) = 0
(6)
+

+
(4)
9
Three equations but five unkowns.
Another free body is needed.
8.1 Internal Forces in Structural Members Example 1, page 4 of 4
10 Free-body diagram of entire beam
4 kN
A
D
C
Ex
E
Cy
Ey
8m
4m
13 Free-body diagram showing correct senses
of internal forces and moment at D
11 Equilibrium equations for entire beam:
+
+ F = 0: E = 0

x
x
MC = 0: (4 kN)(4 m) + Ey (8 m) = 0
(7)
MD = 8 kN·m
D
(8)
E
ND = 0
Solving Eqs. 7 and 8 gives
VD = 2 kN
Ex = 0 and Ey = 2 kN
2 kN
12 Using these results in Eqs. 4, 5, and 6 and solving gives
ND = 0
Ans.
VD = 2 kN
Ans.
MD = 8 kN·m
Ans.
8.1 Internal Forces in Structural Members Example 2, page 1 of 2
2. Determine the normal force, shear force, and moment at a
section passing through B.
3
300 lb/ft
A
C
B
6 ft
1
6 ft
Note that the resultant of the entire distributed
load, (300 lb/ft)(12 ft) = 3,600 lb, does not
act at B. The entire distributed load acts on
the entire 12-ft span; the free body shown
below has 300 lb/ft acting over only a 6-ft
span, and the resultant of this distributed load
acts halfway between B and C.
Pass a section through point B.
4
300 lb/ft
Equations of equilibrium for part of beam to right of B:
+

Fx = 0: NB = 0
A
C
Therefore
B
NB = 0
B
Ans.
Fy = 0: VB + Cy 300 lb/ft)(6 ft) = 0
(2)
MB = 0: MB (300 lb/ft)(6 ft)( 6 ft )
2
+ Cy(6 ft) = 0
(3)
+

Free-body diagram of part of
beam to right of B
300 lb/ft
V
+
2
(1)
C
NB
B
MB
6 ft
Cy
Eqs. 2 and 3 involve three unknowns. An additional free-body
diagram is needed.
8.1 Internal Forces in Structural Members Example 2, page 2 of 2
5
Free-body diagram of entire beam.
300 lb/ft
A
C
Ax
B
Cy
Ay
12 ft
Equation of equilibrium for entire beam:
+
6
MA = 0: Cy(12 ft)  (300 lb/ft)(12 ft)( 12 ft) = 0
2
(4)
Solving Eq. 4 gives Cy = 1800 lb.
Using Cy = 1800 lb in Eqs. 2 and 3 and then solving gives
VB = 0
Ans.
MB = 5,400 lbft
Ans.
8.1 Internal Forces in Structural Members Example 3, page 1 of 2
3. Determine the normal force, shear force, and moment
acting at a section passing through point B on the
quarter-circular rod shown.
70 lb
2
Free-body diagram of portion above B
70 lb
C
C
30 lb
VB
B
MB
B
4 ft
40°
A
NB
O
Pass a section through point B
3
C
30 lb
A
40°
Equilibrium equations for part of rod to right and above
section at B:
+ F = 0: N sin   V cos 40° + 30 lb = 0

x
B
B
(1)
Fy = 0: NB cos  + VB sin 40° + 70 lb = 0
(2)
MO = 0: MB + NB(4 ft)  (30 lb)(4 ft) = 0
(3)
+

70 lb
B

4 ft
O
+
1
30 lb
Choosing point O for summing moments eliminates
both the 70-lb force and the shear VB from the moment
equation because their lines of action pass through O.
8.1 Internal Forces in Structural Members Example 3, page 2 of 2
4
Geometry
C
B
 = 90°  50° = 40°
50°
40°
O
5
Substituting = 40° into Eqs. 1, 2, and 3 and solving
gives
NB = 72.9 lb
Ans.
VB = .0 lb
Ans.
MB = 171.6 lb
Ans.
8.1 Internal Forces in Structural Members Example 4, page 1 of 6
4. Determine the internal torque at sections passing
through points a) B and b) D of the shaft.
2 kip
2 ft
A
2 ft
B
2 kip
6 kip·ft (Torque applied at end)
C
D
8.1 Internal Forces in Structural Members Example 4, page 2 of 6
Part a) point B
1
Pass a section through point B.
2 kip
2 ft
A
2 ft
B
2 kip
6 kip·ft (Torque applied at end)
C
D
8.1 Internal Forces in Structural Members Example 4, page 3 of 6
2
Free-body diagram of part to right of B. This is a better choice
than using the part to the left because the reaction at the support
would have to be calculated, if we used the part to the left.
2 kip
2 ft
B
2 ft
MB
Torque from part
of the shaft to the
left of point B
acting on the part
to the right.
D
3
2 kip
The choice of
sense of the torque
is arbitrary.
C
6 kip·ft (Torque applied at end)
x
Equilibrium equation for part of shaft:
+ Mx = 0: MB  (2 kip)(2 ft)  (2 kip)(2 ft) + 6 kip·ft = 0

Solving gives
MB = 2 kip·ft
Ans.
8.1 Internal Forces in Structural Members Example 4, page 4 of 6
4
Free-body diagram of part to
right of B showing correct sense
of torque at B
2 kip
2 ft
B
2 ft
MB = 2 kip·ft
2 kip
C
D
6 kip·ft (Torque applied at end)
x
8.1 Internal Forces in Structural Members Example 4, page 5 of 6
Part b) point D
5
Pass a section through point D.
2 kip
2 ft
A
2 ft
B
2 kip
C
6 kip·ft
D
8.1 Internal Forces in Structural Members Example 4, page 6 of 6
6
Free-body diagram of part to right of D
D
6 kip·ft
x
MD
7
Equilibrium equation for part of shaft:
+

Mx = 0: MD + 6 kip·ft = 0
Solving gives
MD = 6 kip·ft
Ans.
8.1 Internal Forces in Structural Members Example 5, page 1 of 6
5. Determine the axial force, shear force, and moment at
sections passing through a) C and b) immediately to the left
of roller E, and c) immediately to the right of roller E.
4 kN
Hinge (pin connection)
5 kNm
A
D
C
E
F
B
4m
2m 1m
5m
5m
Part a): point C
1
4 kN
Pass a section through point C.
5 kNm
A
C
Hinge
D
E
F
B
4m
2m 1m
5m
5m
8.1 Internal Forces in Structural Members Example 5, page 2 of 6
2
Free-body diagram of part of beam to right of section. This is a
better choice than the part to the left because only one
unknown reaction (the vertical force at E) occurs on the right
while three (two force components and a moment) occur at A.
VC
C
NC
4 kN
Hinge
D
E
MC
Ey
1m
5m
5m
Equilibrium equations for part of beam:
3
+ Fx = 0: N = 0

C
(1)
Therefore,
NC = 0
Ans.
(2)
MC = 0: MC + Ey(1 m + 5 m)  (4 kN)(1 m + 5 m +5 m) = 0
(3)
+
+

Fy = 0: VC + Ey  4 kN = 0
4
Three equations but four unknowns so an additional equation is needed.
8.1 Internal Forces in Structural Members Example 5, page 3 of 6
5
To calculate the reaction at E, pass a
section immediately to the left of hinge D.
4 kN
Hinge
5 kNm
A
C
D
E
F
B
4m
7
2m 1m
5m
The moment is zero at a hinge
(a "hinge" is just another
name for a pin connection
between two parts of the
beam; no moment is
transmitted at a pin
connection). We will make
use of this fact by summing
moments about D.
5m
6
Free-body diagram of part of beam to right
of section
VD
MD = 0
Hinge
D
E
ND
Ey
5m
5m
4 kN
8.1 Internal Forces in Structural Members Example 5, page 4 of 6
8
+
Equilibrium equation for part of beam:
MD = 0: Ey(5 m)  4(5 m +5 m) = 0
(4)
Note that VD does not appear in this equation because
its moment arm about D is infinitesimal.
Solving Eq. 4 gives
Ey = 8 kN
Substituting Ey = 8 kN in Eqs. 2 and 3 and solving
gives
VC =  4 kN
Ans.
MC = 4 kN·m
Ans.
9
Free-body diagram of part of beam
showing correct senses of forces and
moment at C
Hinge
VC = 4kN
NC = 0
C
4 kN
D
E
MC = 4 kN·m
Ey = 8 kN
1m
5m
5m
8.1 Internal Forces in Structural Members Example 5, page 5 of 6
Part b): point immediately to left of E.
11 Free-body diagram of part of
beam to right of section
VE
10 Pass a section through the beam
immediately to the left of E.
4 kN
Hinge
5 kNm
A
C
D
E
NE
E
ME
Ey = 8 kN
F
B
4m
4 kN
5m
2m 1m
5m
5m
12 Equilibrium equations for part to right of section:
+ Fx = 0: N = 0

E
Therefore,
Fy = 0: VE + 8kN  4 kN = 0
VE =  4 kN
VE = 4 kN
+
E
NE = 0
(6)
Solving gives
4 kN
ME = 20 kN·m
Ans.
+

13 Free-body diagram of part of beam showing correct
senses of forces and moment at section to left of E
NE = 0
(5)
Ans.
ME = 0: ME  (4 kN)(5 m) = 0
(7)
Solving gives
Ey = 8 kN
5m
ME = 20 kN·m
Ans.
8.1 Internal Forces in Structural Members Example 5, page 6 of 6
Part c): point immediately to right of roller E.
15
14 Pass a section through the beam
immediately to the right of E.
Free-body diagram of part of
beam to right of section
4 kN
VE'
4 kN
Hinge
5 kNm
A
C
NE '
E
D
F
ME'
5m
B
4m
2m 1m
5m
5m
16
Equilibrium equations for part to right of section:
+ Fx = 0: N ' = 0

E
Therefore,
Fy = 0: VE'  4 kN = 0
VE' = 4 kN
+
ME' = 20 kN·m
5m
(9)
Solving gives
VE' = 4 kN
NE ' = 0
Ans.
+

17 Free-body diagram of part of beam showing correct
senses of forces and moment at section to right of E
4 kN
NE' = 0
(8)
Ans.
ME = 0: ME'  (4 kN)(5 m) = 0
(10)
Solving gives
ME' = 20 kN·m
Ans.
8.1 Internal Forces in Structural Members Example 6, page 1 of 2
6. Two wooden blocks have been glued together and a
compressive force of 80 N is applied by the clamp to press the
blocks together as the glue dries. Determine the normal force,
shear force, and moment at a vertical section through point A.
A
35 mm
1
Pass a section through point A.
A
35 mm
8.1 Internal Forces in Structural Members Example 6, page 2 of 2
2
Free-body diagram of part of clamp to right of
section. The block is not included in the free body.
VA
A
NA
MA
35 mm
80 N
4
Equilibrium equations for part of clamp:
+ F = 0: N + 80 N = 0

x
A
Fy = 0: VA = 0
(2)
MA = 0: MA + (80 N)(35 mm) = 0
(3)
+

The force from the block points towards
the clamp because the block is in
compression.
+
3
(1)
Solving gives
NA = 80 N
Ans.
VA = 0
Ans.
MA = 2800 N·mm = 2.8 N·m
Ans.
8.1 Internal Forces in Structural Members Example 7, page 1 of 3
7. Determine the axial force, shear force, and moment at C.
100 lb
D
E
C
A
B
100 lb
1
5 ft
Pass a section through point C.
5 ft
D
E
C
A
B
5 ft
5 ft
8.1 Internal Forces in Structural Members Example 7, page 2 of 3
2
Free-body diagram of part of member
above section at C
100 lb
D
E
3
C
FDB
VC
DB is a two-force member so the line of
action of FDB is known to be vertical.
MC
4
5 ft
Equations of equilibrium for part of member:
+ Fx = 0: V = 0

C
Therefore VC = 0
(1)
Ans.
Fy = 0: NC + FDB  100 lb = 0
(2)
MC = 0: MC + FDB(5ft)  (100 lb)(5 ft +5 ft) = 0
(3)
+

5 ft
+
NC
5
Four unknowns but only three equations so an
additional equation is needed.
8.1 Internal Forces in Structural Members Example 7, page 3 of 3
6
Free-body diagram of entire structure
5 ft
5 ft
100 lb
D
E
C
8
+
A
Equations of equilibrium for entire structure:
B
MA = 0: FDB(5ft)  (100 lb)(5 ft +5 ft) = 0
(4)
Ax
Solving gives
Ay
FDB
7
Because DB is a
two-force member, the
line of action of FDB is
known to be vertical.
FDB = 200 lb
Substituting FDB = 200 lb in Eqs. 2 and 3 and solving
gives
NC = 100 lb
Ans.
MC = 0
Ans.
8.1 Internal Forces in Structural Members Example 8, page 1 of 3
8. The spreader bar BCD is used to spread the load acting on the
2,000 lb beam EFGH while it is being lifted. Determine the normal
force, shear force, and moment at the midpoint C of the spreader bar.
Neglect the weight of the bar.
2,000 lb
A
6 ft
B
1
Pass a section through point C.
D
C
2,000 lb
A
F
E
5 ft
G
5 ft
5 ft
H
5 ft
B
E
D
C
F
G
H
8.1 Internal Forces in Structural Members Example 8, page 2 of 3
Free-body diagram
of part of spreader
bar to left of C
TBA
4
3 TBA and TBF are
the tensions in
the cables at B.
B
+ Fx = 0: N + T

BA cos  = 0
C
MC
C
NC
(1)
Fy = 0: TBA sin  TBF VC = 0
(2)
MC = 0: TBF(5 ft) TBA(sin )(5 ft) + MC = 0
(3)
5
Three equations but five unknowns, so at least one more free
body must be used.
+


Equilibrium equations for part of spreader bar to left of C:
+
2
VC
TBF
5 ft
7
Free-body diagram of beam
+
6
TDG
TBF
F
5 ft
G
5 ft
MG = 0: TBF(5 ft+5 ft) + (2,000 lb)(5 ft) = 0
Solving gives
2,000 lb
E
Equilibrium equations for beam:
5 ft
H
5 ft
TBF = 1,000 lb
Another free body is needed, if we are to calculate the
value of TBA in Eqs. 1, 2, and 3.
(4)
8.1 Internal Forces in Structural Members Example 8, page 3 of 3
8
Free-body diagram of
connection A
9
Equilibrium equations for connection A:
+ Fx = 0: T

BA cos  + TDA cos  = 0
2,000 lb
(5)
+

Fy = 0: 2,000 lb TBA sin  TDA sin  = 0
(6)
A


TBA
TDA
11 Substituting = 50.19° in Eqs. 5 and 6 and solving gives
TBA = TDA = 1,302 lb
10 Geometry
Substituting TBA = 1,302 lb and TBF = 1,000 lb into Eqs.
1, 2, and 3 and solving gives
A
6 ft

B
5 ft
C
= tan-1 ( 6 ft ) = 50.19°
5 ft
NC = 833 lb
Ans.
VC = 0
Ans.
MC = 0
Ans.
8.1 Internal Forces in Structural Members Example 9, page 1 of 5
9. The frame shown is pin-connected at D and rests on smooth
surfaces at A and G. Determine the normal force, shear force,
and moment acting at a section passing through point C.
D
E
C
10 m
1
100 kg
B
Pass a section through C.
F
D
G
A
E
1m 2m
3m
2m
3m 1m
10 m
C
100 kg
B
A
F
G
8.1 Internal Forces in Structural Members Example 9, page 2 of 5
Free-body diagram of part CBA
MC
VC
C
LAC

B

T
LAB
A
FA
3
1m2m
Equilibrium equations for part ABC:
+ F = 0: T + N cos  + V cos = 0

x
C
C
(1)
Fy = 0: FA + NC sin VC sin  = 0
(2)
MA = 0: T(LAB) VC( LAC) + MC = 0
(3)
+

NC
4
+
2
Since the floor is smooth,
only a normal force is
present; friction is absent.
Three equations and five unknowns, so at least one
more free body is needed. But first, the angles 
and and distances LAB and LAC must be determined.
8.1 Internal Forces in Structural Members Example 9, page 3 of 5
5
10 m
C
LAC



10 m
 = tan-1 ( 1 m + 2 m + 3 m ) = 59.04°
D
 = 90°  = 30.96°
E
LAB = (1 m) tan  = 1.667 m
C
(1 m + 2 m)
LAC =
= 5.831 m
cos 
B
LAB

A 1m2m
Free-body diagram of entire frame. This free
body will enable us to calculate FA.
6
D
Geometry
B
Weight =
(100 kg)(9.81 m/s2)
= 981 N
Equilibrium equation for entire frame:
+
F
3m
A
7
10 m
MG = 0: FA(8 m + 4 m) + (981 N)(4 m) = 0
(4)
FA
FG
8m
Solving this equation gives
FA = 327.0 N
One more free body is needed, since we now have four
equations but five unknowns.
G
4m
8.1 Internal Forces in Structural Members Example 9, page 4 of 5
Free-body diagram of member
ABCD. This free body will give
us an equation relating FA and T.
Dy
D
9
Forces from the pin
connection at D
Dx
10 m
B
T
A
FA = 327.0 N
LAB = 1.667 m
10 Equilibrium equations for member ABCD:
6m
+
8
MD = 0: (327.0 N)(6 m) + T(10 m 1.667 m) = 0
Solving gives
T = 235.4 N
(5)
8.1 Internal Forces in Structural Members Example 9, page 5 of 5
11 Substituting
12 Free-body diagram showing correct sense of
internal moment and forces at C
T = 235.4 N
NC = 402 N
FA = 327.0 N
 = 59.04°
MC = 196.3 N·m
C
 = 30.96°
VC = 33.6 N
LAB = 1.667 m
T = 235.4 N
A
LAC = 5.831 m
into Eqs. 1, 2, and 3 and solving gives
NC = 402 N
Ans.
VC = 33.6 N
Ans
MC = 196.3 N ·m
Ans
FA = 327 N
8.1 Internal Forces in Structural Members Example 10, page 1 of 8
10. In the floor-beam girder system shown, the four floor panels at the top
are simply supported at their ends by floor beams. The beams in turn
transmit forces to the girder ACI. Determine the axial force, shear force, and
moment in the girder at sections passing through a) point B and b) point J.
Floor panels
D
E
2 kip
F
G
C
B
A
10 ft
3 ft
7 ft
10 ft
Floor beam
(end view)
H
J
I
6 ft
2 ft 2 ft
Girder
(side view)
8.1 Internal Forces in Structural Members Example 10, page 2 of 8
Part a) point B
1
Pass a section through the girder at B.
2 kip
Floor panels
D
A
E
F
B
G
C
Floor beam
(end view)
H
J
I
Girder
(side view)
8.1 Internal Forces in Structural Members Example 10, page 3 of 8
2
Free-body diagram of part of structure to left
of point B, including floor panels DE and EF
D
E
F
3
NB
B
A
FF
MB
VB
FA
13 ft
7 ft
Equilibrium equations for part of structure:
4
+ Fx = 0: NB = 0

Fy = 0: FA VB + FF = 0
(2)
MB = 0: FA(13 ft) + MB + FF(7 ft) = 0
(3)
+

+
5
(1)
Three equations but five unknowns so at least one
more free body is needed.
Force from floor beam F acting
on floor panel EF ("simply
supported" so no moment acts
at F, only a force)
8.1 Internal Forces in Structural Members Example 10, page 4 of 8
6
Free-body diagram of floor panel EF:
E
F
7
FE
8
FF
Obviously, FF = 0 (Just
consider the sum of moments
about E).
Free-body diagram of entire structure (This free body will enable us to calculate F A):
D
E
F
G
B
A
2 kip
H
C
I
Cx
Cy
FA
10 ft
3 ft
7 ft
10 ft
6 ft
2 ft 2 ft
8.1 Internal Forces in Structural Members Example 10, page 5 of 8
9
+
Equilibrium equation for entire structure:
MC = 0: FA(10 ft + 3 ft + 7 ft + 10 ft)  (2 kip)(6 ft + 2ft) = 0
Solving gives
FA = 0.5333 kip
Substituting FA = 0.5333 kip in Eqs. 2 and 3 and solving gives
VB = 0.53 kip
Ans.
MB = 6.93 kip·ft
Ans.
10 Free-body diagram showing correct senses of
internal reactions at B
D
E
F
B
A
VB = 0.53 kip
NB = 0 FF = 0
MB = 6.93 kip·ft
FA
13 ft
7 ft
(4)
8.1 Internal Forces in Structural Members Example 10, page 6 of 8
Part b) Internal reactions at section J
11 Pass a section through the girder at J.
2 kip
D
A
E
F
B
G
C
H
J
I
8.1 Internal Forces in Structural Members Example 10, page 7 of 8
12 Free-body diagram of part of structure to
right of section at J, including floor beam H
FH 13 Force from panel GH
acting on floor beam H
MJ
H
VJ
4 ft
14 Equilibrium equations for part of structure:
+ Fx = 0: NJ = 0

(5)
Therefore
NJ = 0
+

I
+
J
NJ
Ans.
Fy = 0: VJ FH = 0
(6)
MJ = 0: MJ  FH(4 ft) = 0
(7)
15 Three equations but four unknowns so another
free body is needed.
8.1 Internal Forces in Structural Members Example 10, page 8 of 8
Free-body diagram of floor panel GH.
This free body will enable us to
calculate FH.
2 kip
17
+
16
Equilibrium equation for floor panel GH:
MG = 0: (2 kip)(8 ft) + FH(8 ft + 2 ft) = 0
Solving gives
G
H
FH = 1.60 kip
Substituting FH = 1.60 kip in Eqs. 6 and 7 and solving gives
FG
8 ft
FH
VJ = 1.60 kip
Ans.
2 ft
MJ = 6.40 kip·ft
Ans.
FH = 1.6 kip
18 Free-body diagram of part of beam to right
of J showing internal forces and moment
with correct senses.
VJ = 1.60 kip
NJ = 0
J
H
I
MJ = 6.40 kip·ft
4 ft