PROBLEM 12.85
Three steel plates are welded together to form the
beam shown. Knowing that the allowable normal stress
for the steel used is 22 ksi, determine the minimum
flange width b that can be used.
SOLUTION
Reactions:
0: 42 (37.5)(8) (23.5)(32) (9.5)(32) 0
32.2857 kips
E
MAΣ=−+ + − =
= ↑A
0: 42 (4.5)(8) (18.5)(32) (32.5)(32) 0
39.7143 kips
A
MEΣ= − − − =
= ↑E
Shear:
to :AB
32.2857 kips
to :BC
32.2857 8 24.2857 kips−=
to :CD
24.2857 32 7.7143 kips−=−
to :DE
7.7143 32 39.7143 kips− −=−
Areas:
to :AB
(4.5)(32.2857) 145.286 kip ft= ⋅
to :BC
(14)(24.2857) 340 kip ft= ⋅
to :CD
(14)( 7.7143) 108 kip ft−=−⋅
to :DE
(9.5)( 39.7143) 377.286 kip ft−=− ⋅
Bending moments:
0
A
M=
0 145.286 145.286 kip ft
145.286 340 485.29 kip ft
485.29 108 377.29 kip ft
377.29 377.286 0
B
C
D
E
M
M
M
M
=+= ⋅
= += ⋅
= −= ⋅
=−=
Maximum
3all
| | 485.29 kip ft 5.2834 10 kip in. 22 ksiM
s
= ⋅= × ⋅ =
33
min all
3 32
| | 5.2834 10 264.70 in
22
13 1
(19) 2 ( )(1) ( )(1)(10) 428.69 200.17
12 4 12
9.5 1 10.5 in.
M
S
I bb b
c
s
×
= = =
= ++=+
= +=
min
40.828 19.063 264.70
I
Sb
c
==+=
11.74 in.b=
consent of McGraw–Hill Education.
PROBLEM 12.86
For the beam and loading shown, design the cross section of the beam,
knowing that the grade of timber used has an allowable normal stress
of 12 MPa.
SOLUTION
PROBLEM 12.87
Determine the largest permissible value of P for the beam and
loading shown, knowing that the allowable normal stress is
+8 ksi in tension and −18 ksi in compression.
SOLUTION
PROBLEM 12.87 (Continued)
Top, tension:
( 10 )(4.25)
837.25
P−
= −
7.01 kipsP=
Top, comp.:
(20 )(4.25)
18 37.25
P
−=−
7.89 kipsP=
Bottom. tension:
(20 )( 1.75)
837.25
P−
= −
8.51 kipsP=
Bottom. comp.:
( 10 )( 1.75)
18 37.25
P−−
−=−
38.3 kipsP=
Smallest value of P is the allowable value.
7.01 kipsP=
consent of McGraw–Hill Education.
PROBLEM 12.88
Knowing that the allowable normal stress for the steel used is 24 ksi,
select the most economical wide–flange beam to support the loading
shown.
SOLUTION
By symmetry,
AF
RR
=
0: 20 (6)(11) 20 0
yA F
FR RΣ= −− −+ =
50 kips
AF
RR= =
Maximum bending moment occurs at center of beam.
0: (7)(53) (5)(20) (1.5)(3)(11) 0
JJ
MMΣ= − + + + =
221.5 kip.ft 2658 kip in.
J
M= = ⋅
3
max
min all
2658 110.75 in
24
M
S
σ
= = =
Shape
S (in2)
W24 68×
154
W21 62
×
127
W18 76×
146
W16 77×
134
W14 82
×
123
W12 96×
131
Use
W21 62.×
consent of McGraw–Hill Education.
PROBLEM 12.85
Three steel plates are welded together to form the
beam shown. Knowing that the allowable normal stress
for the steel used is 22 ksi, determine the minimum
flange width b that can be used.
SOLUTION
Reactions:
0: 42 (37.5)(8) (23.5)(32) (9.5)(32) 0
32.2857 kips
E
MAΣ=−+ + − =
= ↑A
0: 42 (4.5)(8) (18.5)(32) (32.5)(32) 0
39.7143 kips
A
MEΣ= − − − =
= ↑E
Shear:
to :AB
32.2857 kips
to :BC
32.2857 8 24.2857 kips−=
to :CD
24.2857 32 7.7143 kips−=−
to :DE
7.7143 32 39.7143 kips− −=−
Areas:
to :AB
(4.5)(32.2857) 145.286 kip ft= ⋅
to :BC
(14)(24.2857) 340 kip ft= ⋅
to :CD
(14)( 7.7143) 108 kip ft−=−⋅
to :DE
(9.5)( 39.7143) 377.286 kip ft−=− ⋅
Bending moments:
0
A
M=
0 145.286 145.286 kip ft
145.286 340 485.29 kip ft
485.29 108 377.29 kip ft
377.29 377.286 0
B
C
D
E
M
M
M
M
=+= ⋅
= += ⋅
= −= ⋅
=−=
Maximum
3all
| | 485.29 kip ft 5.2834 10 kip in. 22 ksiM
s
= ⋅= × ⋅ =
33
min all
3 32
| | 5.2834 10 264.70 in
22
13 1
(19) 2 ( )(1) ( )(1)(10) 428.69 200.17
12 4 12
9.5 1 10.5 in.
M
S
I bb b
c
s
×
= = =
= ++=+
= +=
min
40.828 19.063 264.70
I
Sb
c
==+=
11.74 in.b=
consent of McGraw–Hill Education.
PROBLEM 12.86
For the beam and loading shown, design the cross section of the beam,
knowing that the grade of timber used has an allowable normal stress
of 12 MPa.
SOLUTION
PROBLEM 12.87
Determine the largest permissible value of P for the beam and
loading shown, knowing that the allowable normal stress is
+8 ksi in tension and −18 ksi in compression.
SOLUTION
PROBLEM 12.87 (Continued)
Top, tension:
( 10 )(4.25)
837.25
P−
= −
7.01 kipsP=
Top, comp.:
(20 )(4.25)
18 37.25
P
−=−
7.89 kipsP=
Bottom. tension:
(20 )( 1.75)
837.25
P−
= −
8.51 kipsP=
Bottom. comp.:
( 10 )( 1.75)
18 37.25
P−−
−=−
38.3 kipsP=
Smallest value of P is the allowable value.
7.01 kipsP=
consent of McGraw–Hill Education.
PROBLEM 12.88
Knowing that the allowable normal stress for the steel used is 24 ksi,
select the most economical wide–flange beam to support the loading
shown.
SOLUTION
By symmetry,
AF
RR
=
0: 20 (6)(11) 20 0
yA F
FR RΣ= −− −+ =
50 kips
AF
RR= =
Maximum bending moment occurs at center of beam.
0: (7)(53) (5)(20) (1.5)(3)(11) 0
JJ
MMΣ= − + + + =
221.5 kip.ft 2658 kip in.
J
M= = ⋅
3
max
min all
2658 110.75 in
24
M
S
σ
= = =
Shape
S (in2)
W24 68×
154
W21 62
×
127
W18 76×
146
W16 77×
134
W14 82
×
123
W12 96×
131
Use
W21 62.×
consent of McGraw–Hill Education.