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✅ Chapter 1 – Introduction Basic steel design problems, load types, and code-based classifications ✅ Chapter 2 – Tension Members Problems on net area, tensile strength, rupture/yielding checks ✅ Chapter 3 – Compression Members Buckling analysis, slenderness ratio (KL/r), critical load computations ✅ Chapter 4 – Beams Bending stress, shear capacity, and lateral-torsional buckling problems ✅ Chapter 5 – Beam-Columns Combined axial and bending load problems, interaction checks ✅ Chapter 6 – Connections Design problems for bolted and welded connections, strength checks
Typology: Exercises
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PROBLEM 01
Locate the centroid of the shaded area.
Solution
1
= 32 m
2
x
1
= 0 y
1
= 2 m
2
= 4 m
2
x
2
=− 2 m y
2
= 3 m
3
π ( 1 )
2
=0.5 π m
2
x
3
= 2 m y
3
3 π
=3.576 m
T
1
2
3
T
= 32 − 4 −0.5 π
T
=26.429 m
2
T
G
Ay ] 26.429 Y
G
= 32 ( 2 )− 4 ( 3 ) −0.5 π ( 3.576)
G
=1.755 m
T
G
Ax ]
G
= 32 ( 0 ) − 4 (− 2 )−0.5 π ( 2 )
G
=0.18 m
PROBLEM 02
With reference to the plane area, determine the following:
a.) The area of the plane in square millimeters.
b.) The x-coordinate of the centroid.
c.) The y-coordinate of the centroid.
Solution
1
x
1
= 20 ; y
1
2
x
2
= 55 ; y
1
3
x
3
; y
3
4
x
4
; y
3
T
A = 7800 m
2
T
´ x =
ax ]
Y-Axis:
X
3
3
X
=8,797,500 mm
4
Y
Y
c
Y
Y
=117,300 mm
3
[ r
y
y
T
r
x
r
x
=30.76 mm
PROBLEM 01
The single 200 mm x 10 mm steel plate is connected to a 12-mm thick steel plate by four 16 mm diameter
rivets. The rivets used are A502, Grade 2, hot driven rivets. The steel ASTM A36 with F y
= 248 MPa and
F u
= 400MPa. Determine the value of P in all possible modes of failure and the safe value of P that the
connection can resist.
Solution:
Riveted Diameter = 16 mm
Hole = 16 = 1.6 = 17.6 mm
Tension on gross area:
t
y
=0.6 ( 248 )=148.8 MPa
g
= 2000 mm
2
t
g
P = 297600 N =297.6 kN
Tension on net area:
t
u
=0.5 ( 400 )= 200 MPa
Tension on net area Tension on gross area
Net area along section a-a :
e
e
= 1648 mm
2
g
g
= 1700 mm
2
t
e
P = 329000 N =329.6 kN
Bearing on projected area:
p
u
=1.2 ( 400 )= 480 MPa
p
= 640 mm
2
p
p
P = 307200 N =307.2 kN
Shear on rivets
v
= 152 MPa
v
π
v
v
P = 122246 N =122.246 kN
Shear rapture (block shear):
v
v
t
t
v
u
v
= 120 MPa
v
bc
v
= 2172 mm
2
t
u
t
= 200 MPa
t
ab
t
= 824 mm
2
n
n
=1835.2 mm
2
Use A
n
= 1700 mm
2
n
g
= 1700 mm
2
Along second row:
n
n
=1670.4 mm
2
n
n
n
n
n
=1879.2 mm
2
Along third row:
n
n
=1505.6 mm
2
n
n
n
n
n
=1879.2 mm
2
Use A '
n
= 1700 mm
2
(governed by tearing along the first row)
t
u
P = 340 kN
The safe load P =297.6 kN , governed by tension on gross area.
PROBLEM 03
A plate with a width of 400 mm and thickness of 12 mm is to be cconnected to a plate of the same width
and thickness by 34 mm diameter bolts, as shown. The holes are 2 mm larger than the bolt diameter. The
plate is A36 stel with yield strength F y
= 248 MPa. Assume allowable tensile stress on the net area is
0.60F y
. It is required to determine the value of b such that the net width along bolts 1-2-3-4 is equal to the
net width along bolts 1-2-4.
a) Calculate the value of b in millimeters.
b) Calculate the value of the net area for tension in plates in square millimeters.
c) Calculate the value of P so that the allowable tensile stress on the net area will no be
exceeded.
Solution
Any critical path of the top plate shown above, the load is 100% of P , hence no reduction of net area is
required. For the bottom plate, for path 1-2-3, only 75% of P acts on the net area, while for path 1-2-3-4,
100% of P acts on the net area.
Part a:
Net width = Gross width −
holes +
s
2
4 g
Net width (1-2-4) = Net width (1-2-3-4)
2
( 150 − b )
2
2
b
2
2
( 150 − b )
2
b
2
22500 − 300 b + b
2
− 2 b
2
b
2
b =
2
b =19.71 mm
Solution
Since there are two identical plates, let us analyze one plate only using half of the given load,
P = 1000/2 = 500 kN
Direct load:
D
=62.5 kN
Load due to moment:
x
y
( x
2
2
y
x
( x
2
2
T = Pe = 500
= 145 kN ∙ m
2
( x
2
2
2
2
2
2
2
Note: The farther the fastener is from the cg, the greater is load due to moment
Considering bolt 1: ( x = 90 , y = 120 ¿
1 Tx
6
1 Ty
6
1 x
1 Tx
1 y
1 Ty
D
1
1 Tx
2
1 y
2
1
2
2
1
= 212337 N =212.34 kN
PROBLEM 02
The gusset is riveted to a larger plate by four 22-mm rivets in single shear arranged and loaded as shown.
Calculate the stress in the most heavily loaded rivet.
Solution
The gusset shown is riveted to the column flange by 7 20-mm diameter rivet in single shear. Determine
the stress in the most-heavily loaded bolt.
Solution
Direct load:
D
= 20 kN
Moment:
Centroid
n × X
G
x
G
G
=114.29 mm
n ×Y
G
y
G
G
=257.14 mm
e = 200 + 175 − X
G
e =260.71 mm
T = 140 e = 140 (0.26071)
T =36.5 kN ∙ m
( x
2
2
2
2
2
2
2
2
( x
2
2
T 4 x
y 4
( x
2
2
T 4 x
6
T 4 x
=39.818 kN
T 4 y
x 4
( x
2
2
T 4 y
6
T 4 y
=13.272 kN
4
√
( R ¿¿ T 4 x )
2
T 4 y
2
4
2
2
4
=51.89 kN
f
v max
π ( 20 )
2
=165.17 MPa
Effective area = 700
= 5197 m
2
c) Considering the upper weld only:
v
v
v
u
v
= 330 MPa
v
=0.707 tL
v
v
= 2598 mm
2
P = 857414 N =857.4 kN
PROBLEM 02
A double-angle truss member shown consist of two angles 125 mm x 88 mm x 10 mm thick with the 125
mm side welded to a gusset plate. The member is to carry a total tensile force of 848 kN. Using 6-mm
fillet weld with E60 electrode, determine the length of each side fillet weld required for balanced
condition. All steels are A36 with F y
= 250 MPa
Solution
u
= 425 MPa for E60 electrode
Allowable shear on effective area of weld
v
u
=0.3 ( 425 ) =127.5 MPa
Allowable shear on base metal
v
y
=0.4 ( 250 )= 100 MPa
Considering one member only:
= 424 kN
Length of weld on one side , L = L
1
2
Based on weld metal shear:
P =0.707 tL F
v
L = 784 mm
Based on base metal shear:
v
= t × L = 6 L (contact area between the weld and metal)
v
v
L = 706 mm
Use L = 784 mm
For balanced condition:
1
y
1
2
y
2
1
2
1
2
1
2
2
2
2
=250.88 mm
1
=533.12 mm
PROBLEM 03
A plate is lapped over and welded to a gusset plate as shown.
a) Determine the maximum force per millimeter of weld due to moment alone acting at the
centroid of the weld group
b) Determine the maximum force per millimeter of weld due to the given load
Solution
Direct Load:
D
PROBLEM 01
Two A36 L150 x 90 x 10 angles are used with a 10-mm gusset plate to create a top chord of a truss. The
long legs are back-to-back making the short legs unstiffened elements. Determine the axial load capacity
for a length of 2 meters.
SOLUTION
Properties of one L150 x 90 x 10:
Area = 2,315 mm
2
t = 10 mm
I x
= 5,331,434 mm
4
I y
= 1,460,745 mm
4
´ x =20.
´ y =49.
Properties of 2L 150 x 90 x 10
With long legs back-to-back
A` = 2A = 4,630 mm
4
I` x
= 2I x
= 10,662,868 mm
4
r
x
√
I x
I` y
= 2 x [ I y
2
]
I` y
= 2 x [1,460,745 + 2,315(20.36 + 5)
2
I` y
= 5,899,190 mm
4
r
y
√
I y
=35.69= r
min
Check local buckling. The 90-mm legs of this double angle member are unstiffened compression
elements
b = 90, t = 10 mm
b/t = 9
H= 200
√
y
Since ¿ t < H /
√
y
, the combination can fully develop compressive stress.
r
√
2 π
2
r
r
Since
r
r
3 ( KL / r )
c
( KL / r )
3
c
3
3
3
a
(
( KL / r )
2
c
2
)
y
a
(
2
2
)
(
)
a
=122.7 MPa
[P = F a
A] P = 122.7(4,630)
P = 568,101N = 568.101 kN
PROBLEM 02
A double angle compression chord member for a truss consist of 2 L200 x 100 x 12, having a short legs
back-to-back in a 10-gusset plate. The member is braced in the plane of the truss every 2.13 m, but only at
the end of the transverse and the spacing of connectors is close enough that the double angle member
reaches its maximum axial load strength. Use A
