Set 16-Code, Fundamentals,
Part 5- Structural
Bolted ＆Welded Connections
Steel Design
Set 16
Problem 1 (CE May 1994)
The “Slenderness Ratio” of a column is generally defined as the ratio of its:
A. Unsupported length to its maximum radius of gyration
B. Unsupported length to its minimum radius of gyration
C. Length to its minimum width
D. Length to its moment of inertia
Problem 2 (CE May 1994)
The following are types of welding inspection:
A. radiographic and ultrasound inspection
B. Penetrant inspection
C. Magnetic particle inspection
D. All of the above
Problem 3 (CE November 1994)
The following are welding process:
A. SMAW Shielded metal arc welding
B. GMAW gas metal arc welding
C. SAW submerged arc welding
D. All in the list
Problem 4 (CE November 1995, May 1997)
In Shield metal Arc Welding designation, E70XX, the E denotes the electrode.
The next two indicate the strength in ksi, and the two X’s represent the number
indicating:
A. Use of the electrode
C. method of welding
B. Method of welding
D. angle of welding
Problem 5 (CE November 1995)
In computing the net area for a plate with a bolt hole, width of a rivet or blot hole
shall be taken as ______mm greater than the nominal dimension of the hole
normal to the direction of the applied stress.
A. 2.40
B. 1.80
C. 1.60
D. 2.00
Problem 6 (CE May 1997)
The AISC allowable tension stress of members except pin-connected members
are 0.60 F y for gross section area or 0.50
for netF. uOn the section area
across the pin hole of an eye bar, the allowable AISC stress in terms of
is:
A. 0.55
B. 0.45
C. 0.60
Set 16-Code, Fundamentals,
D. 0.65
Part 5- Structural
Bolted ＆Welded Connections
Steel Design
Problem 7 (CE November 1996)
In structural steel design, structures carrying liveload which induce impact, the
assumed live load shall be increased sufficiently to provide for it. In not otherwise
specified, the increase for supports of elevators shall be:
A. 50%
B. 75%
C. 33%
D. 100%
Problem 8 (CE May 1994)
The allowable stress in structural steel in bending of non- compact section is:
A. O.60
C. 0.40
B. 0.70
D. 0.66
Problem 9 (CE May 1994)
The slenderness ratio, KL/r of compression members shall not exceed
A. 200
B. 0.45
C. 300
D. 240
Problem 10(CE May 1994)
In general, tension and compression on extreme fibers of compact or rolled or
built up members, symmetrical about and loaded in the plane of their minor axis
is:
A.
F
b
 0.75 F y
B.
F
b
 0.60 F y
C.
F
b
 0.40 F y
D.
F
b
 0.66 F y
Problem 11(CE May 1994)
To prevent local buckling, AISC specifies the value of this ratio not to be exceed a
certain value, e.g., 76sqrt( )where
is in ksi:
A. d/t
B. L/r
C. b/t
D. sqrt(I/A)
Problem 12(CE November 1994)
In built up section, intermediate stiffeners are required when the value of this ratio
is exceed e.g., ratio greater than 260 for the unstiffened webs:
A. h/t
B. L/r
C. d/t
D. b/t
Problem 13 (CE November 1995)
Unstiffened structural steel elements subjected to axial compression due to
bending shall be considered as fully effective when the ratio of the width to
thickness is not greater than _____ for compression flanges of beams and
Fy
stiffeners on plate girders.
A.
333
F
y
B.
625
F
y
C.
F b  0.60 F y
D.
250
F
F
y
y
Problem 14(CE May 1996)
According to the NSCP, stiffened structural steel elements subject to axial
compression or to uniform compression due to bending as in the case of the
flange of a flexural member, shall be considered as fully effective when the ratio
of width
Set 16-Code, Fundamentals,
Part 5- Structural
Bolted ＆Welded Connections
Steel Design
To thickness is not greater than _____ for flanges of square and rectangular box
sections of uniform thickness:
A.
664 sqrt
F  B.
y
250 sqrt
F 
C.
y
832 sqrt
F 
D.
y
620 sqrt
F 
y
Problem 15(CE November 1996)
Web of beams and welded plate girders shall be proportioned that the compressive
stress at the web toe of the fillets resulting from concentrated loads not supported by
bearing stiffeners, shall not exceed to 0.75 . If R is the concentrated load or reaction in
newton, t is the thickness of the web in mm, N is the length of the bearing (not less than k
for end reactions) in mm and k is the distance from the outer face of the flange to web toe
of fillet in mm. The governing formula for interior load is:
R
 0.75 F y
A. t  N  2k 
R
 0.75 F y
C. 2t  N  k 
R
 0.75 F y
B. 2   N  2k 
R
 0.75 F y
D. t  N  k 
Problem 16 (CE November 1996)
According to AISC, the minimum
A
g
must ≥ T/(0.6
) where F y is the yield
strength of steel and T is the load. Given the ultimate strength
specifications required area An ≥:
A.
T 0.6 F y
B.
T
0.5F
u
C.
T
0.65 F
u
D.
F
u
T
0.55 F
, the same
u
Problem 17(CE November 1995)
In the design of the beam-columns, structural members whose loading in compression
and bending are both significant, the AISC specifies cm which is a modification or
reduction factor which keeps the estimated moments caused by deflection from being
too large. For columns in frames subjected to joint translation and sideways, or flames
that depends upon the bending stiffness of their members for lateral stability, the value of
F
y
F
y
cm is: Where
ends.
M M
1
2
is the ratio of the smaller moment to the larger moment at the
B. 0.6-0.4 M 1 M 2 >0.4
A. 0.90
C. 1.0
D. 0.85
Problem 18
A 10-mm gusset plate 300mm wide is bolted to a 20-mm thick column by four 25 mm bolt
arranged in a straight line along the 300 mm side. For the gusset plate,
F
u = 400 MPa. Allowable shearing stress on blots is 160 MPa. What is
= 250 MPa,
the capacity of the gusset plate?
A. 314 kN
B. 450kN
C. 400kN
D. 600kN
Problem 19 (CE May 1997)
Set 16-Code, Fundamentals,
Part 5- Structural
Bolted ＆Welded Connections
Steel Design
Base on the AISC Specifications, the effect of net area Ae at the root of the thread of a
threaded tension member is Ae  0.7854( D  0.9743 n  where D is the nominal outside
diameter of the threads and n is the number of threads per 25.4. if the load is 25kN and
the number of threads per inch of the threaded tension member is 4,
=250Mpa. The
diameter sufficient to carry the load is:
A. 66mm
B. 60mm
C. 70mm
D. 55mm
Situation: Problem 20-22 (CE May 1998)
A plate with width of 300 mm and Thickness
of 20mm is to be connected to two plates of
The same width with half the thickness by
25 mm diameter rivets, as shown. The
rivet holes have a diameter of 2mm larger
than the rivet diameter. The plate is A36 steel
F
y
with yield strength
= 248 MegaPascals,
allowable tensile stress of 0.60
bearing stress of 1.35
and allowable
. The rivets are A502,
Grade 2 hot-driven rivets with allowable shear
Figure
stress of 150 MegaPascals.
Problem 20 
Which of the following most nearly gives the maximum load, in KiloNewtons that
can be applied to the connection wothout exceeding the allowable tensile stress
in the plates:
A. 750
B. 780
C. 700
D. 730
Problem 21 
Which of the following most nearly gives the maximum load, in kilonewtons, that
can be applied to the connection without exceeding the allowable shear stress in
the rivets:
A. 640
B. 590
C. 550
D. 700
Problem 22 
Which of the following most nearly gives the maximum load, in KilonNewtons,
that can be applied to the connection without exceeding the allowable bearing
stress between the plate and the rivets:
A. 670
B. 650
C. 620
D. 700
Situation: Problem 23 to 25
A steel hanger consist of a plate 125mm x 6 mm held at the top of by 20-mm
rivets in single shear, The rivets are arranged in diamond form so that the first net
section has one hole and the second net section has two holes. Allowable
stresses are 110 MPa for tension in plates, 83MPa for shear on rivets, and
Set 16-Code, Fundamentals,
Part 5- Structural
Bolted ＆Welded Connections
Steel Design
F
F
y
y
165MPa for bearing between the rivets and the plate. Hole diameter is 3mm
larger than the rivet diameter.
Problem 23 
Which of the following most nearly gives the load on the hanger so that the
allowable tensile stress in the plates will not be exceeded.
A. 67. 32kN
C. 58. 17kN
B. 72.14kN
D. 92. 56kN
Problem 24 
Which of the following most nearly gives the load on the hanger so that the
allowable shearing stress in the rivets will not be exceeded.
A. 152kN
C. 98kN
B. 104kN
D. 125kN
Problem 25 
Which of the following most nearly gives the load on the hanger so that the
allowable bearing stress will not be exceeded.
A. 65.5kN
C. 45.2kN
B. 85.7kN
D. 79.2kN
Situation: Problem 26 to 28
The two plates shown in Figure ST-02
joined together by four 25-mm rivets.
Allowable shearing stress for the rivets
is 70MPa, allowable tensile stress on
net area of the plates is 100MPa, and
the allowable bearing stress on contact
area between the rivet and the plate is
140MPa.
Problem 26 
Which of the following most nearly gives the maximum value of P such that the
allowable shearing stress in the rivets will not be exceeded.
A. 150kN
C. 137kN
B. 120kN
D. 152kN
Problem 27 
Which of the following most nearly gives the maximum value of P such that the
allowable tensile stress in the plate will not be exceeded.
A. 104kN
C. 150kN
B. 160kN
D. 120kN
Problem 28 
Which of the following most nearly gives the maximum value of P such that the
Set 16-Code, Fundamentals,
Part 5- Structural
Bolted ＆Welded Connections
Steel Design
Allowable bearing stress will not be exceeded.
A. 140kN
C. 133kN
B. 123kN
D. 150kN
Problem 29(CE November 1997)
A steel plate is 360mm wide and 20mm thick with four bolt holes cut into a plate
as shown in the figure. The general expression to the specification method for
computing the net area is:
A
n

 T B  summation( H )  summation( S 4 g
2

Where T=thickness of the plate, B is the width, H=diameter of the holes, S= pitch
and g=Gage.
Which of the following most nearly gives
the critical net area ( An ) of the section
In accordance with the 1992 National
Structural Code of the Philippines:
A.
A
= 4125 sq.mm.
B.
A
= 4625 sq.mm.
C.
A
= 5625 sq.mm.
D.
A
= 5125 sq.mm.
n
n
n
n
Situation: problem 30 (CE November 1998)
A W16 x 58 is connected to W8 x 31 as shown in the figure. The material is A 36
steel with
= 248MPa. The allowable bearing stress is 1.35
. The rivets are
A502 grade 2 hot driven rivets with allowable shearing stress of 120MPa. The
support is to be designed using the full strength of the A16 x 58 beam based on
gross section. The properties of the section are:
W16 x 58: Total depth, d=403mm; Thickness of web: 10mm
W8 x 31: Flange thickness= 11mm
F
y
F
Set 16-Code, Fundamentals,
Part 5- Structural
Bolted ＆Welded Connections
Steel Design
Problem 30 
Which of the following most nearly gives the minimum diameter of the rivets
without exceeding the allowable shearing stress in the rivets.
A. 20mm
C. 23mm
B. 25mm
D. 28mm
Problem 31 
Which of the following most nearly gives the minimum diameter of the rivets
without exceeding the allowable bearing stress in steel.
A. 30mm
C. 25mm
B. 22mm
D. 36mm
Problem 32 
Which of the following most nearly gives the required diameter of the rivets:
A. 23mm
C. 30mm
B. 28mm
D. 25mm
Situation: Problem 33 to 35 (CE May 2000)
In the connection shown in Fgure ST-23, a load
transmits 200kN load acting at an eccentricity of
200mm. For this problem a=b=250mm. The load
is transmitted to the column by the plates and 822 mm diameter rivets. The plates are adequate
to transmit the load to the rivets. The connection
can be analyzed by replacing the given load with
an equivalent loading composed of a vertical force
alone acting through the centroid of the rivets
and the moment.
Problem 33 
Which of the following most nearly gives the maximum shear stress in the rivets
in MegaPascas for equivalent vertical force alone acting on the centroid.
A. 90
B. 145
C. 120
D. 65
Problem 34 
Which of the following most nearly gives the maximum shear stress in the rivets
in MegaPascals for equivalent moment alone.
A. 700
B.74
C. 720
. 760
Problem 35 
Which of the following most nearly gives the maximum shear stress in the rivets
in MegaPascals.
A. 170
C. 110
B. 160
D. 130
Set 16-Code, Fundamentals,
Bolted ＆Welded Connections
Situation: Problem 36 to 38
The gusset is riveted to a larger plate by
four-22mm rivets arranged and loaded as
shown in Figure ST-9.
Problem 36 
Which of the following most nearly gives
The direct load on each rivet in KiloNewton.
A. 8
C. 12
B. 15
D. 10
Part 5- Structural
Steel Design
Problem 37 
Which of the following most nearly gives the force in KiloNewton on the heavily
loaded rivet.
A. 15
C. 12
B. 18
D. 16
Problem 38 
Which of the following most nearly gives the force in KiloNewton on the lightly
loaded rivet.
A. 13
B. 9
C. 11
D. 15
Problem 39
Which of the following most nearly gives the throat of a 9-mm fillet weld:
A. 4.5mm
C. 6.4mm
B. 7.4mm
D. 12.7mm
Problem 40
What is the capacity of a 9-mm fillet weld of E70 electrode in a length of 250mm?
A. 260kN
B. 250kN
C. 180kN
D. 230kN
Problem 41
A double-angle truss member consist of two angles 125mm x 88mm x 10mm
thick with the 125mm side welded to a gusset plate. The member is to carry a
total tensile force of 848 kN. Using 6-mm fillet weld with E70 electrode, determine
the total length of weld required.
Use
=250 MPa.
See accompanying
figure.
A. 856mm
B. 345mm
C. 691mm
D. 1382mm
Set 16-Code, Fundamentals,
Part 5- Structural
Bolted ＆Welded Connections
Steel Design
Problem 42(CE May 1994)
The allowable load in shear for an 8.5mm weld having an allowable shearing
strength of 124MPa for a total length of 306mm is about:
A. 228000N
C. 1054N
B. 745N
D. none of the above
Situation: Problem 43 to 45
In the connection shown in Figure ST-01, a load
transmits 40kN load acting at an eccentricity of
200mm. The load is transmitted to the column by
the plates and 8mm weld. The plates are adequate
to transmit the load to the weld. The connection can
be analyzed by replacing the given load with an
equivalent loading composed of a vertical force
alone acting through the centroid of the rivets and
a moment.
Problem 43 
Which of the following most nearly gives the maximum shear stress in the welds
in MegaPascals for equivalent vertical force alone acting on the centroid.
A. 15
Problem 44 
B. 14
C. 18
D. 21
Which of the following most nearly gives the maximum shear in the welds in
MegaPascals for the equivalent moment alone.
A. 38
B. 46
C. 26
D. 22
Problem 45 
Which of the following most nearly gives the maximum shear stress in the weld in
megaPascals.
A. 63
B. 39
C. 42
D. 52
Answer Key
Set 16-Code, Fundamentals,
Part 5- Structural
Bolted ＆Welded Connections
Steel Design
Solution Fundamentals, Bolted & Welded Connecti ns
o
Solutions to Set 16
5
According to the section 4.14.4 of NSCP, in computing the net area for a
plate with a bolt hole, width of a rivet or bolt hole shall be taken as 1.6mm
greater than the nominal dimension of the hole normal to the direction of the
applied stress.
6
According to the Section 4.14.5 and 4.5.1.1 of NSCP, the allowable stress for
pin-connected members is 0.45
.
7
The increase in values for different supports as provided in Section 4.3.3 of
NSCP are the following:
For support of elevators…………………………..…100%
For cab operated traveling crane support
Girders and their connections ……………………25%
For pendant operated traveling crane support
girders and their connections……………….…….10%
For support of light machinery, shaft
motor driven, not less than ……………………….20%
For support of reciprocating machinery or
power driven units, not less than…………………50%
For hangers supporting floors and balconies………..33%
8
According to Section 4.5.1.4.4 of NSCP, the allowable bending stress on
non-compact sections is 0.6
9
According to Section 4.8.4 of NSCP, the slenderness ration for companion
shall not exceed 200.
10
F
y
According to section 4.5.1.4 of NSCP, Tension and compression on extreme
fibers of compact hot-rolled or built-up members (except hybrid girders and
member of A514 steel) symmetrical about, and loaded in, the plane of their
minor axis, the allowable stress is 0.66
12
h/t (See section 4.10.5.3. of NSCP)
Set 16-Code, Fundamentals,
Part 5- Structural
Bolted ＆Welded Connections
Steel Design
13
250
F
14
620 sqrt
y
See Section 4.9.1.2
F  , See Section 4.9.2.2 of NSCP
y
15
According to Section 4.10.10 of NSCP, the governing formula is:
R
 0.75 F y
t  N  2k 
16
According to Section 4.5.1.1 of NSCP for structural steel in tension, except
for pin-connected members,
A
g
nor 0.5
F
u
F
t
shall not exceed to
on the effective area
Thus, T≤(0.5 F u ) or
A
n
on the gross area
A
n
≥T/(0.50 F u )
17
According to Section 4.6.1 of the NSCP(1.) for compression members subject
 0.85
to joint translation(sideway) C m
18
Tension on gross area:
F
F
 0.6 F y  150 MPa
t
P
F xA
t
g
 15030010   450,000 N
Tension on net area:
F  0.5 F  200MPa
A  300  25 x410  2000 mm
85% A  2550 mm  2000 mm
t
u
2
n
2
2
g
P  200 x 2000  400,000 N
Bearing on plate:
F
p
 1.5 F u  600MPa
P  600(25 x10) x 4  600,000 N
Shearing on bolts:
E
v
 160Mpa
P  160 4 

25
2 
x 4  314,159 N

Thus, the capacity of the plate is 314, 159kN
Set 16-Code, Fundamentals,
Bolted ＆Welded Connections
19
The tensile capacity T of the thread is given by:
T  F t Ae
 0.6 F y
F
t
F
y
(NSCP Section 4.5.1.1)
=250MPa
t
 0.6(250)  150MPa
e
 0.7854( D 
F
A
0.7854 n
2
in square inch
Part 5- Structural
Steel Design
N=4 (number of threads)
T-250kN
250,000=150
A
e
1in 25.4mm  2.58in
2.58  0.7854 D 0.9743 4
2
 1666.67 mm x
e
2
A
2
2
D=2.06 in x (25.4mm/in)=52.3mm say 55mm
20 to 22
Allowable tensile stress in plates:
F
F
t
 0.6 F y  0.60( 248)
t
 148.8MPa
P
A
n
F xA
t
n
2
=4920 mm
P=148.8(4920)=732,096N
P=732kN
P  F v Av
F
v
 150MPa
A = total area sheared off(double shear)
A  2 A x4  2 425 x4  3927 mm
v
2
v
b
P  150(3927)  589,050 N  589kN
2
P
F A
F  1.35 F
b
b
b
y
 1.35(248)  334.8MPa
=bearing area of four rivets=25(20)x4=200
A
b
P=334.8(2000)=669,600N=669.6kN
Set 16-Code, Fundamentals,
Part 5- Structural
Bolted ＆Welded Connections
Steel Design
23 to 25
Based of tensile stress in plates:
Tearing along rivet A
A
 (125  23)6  612 mm
P
F A
n
t
2
n
P  110(612)  67,320 N
Tearing along rivets BC:
A
P
n
 (125  23)6  474 mm
net

2
F A
t
n
P  P / 4  110( 474)  69,520 N
Tearing along rivet D:
A
n
 (125  23)6  612 mm
2
P  3P / 4  110(612); P  269,280 N
Thus, P=67,320 N=67.32kN
2
mm
Based on shearing of rivets:
P
F A
v
v
 83 x

4
20 (4)  104,300.9 N  104.3kN
2
Based on bearing:
P
F A
p
26 to 28
p
 165 x(20)(6) x 4  79000 N  79.2kN
Based on shearing on rivets:
P
F A
v
s
 
2


P  70  25 x 4
4

P  137445 N  137.445kN
Based on tearing on
Net area of plates:
For the 14-mm thick plate:
85% A
g
 1547 mm
2

Tearing along rivet “a”:
A
 (130  27)(14)  1442 mm
P
FA
net
t
net
2
 100(1442)  144,200 N  144.2kN
Tearing along rivets “b” and “c”:
net
 (130  27 x 2)(14)
net
 1064 mm
A
A
P  P/4 
2
F A net
t
3P / 4  1000(1064)
P  141867 N  141.867kN
Set 16-Code, Fundamentals,
Part 5- Structural
Bolted ＆Welded Connections
Steel Design

2
For the 10-mm thick plat: 85%
Ag  1275 mm
Tearing along rivet “d”
A
 (150  27)(10)  1230 mm
P
FA
net
t
net
2
 100(1230)  123,000 N  123kN
Tearing along the rivets “b” and “c”

net
 (150  27 x 2)(10)
net
 960 mm
A
A
P P/4 
2
FA
t
net
3P / 4  100(960)
P  128000 N  128kN
Thus the safe P=123kN
Based on bearing:
(The thinner plate is more critical in bearing)
P
F A
p
b
 1402510x 4  140000 N  140kN
29
A  t b   H    S
A  tw
2
4g
n
n

n
Solving for the critical net width,
W
n
:
Path ABEF:
S=90-60=30mm
G=90(3)=270mm
2
30
W n  360  25(2) 
W
n
4(270)
 310.83mm
Path ABCEF:
2
W
n
 360  25(3) 
45
4(90)
2

15
 290.94mm
4(180)
Path ABCDEF:
2
W
Use
n
 360  25(4) 
W
n
 280.25mm
45
4(90)
2
x2 
60
4(90)
 281.25mm
then; An  txW n  20 x 281.25  5625mm
2
Set 16-Code, Fundamentals,
Bolted ＆Welded Connections
Part 5- Structural
Steel Design
30 to 32
The full strength (shear) of W16x58 is:
F
v

V
dt
 0.4 F y
w
V
 0.4(248); V  399776 N
403(10)
Shear stress in the rivets:
(Considering the 4 rivets in double shear connected to W16 x 58)
F

b
V 4
Dt
 1.35 F y
w
399776 / 4
 1.35(248); D  29.85mm
D(10)
From the previous results, the diameter of the rivets must be 30mm.
33 to 35
Analyze one side of the bracket:
Direct load on one rivet:
R
R
D
 P/4
D
 25kN
Stress due to direct load:
R
R
D
 P/4
D
 25kN
T=P e=100(200)
T=20,000kN-mm
 X  Y   (125  125 ) x4  125,000 mm
20,000(125)
T

 20kN
R 
125,000



 x y 
2
2
2
2
2
y
x
R
2
2

y
20,000(125)
T

 20kN
125,000
  
 x y 
x
2
2
20  20  28.284kN
2
R
2
Set 16-Code, Fundamentals,
Bolted ＆Welded Connections
R 28.284(1000)

2

A
22
4
Stress  74.4 MPa
Stress 
 
The most stressed rivets are rivets B and C:
R
45  20
S
max

S
max
2
2
 49.24kN
49.24kN
22
4

2
 129.5MPa
36 to 38
Direct load on each rivets;
Part 5- Structural
Steel Design
R
R
D
 40 / 4
D
 10kN
R
x

R
y

T
y
2
  x 
y
T
  x  y
2


2


x
2
T=40(100)=4000kN-mm
  x  y   40 120  40 120   32,000
R R R R 0
2
2
1x
2
2x
3x
2
2
2
4x
The most heavily loaded rivets are rivets 1 and 4.
R
4000(120) / 32,000  15kN 
R

1y
1
2
R D  R2 y
2

R
4y
10  15
2
2
 18.03kN 
R
4
The most lightly loaded rivets are rivet 2 and 3.
R
R
2y
2
 4000(40) / 32,000  5kN 

RD
2
2
 R2 y 
R
10  5
2
2
3y
 11.18kN 
R
3
39
Throat=t sin 45°=0.707t=0.707(9)=6.36mm
40
P
F
V
0.707tL; F V  0.3 F u ; F u  70ksi  482.5MPa
P  0.3(482.5)(0.707)(9)( 250)
P  230,261N  230.26kN
Set 16-Code, Fundamentals,
Bolted ＆Welded Connections
41
F
u
 70ksi  70,000 psix
0.101325MPa
 482.5MPa
14.7 psi
Part 5- Structural
Steel Design
Length of the weld on one side, L  L1  L2
Based on weld metal shear:
P
AF
F  0.45 F (NSCP Table 4.5.3)
F  0.4(250)  100MPa (contact area between the weld and metal)
A  2tx1  12L
V
V
V
Y
V
V
424,000=12L(100); L=353mm
Use L =691 mm on one side only.
Therefore; the total length of weld required is 2L=1382 mm
42
P  0.707tL F V  0.707(8.5)(306)(124)  228024.5 N
43
Stress due to the vertical force alone:
f
v

40,000
 17.68MPa
0.707(8)( 2 x 200)
Maximum stress to moment: (at B)
T  Pe  40,000(200)  8,000,000 N  mm
2
2
J   L L 12  x 


y
2


J  2 x 200 200 12  0  100
2
2
J  5,333,333 mm
F
F
BX
BX
T
2

3
6
8 x10 (100)

J
5333333
 150 N mm
Y
6
F
BX
Tx 8 x10 (100)


J
5333333
Set 16-Code, Fundamentals,
Part 5- Structural
Bolted ＆Welded Connections
F
BY
 150 N / mm
F
f
V
Steel Design
F BX

2
2
 F BY  212.13N / mm
212.13(1)
 37.51MPa
0.707(8)(10
Stress per mm length at B,
Maximum stress in the weld (at B)
R
D

40,000
 100 N / mm
2(200)
F BY   F BX  
10150  150
R   R D 

R
2
2
2
2
R  291.55 N / mm
Stress per 1 mm length at B= 291.55(1)
 51.55MPa
0.707(8)(1)
f
v
Set 17- Columns, Beams,
Part 5-Structural
Combined Stress, Plastic Design
Steel Design
Problems-Columns, Beams, Combined Stress, Plastic Design
Set 17
Problem 1 (CE November 1994)
The value of the Euler buckling load of a column fixed at the top and bottom and
of length L with modulus of elasticity E and moment of inertia I subjected to a
central axial compressive load:
A.
4 EI / L
B.
0.25 EI  L
2
2
0.5 EI  L
 EI  L
2
2
C.
2
2
D.
2
2
Problem 2 (CE May 1999)
A wide flange section for a 5 m long column (hinged at both ends) has the
2
following properties:
mm
Cross sectional area= 8000
Radius of gyration,
Radius of gyration,
r
r
x
 100mm
y
 50mm
Modulus of elasticity, E= 200,000Mpa
Which of the following most nearly gives the Euler critical load of the column.
A. 1230kN
B. 1970kN
C. 2140kN
D. 1580kN
Problem 3
According to Section 4.5.1.3.1 of NSCP, on gross section of axially loaded compression
members whose cross-sections meet the provisions of the code when KL/r, the largest
effective slenderness ratio of any unbraced segment, is less than C c :



KL
/
r

F
F1 
2 E
2
Cc 
F
a


2
y
2C c
2
where
5 3KL / r 
FS  

3
8C c
y
 FS

KL / r 
3
8C c
3
On gross section of axially loaded compression member, when KL/r exceeds Cc:
F
a
12

23

2
E
KL / r 
2
A structural steel column with Fy=250MPa having an unbraced length of 3 m is to
carry a total axial load of 1800kN. E of steel = 200GPa. Which of the following
sections is most economical(lightest) for the given loads. K=1
A. A  13800 mm ; Ix  293.63x10
mm ; Iy  67.59 x10 mm
B. A  115550 mm ; Ix  177.04 x10 mm ; Iy  39.14 x10 mm
C. A  16050 mm ; Ix  452.60 x10 mm ; Iy  107.29 x10 mm
D. A  15675 mm ; Ix  391.73x10 mm ; Iy  107.28 x10 mm
2
6
2
4
6
6
4
4
6
4
2
6
4
6
4
2
6
4
6
4
Set 17- Columns, Beams,
Part 5-Structural
Combined Stress, Plastic Design
Steel Design
Problem 4 (CE November 1994)
A steel column has the following properties: Modulus of Elasticity E=200,000MPa,
6
4
yield strength Fy=200Mpa,
length
L=12m,
moment
of
Inertia
I=37.7
x
10
mm
2
8000 mm
And area=
. The allowable stress is closest to:
A. 120Mpa
B. 67.4MPa
C. 33.7MPa
D. 91.1MPa
Situation: Problem 5 to 7 (CE November 1998)
According to Section 4.5.1.3.1 of NSCP, on gross section of axially loaded compression
members whose cross- sections meet with the provisions of the code, when KL/r, the
largest effective slenderness ratio of any unbraced segment, is less than Cc:
2

KL r  F y
3

F a  1 
2
 FS
5 3KL r 
2C c 
FS  


3


KL / r 
3
Where
2 E
2
C
c

Fy
8C
c
8C c
On gross section of axially loaded compression member, when KL/r exceeds Cc:
12 E
2
Fa 
23
KL ? r 
2
A column 3 m long and pinned at both ends is carries an axial load of 190kN. The column
is made up of 2 angles of unequal legs with long legs back to back and separated by a
guest plate whose thickness is 11mm. Use A36 steel with Fy=248Mpa and
E=200,000MPa. Three sections are being considered, as follows, with their respective
properties relevant of this problem (length units are in meters):
Section
A
Rx
Ry
2L 125 x 75 x 12
0.00454
0.0390
0.0160
2L 150 x 90 x 10
0.00463
0.0480
0.0195
2L 150 x 90 x 12
0.00550
0.0500
0.0251
Problem 5 
Which of the following sections gives the largest allowable compressive stress.
A. 2L 125 x 75 x 12
C. 2L 150 x 90 x 12
B. 2L 150 x 90 x 10
D. Not enough data
Problem 6 
Which of the following sections gives the smallest allowable compressive stress.
A. 2L 125 x 75 x 12
C. 2L 150 x 90 x 12
B. 2L 150 x 90 x 10
D. Not enough data
Problem 7 
Which of the following sections gives the most economical(lightest) section for the given
load.
A. 2L 125 x 75 x 12
C. 2L 150 x 90 x 12
B. 2L 150 x 90 x 10
D. Not enough data
Set 17- Columns, Beams,
Part 5-Structural
Combined Stress, Plastic Design
Steel Design
Problem 8 (CE November 1994)
A 10m steel wide flange beam is laterally supported. The width of the web is the beam is
700mm and web thickness is 10mm. If Fy is equal to 250MPa, the allowable bending
stress is:
A. 180MPa
B. 150MPa
C. 165MPa
D. 200MPa
Problem 9 (CE May 1995)
kN
mm
2
The structural I-beam supporting a floor carries a floor load of 4.6
. The beam
span 6.0m and are simply supported at their ends. Determine the centerline spacing if
the allowable
stress
in the beam is 120MPa and the section modulus
3
3
is 534 x10 mm .
A. 3.75m
B. 3.45m
C.4.0m
D.3.0m
Problem 10(CE November 1995)
A BW 75 x 287 steel I beam has the following dimensions:
Total beam depth, H=775mm
Web thickness, tw=19mm
Flange width, B=360mm
Flange thickness, tf=32mm
The moment of inertia Ix about the principal axis as:
4
A. 375,086
cm
375,086 mm
B. 375,086
mm
6
4
C.
D. None in the list
Problem 11 (CE May 1996)
A structural steel angle 203x152x25.4 carries a moment of 10 kN-m. The longer leg is
parallel to the y-axis and the shorter to the x-axis. The inertial properties of the angle are
as follows:
6
4
I
x
 33.6 x10
mm
Moment of inertia about the x-axis,
Moment of inertial about the y-axis,
Product of inertia,
I
xy
I
y
 13 .4 x10
6
 16 .1x10
mm
6
4
mm
4
The stress at point O in MPa which is at x-distance of -41.9mm and y-distance
+135.7mm from the centroid is:
A. 44.92 compression
C. 44.92tension
B. 17.31 tension
D. 17.31 compression
Problem 12 (CE November 1997)
A floor is made up of a 150mm thick concrete slab and 50mm thick floor finish both of
which has a unit weight of 24 kN/cu.m. The floor also carries a ceiling whose weight is
720Pa and a live load of 2,400Pa. The floor is supported by simply supported beams
with a span of 10m spaced at 4.6m on centers, with yield strength (Fy)=248MPa. The
allowable shear stress specified in the 1992 National Structural Code of the Philippines is
0.40Fy. Considering the weight of the beam, which of the following sections is the most
economic (lightest) section for the given load, assuming that shear stress governs the
design.
A. W12 x 19; beam weight= 278N/m, depth=309mm, web thickness=6mm
B. W10 x 33; beam weight= 484N/m, depth=247m, web thickness=7mm
C. BW300 x 41; beam weight =407N/m, depth=300mm, web thickness=6mm
D. BW300 x 47; beam weight=460N/m, depth=300mm, web thickness=6mm
Problem 13 (CE November 1997)
A simply supported steel beam has a span of 6m and is subjected to a vertical
concentrated load of 150kN acting at midspan. An allowance of 100% for impact is made
for the concentrated load only. The beam is laterally supported only at the supports. The
1992 national Structural Code of the Philippines states that the allowance tensile stress
for laterally unsupported members subject to bending is given by the equation:
F
 2 / 3  F y
B

C
b
L
R t  / 10,550,000 C  F
2
b
y
Where
may be taken as 1.0 since the maximum moments does not occur at
the supports. Rt Is the radius of gyration comprising the compression flange and
1/3 of the compression web area, taken about the axis in the plane of the web.
Steel is A36 with yield strength (Fy)=248MPa. Which of the following most nearly
gives the most economic (lightest) section that will limit the flexural stress to the
3
allowable value:
 75mm,

R
t
S
x
0.00239m
A. W18x76; beam weight =1108N/m,
B. W21x62; beam weight=911N/m,
R
C. W24x68; beam weight =1000N/m,
D. W27x84; beam weight= 1234N/m,
t
 53mm, S x  0.0023m
3
R
t
 63mm, S x  0.0035m
R
t
 57mm, S x  0.00253m
3
3
Situation: Problem 14 to 16 (CE May 1998,May 2000)
A simply supported beam with span of 8 m is subjected to a uniform vertical
downward load equal to 50 kN/m acting on the plane of the minor axis of the
beam section, which includes the beam weight. The beam is restrained against
lateral buckling of the top and bottom flanges for the entire span. The material is
A36 still with Fy=248megaPascals and modulus of elasticity=20GigaPascals.
The allowable flexural stress for laterally braced compact sections is 0.66 Fy. The
allowable deflection 1/360 of the span. Three compact sections are being
considered, as follows, with their respective properties relevant of this problem
(length units are in meters):
Section
Moment of Inertia (Ix)
Beam depth (D)
W 24x55
0.000558
0.598
W 21x62
0.000554
0.533
W 21x68
0.000616
0.537
(Note: The symbol x^y means raising the quantity or expression x to the exponent
y.)
Problem 14
Which of the following most nearly gives the minimum section modulus (Sx), in
meter^3, such that the maximum flexural stress will not be exceeded:
A. 0.0018
B. 0.0015
C. 0.0024
D. 0.0021
Set 17- Columns, Beams,
Part 5-Structural
Combined Stress, Plastic Design
Steel Design
Problem 15 
Which of the following most nearly gives the minimum moment of Inertia (Ix), in
meters^4, such that maximum deflection will not be exceeded:
A. 0.00065
B. 0.00050
C. 0.00060
D. 0.00055
Problem 16 
Assuming that shear stress is not critical, which of the sections being considered
is the most economic section that is adequate for the given load:
A. W 21x62
C. W 21x68
B. W 24x55
D. None of the sections are adequate
Situation: Problem 17 to 19
A W 14x500 is used as a beam to support a concrete floor system. The floor is to
carry a total load of 120KPa. The beam is simply supported over a span of 7m.
the properties of the section are as follows:
Depth, d=498mm
Thickness of web=56mm
Moment of Inertia,
Section modulus,
I
S
x
 3417 x10
6
 13730 x10
3
x
4
mm
3
mm
Weight= 7.32 kN/m
Assume the beam is laterally supported over its length and that allowable stress in
bending is 0.66Fy and in shear on gross section is 0.4Fy. Use A36 steel with
Fy=250MPa. Allowable deflection is L/360.
Problem 17
Which of the following most nearly gives the center-to-center spacing of the beams
without exceeding the allowable shear stress.
A. 7.21m
B. 6.53m
C. 7.21m
D.8.51m
Problem 18
Which of the following most nearly gives the center-to-center spacing of the beams
without exceeding the allowable bending stress.
A. 2m
B. 2.5m
C. 3m
D. 3.5m
Problem 19 
Which of the following most nearly gives the center-to-center spacing of the beams
without exceeding the allowable deflection.
A. 3.48m
B. 4.21m
C. 2.87m
D. 6.23m
Situation: problem 20 to 23
A w 27x178 is used as a beam that is simply supported at its ends. The properties of the
section is as follows:
3
4
I  2,909.458x10 mm
S  8,237.4 x10 mm
r  293.64mm
x
Weight, W=265.4kg/m
3
3
x
Depth, d=706.4mm
Area, A  33,742 mm
2
x
Set 17- Columns, Beams,
Part 5-Structural
Combined Stress, Plastic Design
Steel Design
Web, thickness
t  18.4mm
b  357.8mm
t  30.2mm
r  94.49mm
w
f
Flange width,
Flange thickness,
f
t
3
I  231,008x10 mm
S  1,291.3x10 mm
r  82.74mm
F  250MPa
4
y
3
3
y
y
y
Problem 20 
Which of the following most nearly gives the allowable bending stress of the beam over
an unsupported length of 4m.
A. 185MPa
B. 125MPa
C. 150MPa
D. 125MPa
Problem 21 
Which of the following most nearly gives the allowable bending stress of the beam over
an unsupported length of 7m.
A. 150MPa
B. 165MPa
C. 185MPa
D. 125MPa
Problem 22 
Which of the following most nearly gives the allowable bending stress of the beam over
an unsupported length of 9m.
A. 160MPa
B. 125MPa
C. 140MPa
D. 155MPa
Problem 23
Which of the following most nearly gives the allowable bending stress of the beam over
an unsupported length of 12m.
A. 90MPa
B. 105MPa
D. 115MPa
D. 130MPa
Problem 24 (CE November 1997)
A simply supported beam has a span of 4.5m. The beam is subjected to an axial tensile
force of 250kN and a vertical concentrated load (P) acting at midspan. The member is
fully laterally supported for its entire length. The effect of any bolt holes can be neglected.
The beam consists of 2-150mm x 90mm x 12mm angles, with long legs back-to-back
and spaced 10mm apart. For this problem, the relevant properties of one 150mm x
90mm x 12mm angle are:
3
Area( A)  2751mm
2
Elastic section modulus
S   0.000063m
x
Steel is A36, yield strength (Fy)=248MPa. The allowable tensile stress Fa=0.60Fy and
the allowable flexural stress Fb=o.66Fy. Neglecting member weight; which of the
following nearly gives the maximum value of the vertical load (P) that member can
support:
A. P=72kN
B. P48MPa
C. P=13kN
D. P=96kN
Situation: problem 25 to 27 (CE November 1999)
According to Section 4.6.1 of NSCP, members subjected to both axial
compression and bending shall be proportioned to satisfy the following
requirements:
Set 17- Columns, Beams,
Part 5-Structural
Combined Stress, Plastic Design
Steel Design
f
F
C f


1  f 

'  F
 F 
f
f
f
a

mx
a

bx
a
bx
my
1
by
a
bx
ex
a
C f

f 

1  '  F
 F 


by
ey
by
(1)
(2)
In formulas (1), (2) and (3), the subscripts x and y, combined with subscripts b, m,
and e, indicate the axis of bending about which a particular stress or design
property applies, and
Fa=allowable axial compressive stress if axial force alone existed, MPa
Fb=allowable bending stress if bending moment alone existed
12 E
2
F'
e

23
Kl b r b
Euler stress divided by a factor of safety, MPa the
'
expression by F el l b is the actual unbraced length in the
plane of bending and r b is the corresponding radius of
gyration, K is the effective length factor in the plane of
bending. As in the case of F a , F b, and 0.6Fy, F’e may
be increased 1/3 in accordance to Section 4.5.6.
2
Fa=computed axial stress, MPa
Fb=computed bending stress, MPa
Cm=a coefficient whose value is as follows:
1. For compression members in frames subject to joint translation (sideway),
Cm=0.85.
2. For restrained compression members in frames braced against joint translation
and not subject to transverse loading between their supports in the plane of
bending,
M
C  0.6  0.4
M but not less than 0.4
3.
1
m
2
Where M1/M2 is the ratio of the smaller to larger moments at the ends of that
portion of the member unbraced in the plane of bending under consideration.
M1/m2 is positive when the member is bent in reverse curvature, negative when
bent in single curvature.
3. For compression members in frames braced against joint translation in the
plane of loading and subjected to transverse loading between their supports, the
value of Cm may be determined by rational analysis. However, in lieu of such
analysis, the following values may be used:
A. For members whose ends are restrained, Cm=0.85
B. For members whose ends are unrestrained, Cm=1.0
Set 17- Columns, Beams,
Part 5-Structural
Combined Stress, Plastic Design
Steel Design
Steel column 3.6m long and hinged at both ends is used to carry an axial load of
800kN. The column is subjected to end moment (single curvature) with M 1  90% M 2
Fy=248MPa, E=200GPa. The allowable axial stress Fa=115MPa and the
allowable bending stress Fb=149MPa. The properties of the section are as
follows:
A  0.013m ; S x  0.00012m ; r b  9mm
2
3
Problem 25 
Which of the following most nearly gives the computed (actual) axial stress of the
column in MPa.
A. 61
B. 87
C. 73
D. 54
Problem 26 
Which of the following most nearly gives the computed (actual) bending stress of
the column in MPa.
A. 91
Problem 27 
B. 78
C. 66
D. 42
Which of the following most nearly gives the moment capacity of the column in
kN-m.
A. 3
B. 8
C. 10
D. 23
Situation problem 28 to 30 (CE November 2000)
A simply supported steel beam 6m long carries a uniform load of 32kN/m and an
axial compressive force of 320kN. The properties of the steel section is as
follows:
2
Area
A  14700mm
S
Section Modulus,
Flange width,
b
Flange thickness,
f
 1921x10
3
x
3
mm
 280mm
t
f
 16mm
Overall depth, d=390mm
 19mm
Web thickness, t w
According to Section 4.6.1 of the NSCP, for members subject to axial
compression and bending,
fa
f
 b  1.
0.6F
y
F
b
Fa=computed axial stress, MPa
Fy=yield strength of steel=248MPa
fb=computed bending stress, MPa
Fb=allowable bending stress= 0.66Fy.
Problem 28 
Which of the following most nearly gives the computed axial stress in the beam due to
axial force alone acting on the beam, in Megapascals.
A. 22
B. 27
C. 16
D. 32
Set 17- Columns, Beams,
Part 5-Structural
Combined Stress, Plastic Design
Steel Design
Problem 29 
Which of the following most nearly gives the computed bending stress in the
beam due to the uniform load alone acting on the beam, in MegaPascals.
A. 64
B. 84
C. 70
D. 75
Problem 30 
Which of the following most nearly gives the value the interaction equation.
A. 0.5
B. 0.4
C. 0..7
D. 0.6
Problem 31 (CE November 1994)
The shear capacity V if a steel wide flange section, 600mm deep whose web is
9mm thick, is closest to: (Assume Fy=200MPa)
A. 486kN
B. 864kN
C. 432kN
D. 684kN
Problem 32
A wide flange beam of depth 800mm carries a total shear of 2000kN. If
Fy=250Mpa, the required web thickness is:
A. 20mm
B. 30mm
C. 25mm
D. 35mm
Problem 33 (CE May 1995)
A structural steel I-beam is subjected to a shear of 90kN. The top and bottom
flanges are 12mm by 150mm, while the web is 9mm by 300mm. The average
shearing stress is V or 33.33Mpa. The moment of Inertia I of the section is
dt
6
w
4
108x10 mm
And the first moment of the area Q above the neutral axis is
382,050mm^3. However, the maximum shearing stress is:
A. 35.37MPa
Problem 34
B. 14.29Mpa
C. 33.33MPa
D. 7.77MPa
An A36 ateel girder (Fy=248MPa) has the following properties:
b
f
 250mm, t f  15mm, d  300mm, t w  15mm
What is the shear capacity of this beam in kN?
A. 523
B. 386
C. 446
D. 495
Problem 35 (CE May 1996)
A composite beam system is composed of structural steel sections on which a
reinforced concrete floor has been attached. The beams have a span of 8.0m.
and spacing of 2.4m. The steel section flange width is 180mm and the slab
thickness is 120mm. The effective flange width of the composite section
according to AISC specification is:
A. 2.1m
C. 2.4m
B. 2.0m
D. 2.58m
Set 17- Columns, Beams,
Part 5-Structural
Combined Stress, Plastic Design
Steel Design
Problem 36 (CE May 1997)
A W 410 x 74.4 supports a slab with a thickness of 150mm. The effective width of
the slab was computed to be 2000m. The structural steel section of composite
system has the following properties:
Beam depth, d
=413mm
Flange width, bf
=180mm
Cross-sectional area, A
=9480mm^2
Moment of Inertia, I
=274.3x10^6mm^4
Where is the new location of the neutral axis from the top of the slab in mm. Use
n=10 to transform the area of concrete to steel.
A. 234.50mm
C. 562.43mm
B. 142.59mm
D. 152.20mm
Problem 37(CE May 1999)
A composite section consist of a 100-mm thick slab, 1000mm wide attached on
top of wide flange section having a cross-sectional area of 10000mm^2.
Determine the required number of shear connectors for the composite section if
each has a capacity of 5kN. Use f '  27 MPa, F y  250MPa.
c
A. 20
B. 15
C. 23
D. 12
Situation: Problem 38 to 40
A w21 x 57 of A36 steel transfers end reaction
196kN to a wall built of solid brick by means of
a solid plate of A36 steel. Assume type S
mortar and a brick with Fp=1200kPa. The N
dimension of the plate is limited to 250mm.
K1 for the beam is 22mm.
Problem 38 
Which of the following most nearly gives
required area of the plate in square mm.
A. 163,000
C. 152,000
B. 169,000
D. 185,000
Problem 39 
Which of the following most nearly gives the value of “B” rounded-off to the
nearest 10mm.
A. 720
C. 660
B. 610
d. 580
Set 17- Columns, Beams,
Part 5-Structural
Combined Stress, Plastic Design
Steel Design
Problem 40 
Which of the following most nearly gives the required plate thickness in mm.
A. 36
B. 43
C. 53
D. 25
Problem 41 (CE November 1995)
The yield stress of the steel beam is 250 N/mm^2. The beam must be designed
for a plastic moment capacity of 288kN-m. The required plastic section modulus
is:
A. 2300cm^3
C. 4600cm^3
B. 1150cm^3
D. 3450cm^3
Situation: problem 42 to 45
The section of a steel beam is as shown in Figure ST-08. Use Fy=250MPa.
Problem 42 
Which of the following most nearly gives the area of
the section is square meter.
A. 0.019
B. 0.024
C. 0.015
D. 0.012
Problem 43 
Which of the following most nearly gives the
location of the plastic neutral axis from the bottom
of the section in millimeters.
A. 95
C. 85
B. 90
D. 80
Problem 44 
Which of the following most nearly gives
plastic section modulus of the section in m^3.
A. 0.0007
C. 0.0008
B. 0.0006
D. 0.0009
Problem 45
Which of the following most nearly gives
the plastic mment capacity of the section in kN-m.
A. 150
C. 190
B. 170
D. 200
Set 17- Columns, Beams,
Part 5-Structural
Combined Stress, Plastic Design
Steel Design
Situation: Problem 46 to 48
For the steel section shown:
Problem 46 
Which of the following most nearly gives
the section modulus of the section.
A. 4,900,000mm^3
B. 4,300,000mm^3
C. 3,400,000mm^3
D. 5,100,000mm^3
Problem 47 
Which of the following gives the plastic section modulus of the section.
A. 5,560,000mm^3
C. 4,550,000mm^3
B. 5,120,000mm^3
D. 6,130,000mm^3
Problem 48 
Which of the following most nearly gives the shape factor of the section.
A. 1.8
B. 1.1
C. 1.3
D. 1.5
Problem 49 (CE May 1999)
A fixed ended beam, 6m long carries a uniformly distributed load of w throughout
the entire span. If the plastic moment capacity of the beam section is 400 kN-m,
which of the following most nearly gives the value of w in kN/m.
A. 153
B. 198
C. 178
Problem 50
Determine the plastic moment for the
beam loaded as shown.
A. 120kN-m
B. 100kN-m
C. 110kN-m
D. 130kN-m
Problem 51
Determine the design plastic moment for
the beam loaded as shown.
A. 94kN-m
D. 204
B. 76kN-m
C. 83kN-m
D. 88kN-m
Set 17- Columns, Beams,
Part 5-Structural
Combined Stress, Plastic Design
Steel Design
Situation: problem 52 to 54
Situation 9- The members of the rectangular
portal frame shown is Figure ST-01 have
uniform cross-sections. It is required to
determine the plastic moment on three
possible modes of failure. Use a load
factor of 2.
Problem 52 
Which of the following most nearly gives the plastic moment Mp through the
formation of plastic hinges at B, C, and D (beam mechanism), in kN-m.
A. 167
C. 180
B. 123
D. 215
Problem 53 
Which of the following most nearly gives the plastic moment Mp through the
formation of plastic hinges at B and C (sideway mechanism), in kN-m.
A. 89
C. 165
B. 111
D. 120
Problem 54 
Which of the following most nearly gives the plastic moment Mp through
formation of plastic hinges at C and D (collapse or combined mechanism), in
kN-m.
A. 240
C. 268
B. 345
D. 220
Set 17- Columns, Beams,
Part 5-Structural
Combined Stress, Plastic Design
Steel Design
2
F
Euler critical stress,
3
a


2
 200000
2
E

L r  5000 50
2
P  F a A  197.4(8000)  1579200 N  1579.2kN
C
c
2
F

2
E
2

2
( 200000)
250
y
 125.66
Try the lightest: Choice B:
6
I
39.14 x10  58.21
r min  A 
11550
Kl / r  (1)(3000) / 58.21  51.54
Since
Kl r  C c
51.54  1.81
125.66
3
5 3(51.54)
FS  

3 8(125.66) 8



3
2


51
.
54
1 
 250  126.5MPa

Fa 
2
 1.81
 2 125.66 
P  F a xA  126.5(11550)  1461075 N

P  1461kN  1800kN
NOT ADEQUATE
Try Choice A:
6
I
67.59 x10  69.98mm
r min  A 
13800
KL / r  (1)(3000) / 69.98  42.87
2
 197.4MPa
KL r  C c
Since
5 342.87 
FS  

3 8125.66  8



42.87  1.79
125.66
3
3
2


42
.
87

 250  131.54 MPa
F a  1 
2
 1.79
 2 125.66 
P  FaxA  131.54(13800)  1815252 N

P  1815kN  1800kN
ADEQUATE
Set 17- Columns, Beams,
Part 5-Structural
Combined Stress, Plastic Design
Steel Design
4
The reference for this problem is Section 4.5.1.3.1 of NSCP:
6
I
r
 37.7 x10  68.65mm
A
8000
KL r  (1)(12000)(68.65)  174.8
C
c

2
F
2
E

2
2
(200000)
200
y
 140.5
E
SinceKL / r  C .then F  12
23 KL r 
12 (200000)  33.7 MPa
F 
23 174.8
2
a
c
2
2
2
a
5 to 7
C
c

2
F
2
E

2
y
2
( 200000)
248
For 2l 125 x 75 x 12
KL r  1(3) / 0.0195  153.8  C c
 126.16
12 E  12 (200000)  29.29MPa
F 
23 187.5
23 KL r 
2
2
2
a
2
For 2L 150 x 90 x 10
KL r  1(3) / 0.016  187.5  C c
12 E  12 (200000)  43.54MPa
F 
23 153.8
23 KL r 
2
2
2
a
2
For 2L 150 x 90 x 12
KL / r  1(3) / 0.0251  119.52  C c
KL r   5  3(119.52)  119.52  1.916
3 8(126.16) 8126.16
8C c


KL r   F  119.52  248

 1
 1
 71.35MPa





2C c  FS  2 126.16  1.916
3
5 3( KL / r )
FS  

3
8Cc
3
2
F
2
3
2
y
a
2
2
Thus, the section with the largest allowable compressive stress is 2L 150 x 90
x 12 with Fa=71.35MPa
The section that has the smallest allowable compressive stress is 2L 125 x 75
x 12 with Fa=29.29MPa.
Set 17- Columns, Beams,
Part 5-Structural
Combined Stress, Plastic Design
Steel Design
For 2L 125 x 75 x 12 (Fa=29.29MPa)
P=Fa x A= 29.29 (0.00454 x 1000^2)= 133kN
For 2L 150 x 90 x 12 (Fa=71.35MPa)
P=71.35(0.0055 x 1000^2) = 392.4kN
Thus, the most economical section is 2L 150 x 90 x 10.
8.
Reference: NSCP Section 4.5.1.4:
Check for compactness of the section:
d
t
700
 70
10

w
1680
F
b
f
b
1680
d

250
y
Since
F

t
 106.25
1680
F
w
, then F b  0.66 F y
y
 0.66(250)  165MPa
9

120 
MC M

I
S
M
; M  64080000 N  mm  64.08kN  m
3
534 x10
w 6
M  wL ;64.08 
; w  14.24kN / m
2
2
8
8
W=floor pressure x spacing between beams
14.24 = 4.6 x S; S= 3m
10
I
x

360
775  170.5 711 x2
3
3
12
12
I x  3750858585.75 mm
4
I  375,086cm
x
4
1cm /10mm
4
Set 17- Columns, Beams,
Part 5-Structural
Combined Stress, Plastic Design
Steel Design
11
The stress at a certain point O is given by:
M I
M I
M
I
M
I
x

y
f
I I  I xy
I I  I xy
M  0, M  10 x10 N  mm
(0)(33.6 x10 )  (10 x10 )( 13.4 x10 )
 41.9
f 
(33.6 x10 )(16.1x10 )  (13.4 x10 )^ 2
(0)(33.6 x10 )  (10 x10 )( 13.4 x10 )
135.7 

(33.6 x10 )(16.1x10 )  (13.4 x10 )^ 2
f  15.54  60.45  44.91MPa(tension)
a

y
x
x
xy
x
y
2
x
y
xy
2
y
x
y
6
y
x
6
o
6
6
6
6
6
6
6
6
6
6
6
o
12
Fv=0.40Fy=0.40(248)=99.2MPa
Solving for the load intensity, w:
Weight of concrete, 24(0.15+0.05)(2.4)
=22.080kN.m
Ceiling weight, 7.20(4.6)
=3.312kN/m
Live load, 2.4(4.6)
=11.040kN/m
Total =36.432kN/m
Try the lightest beam section:
Choice D: W12(19;Wt=278N/m, d=309mm, tw=6mm
Total w = 36.432+0.278 = 36.71kN/m
Maximum shear, V=R=wL/2 = 36.71(100/2 = 183.55kN
f
v

V
dt

w
183.33x1000
 99MPa  F v (OK )
309(6)
Thus the lightest beam is Choice letter D: W12 x 19
13
Maximum bending moment:
PL
M 

4
wL
2
8
2
300(6) w(6)
M 

4
8
M  450 _ 4.5w(kN  m)
Actual bending stress:
M
f 
S
b
x
Set 17- Columns, Beams,
Part 5-Structural
Combined Stress, Plastic Design
Steel Design
Allowable tensile stress:
F  0.6F
t
y
 0.6(2480  148.8MPa
Try first the lightest section:
W 21x62; w  911N / m, r t  53mm, S x  0.0023m
3
Allowable compressive stress:
Actual bending stress (tension and compression)
M=450+4.5(0.911)=454.1kN-m
f
b

Since
454.1
 197,435kPa  197.435MPa
0.0023
f
is greater than
b
F and F
t
b
the section is not adequate.
r
Try W24 x 68;w=1000N/m,
 57mm, S x  0.00253m
3
t
Allowable compressive stress:
F
F
b
b
 2 / 3  F y

 2 / 3  248

L r t  / 10,550,000(1) C  F
6000 / 57 / 10,550,000(1)(248)  100.74MPa
2
y
b
2
Actual bending stress(tension and compression)
M  450  4.5(1)  454.5kN  m
454.5
f b  0.00253  179,644.3kPa  179.644MPa
Since
f
b
is greater than
F and F
t
Try W18x76; weight=1108N/m,
r
the section is not adequate.
b
 57mm, S x  0.00253m
3
t
Allowable compressive stress:
 2 / 3  F y

F b  2 / 3  248
F
b
L r t  / 10,550,000 C  F
6000 57 / 10,550,000(1)(248)  100.74MPa
2
b
y
2
Actual bending stress(tension and compression)
M=450+4.5(1.108)=454.986kN-m
f
b

454.986
 190,371kPa  190.371MPa
0.00239
In this section, fb<Ft but fb>Fb. The section may be adequate but among the
choices, this section is much better.
Set 17- Columns, Beams,
Part 5-Structural
Combined Stress, Plastic Design
Steel Design
14 to 16
For simply supported beam with a uniform loading;
M  wL
8
2
50(8)

2
8
 400kN  m
The allowable flexural stress is:
F  0.66F
 0.66(248)  163.68MPa
M

 Fb
f
b
The actual flexural stress is
Sx
b
y
6
163.68  400 x10
S
S  2,443,792.8 mm x 1m / 1000mm  0.00244m
x
3
3
3
x
According to the problem, the maximum deflection is
L/360=8000/360=22.2mm
For simply supported beam with uniform loading, the maximum deflection is at
the midspan and is given by the formula:

 5wL
4
384 EI
4
4
5(50)(80) x1000
22.22 
384(200000) I
I  600.06 x10
6
mm x 1 / 1000  0.0006m
4
4
4
The required section must have
Try W 21 x 68, with D=0.537m
S
x

S
x

I
I
0.00066


 0.0023  0.00244(OK )
C D / 2 0.537 / 2
Therefore, the most adequate section is W 21x 68.
17 to 19
3
0.00244m
and
I
x

4
0.0006m
Spacing based on shear:
The shear capacity of the section is:
F
v

V
dt
 0.4F y
w
V  0.4(248)( 498)(56)  2766489.6 N  2766.5kN
This shear serves as the reaction R at the ends. Since the beam is loaded
uniformly, then
R
wL w(7)

 2766.5; w  790.4kN / m
2
2
Set 17- Columns, Beams,
Part 5-Structural
Combined Stress, Plastic Design
Steel Design
Since the beam weighs 7.32kN/m, then the total load it can carry is
790.4-7.32=783.08kN/m= W net
W
net
=120 x S = 783.08; S=6.53m
Spacing based on bending:
The moment capacity of the section is:
F
b

Mc M

 0.66F y
I
Sx
M
3
 0.66(2480, M  2247326400 N  mm
13730 x10
M  2247.33kN  m
w(7)
M  wL 
2
2
 2247.33, w  366.91kN / m
8
8
W net  366.91  7.32  359.59kN / m
W
net
 120 xS  360; S  3m
Spacing based on deflection:
  5wL
4
384 EI
 7000 / 360  19.44mm
4
5w(7000)
 19.44
384(200000)(3417 x10 )
6
W  424.95 N / mm  424.95kN / m
W
W
net
net
 424.95  7.32  417.63kN / m
 120 xS  417.63; S  3.48m
20 to 23
Check for compactness:
(This may not be necessary anymore since most rolled shapes are compact
sections)
b
2t

f
F
f
d
170
t
y

w
1680
F
y
706.4 1680

18.4
250
38.4  106.3(OK )
357.8
170

2(30.2)
250
5.9  10.7(OK )
Thus the section is compact.
L
u

138000
2
; A f  b f t f  357.8(30.2)  10805.56mm
d Af F y


Set 17- Columns, Beams,
Part 5-Structural
Combined Stress, Plastic Design
Steel Design
L
u
L
c

138000
 8444mm
(706.4 / 10805.56)( 250)

200b
F
y
f

200(357.8)
250
 4526mm
For an unsupported length of 4m (L=4000mm)
SinceL 
L andL  L
F  0.66F  0.66(250)  165MPa
u
c
b
y
For an unsupported length of 7m (L=7000mm)
SinceL 
L andL  L
F  o.6F  0.6(250)  150MPa
c
b
u
y
For an unsupported length of 9m (L=9000mm)
SinceL  Lc andL  Lu
Check if
703000C
F
C 1
b

y
L
r

t
3520000C
F
b
y
b
(for simply supported beam)
L
r

t
9000
 95.25
94.49
703000C  703000(1)  53
250
F
3520000C  3520000(1)  118.66
250
F
L
Since, 703000C 
 3520000C
F
r
F




L r t  
2
2 250 95.25 
F


250


F  3 
F
3

(
1
)
10.55x10 C 
10.55x10 


b
y
b
y
b
b
y
t
y
2
2
y
6
b
y
6
b
83,000C 
83,000(1)
 141.0MPa
L(d A )
9000(706.4 / 10805.56)
Therefore, F  141.07 MPa
F

b
b
f
b
Set 17- Columns, Beams,
Part 5-Structural
Combined Stress, Plastic Design
Steel Design
For an unsupported length of 12m (L=12000mm)
L
r

t
12000
 127
94.49
3520000C
r
F
1170x10 C  1170x10 (1)  72.54MPa
F 
L r t 
127
Since,
L

b
t
y
3
3
b
2
b
2
83,000C 
83,000(1)
 105.8MPa
L d A 
12000(706.4 / 10805.56)
Therefore, F  105.8MPa
F
b

b
f
b
Verify also:
24
The beam is subjected to a combined axial
and bending stress. The relevant equation is:
f
F
a
t

f
F
b
1
b
Where:
f
a
F
a
f
b
F 250,000

 45.44MPa
= actual axial stress = A 2751(2)
= allowable axial tensile stress =
0.60F
= actual bending stress = M/Sx
Where M= (P/2)(2.25)=1.125P(kN-m)
y
 0.60(248)  148.8MPa
2
1.125P(1000)
f 
2(0.000063x1000 )
3
b
F
b
= 8.928P (MPa) with P in kN
= allowable bending stress = 0.66Fy = 0.66(248)= 163.68MPa
45.44 8.929 P

 1; P  12.73kN
148.8 163.68
25 to 27
f
a

P

A
800kN
2
 61538.46kPa  61.54MPa
0.013m
f F  61.54 115  0.535  0.15
C  0.6  0.4M M 
a
a
1
m
2
Set 17- Columns, Beams,
Part 5-Structural
Combined Stress, Plastic Design
Steel Design
M   0.96
12 E  12 (200,000)
' 
F 23 
Kl b r b  231(3600 / 94)
f
C f
C f


1


F  f 
f 

1  '  F
1  '  F
 F 
 F 


0.96 f
61.54
C
m
 0.6  0.4 0.9M 2
2
2
2
2
e
mx
a
a
2
my
bx
a
by
a
bx
by
ex
115
f
b

ey
b
61.54 

1 
149
 702.16 
 65.83MPa
1
 702.16
Also:
f
0.6F

a
y
f
F
f
bx

bx
f
F
61.54
 b  1;
0.6(248) 149
Thus,
f
b

f
b
1
by
f
b
 87.38MPa
 65.83MPa
M
S
by
M

3
 65.83
0.00012(1000)
x
M  7,89,600 N  mm  7.9kN  m
28 to 30
Axial stress:
f
a

P 320000

 21.77 MPa
A 14700
Bending stress:
M  wL
8
2
2
32(6)

8
 144kN  m
6
f   144 x10
S 1921x10
M
3
b
 74.96MPa
x
Interaction equation:
21.77
74.96

 0.604
0.6(248) 0.66(248)
Set 17- Columns, Beams,
Part 5-Structural
Combined Stress, Plastic Design
Steel Design
31
According to Section 4.5.1.2.1 of NSCP;
0.4(2000 
32
V
;V  432000 N  432kN
600(9)
F  0.4F
v
y

V
dt
w
F
v

V
dt
 0.4(250)  100
w
2,000,000
8000t
 100; t w  25mm
w
33
f
v

VQ 90,000(382,050)

 35.375MPa
6
It
(
9
)
108x10
34
F  0.4F
v
F
v

V
dt

w
y
 99.2MPa
V
 99.2;V  446,400 N  446.4kN
300(15)
35
According to Section 4.11 of NSCP, the
effective width of the concrete flange shall
be taken as not more than 1/4 the span of
the beam, and the effective projection
beyond the edge of the beam shall not
be taken as more than 1/2 the clear
distance to the adjacent beam nor more
than 8 times the slab thickness.
(1)  b  L / 4  8000 / 4  2000..  2m
(2)b  b f  2b1
b
1
 1 / 2(2220)  1110
or b1  8(120)  90
b  180  2(960)  2100mm  2.1m
Use b=2m
Set 17- Columns, Beams,
Part 5-Structural
Combined Stress, Plastic Design
Steel Design
36
Transforming the concrete to steel, its equivalent width is 2000/n=200mm
Then;
AY G 
A
39480Y
slab
G

A
steel
 200(150)  9480  39480mm
2
 (200)(150) x75  9780 x356.5;Y G  142.59mm
37
Where Ac is the actual area of effective concrete flange and As is the
area of steel beam.According to Section 4.11.4, the total horizontal shear to
be resisted by shear connectors at the junction of the steel beam and the
concrete slab shall be taken as the smaller value of the following:
V
V

h
b

0.85 f ' A
2
0.85 f ' A
c
2
AF
c
c
c

and V h 
AF
s
y
2
0.85(27)1000(100)
 1,147,500 N
2
10,000(250)
 1,250,000 N
2
2
Thus, useV h  1,147,500 N  1,147.5kN
1,147.5
 22.95say 23
Number of shear connectors = 50
V

h
s
y

38 to 40
A
Area of plate,
R
F

p
2
196,000
 163,333mm
1.2
A=B x N= B(250)=163,333;B=653 mm say 660mm
Set 17- Columns, Beams,
Part 5-Structural
Combined Stress, Plastic Design
Steel Design
n
t
B
660
 k1 
 22  308mm
2
2
3f n
F
2
p
b
f
t
3(1.188)(308)
186
2
p

196000
 1.188MPa
250(660)
F  0.75F
b
 42.63mmsay43mm
y
 (0.75)( 248)  186MPa
41
The plastic moment capacity is given by the relationship Mp=Fy x Z, where Z
is the plastic section modulus of the section.
6
Then,
288x10
 (250) Z ; Z  1152000 mm  1152cm
3
3
42 to 45
100(200)  60(140)  11600mm  0.0116 mm
2
Area =
The plastic NA divides the
section into two equal areas.
Area below NA=11600/2
Area below NA = 5800
60(40)+2[20(x)]=5800
X=85mm
A. 115-10=105mm
B. 115/2=57.5mm
C.85/2=42.5mm
D.85-20=65mm
Plastic section modulus:
 Ay  60(20)(105)  2 x(115)(20)(57.5)
Z=
+60(40)(65) + 2 x (85)(20)(42.5)
Z=691,000mm^3=0.000691m^3
Mp=FyZ=250 x 691,000 = 172.75 x 10^6N-mm
Mp=172.75kN-m
2
46 to 48
2
A1  300(40)  12000mm
A  300(400  12000mm
A  400(40)  16000mm
A  A  A  A  40000mm
A Y   Ay
2
2
2
3
T
T
1
2
2
3
G
Set 17- Columns, Beams,
Part 5-Structural
Combined Stress, Plastic Design
Steel Design
40000Y  12000(360)  12000(190)  16000(20)
Y  173mm
G
G
a  360  173  187 mm
b  190  173  17 mm
c  173  20  153mm
I
g

 I
3
g

Ad
2
  300(40)
12
3
2
 12000(187)
40(300)

12
3
2
 12000(17) 
I
g
400(40)
12
 891,373,333mm
2
 16000(153)
4
I
; c  380  173  207
C
3
891,373,333
S 
 4,306,151mm
207
S 
Plastic Section Modulus:
The plastic N.A. divides the section
2
A  40,000mm
A
 20,000
A
T
aboveN. A.
belowN . A.
300(40)  40a  20000
a  200mm, b  100mm
Y
Y
Y
Y
1
 200  20  220mm
2
 200 / 2  100mm
3
 100 / 2  50mm
4
 100  20  120mm
Z   Ay  300(40)( 220)  200(40)(100)  100(40)(50)  400(40)(120)
3
Z  5,560,000mm
49
Internal work=external work
Mp(40)=1/2(w)(6)(30)
400(4)=9w
W=177.78kN/m
Set 17- Columns, Beams,
Part 5-Structural
Combined Stress, Plastic Design
Steel Design
50
Internal work = external work
M
M
p
 2  5   140(6 )
p
 120kN  m
51
For collapse mechanism 1:
M
M
p
 (4  6 )  90(8 )  80(2 )
p
 88kN  m
For collapse mechanism 2:
M
M
p
   6   90(2 )  80(5 )
p
 82.86kN  m
Thus, the design plastic moment
Mp=88kN-m
52 to 54
Beam Mechanism
M
p
   2     111(6 )
Mp  180kN  m
Sidesway mechanism:
M
p
 (   )  40(6 )
Mp  120kN  m
Combined Mechanism:
Mp (2     )  40(6 )  120(6 )
Mp  240kN  m
Set 18-Miscellaneous &
Part 5-Structural
Recent Board Problems
Steel Design
Situation: Problem 1 to 3:
A riveted lap joint is shown in Figure ST-17. The rivet have 16mm diameters, and
the permissible stresses are 124MPA
for shear in rivets, 296MPa for
bearing and 165MPa for tension
(on net area) in plates. The plates
are 10mm thick.
Assume hole diameter be 1.5mm
larger than the rived diameter.
Problem 1
Which of the following gives the value of P so that the allowable bearing stress in
rivets will not be exceeded.
A. 285.7kN
C. 236.8kN
B. 214.5kN
D. 257.7kN
Problem 2
Which of the following gives the value of P so that the allowable tensile stress in
plates will not be exceeded.
A. 302.5kN
C. 298.3kN
B. 254.8kN
D. 280.5kN
Problem 3
Which of the following gives the safe value of P.
A. 124.6kN
C. 280.5kN
B. 114.3kN
D. 236.8kN
Situation: problem 4 to 6 (CE May 2002)
A plate with width of 400mm and thickness of 12mm is to be connected to a plate
of the same width and thickness by 34mm diameter blots, as shown in Figure
ST-13. The hole are 2mm larger than the bolt diameter. The plate is A36 steel
with yield strength Fy=248MegaPascals. Allowable tensile stress is 0.60Fy. It is
required to determine the value of b so that the net width along bolts 1-2-3-3 is
equal to the net width along bolts 1-2-4.
Set 18-Miscellaneous &
Part 5-Structural
Recent Board Problems
Steel Design
Problem 4
Which of the following gives the
value of b in millimeters.
A. 28.6
B. 37.4
C. 52.1
D. 19.7
Problem 5 
Which of the following gives the
value of the net area for tension in
plates in square millimeters.
A. 3624
B. 3214
C. 3867
D. 4178
Problem 6 
Which of the following gives the value of P so that the allowable tensile stress on
net area will not be exceeded.
A. 575
B. 687
Situation: Problem 7 to 9
An axially loaded connection is
shown in Figure ST-71. The plate
is A36 steel with thickness of 6mm.
The rivets are 28mm diameter
with allowable shearing stress of
190MPa. Allowable bearing stress
on projected area between rivet
and plate is 370MPa and the
allowable tensile stress on plates
on net area is 150MPa. Hole
diameter is 1.5mm larger than
C. 539
D. 424
the rivet diameter.
Problem 7 
Which of the following gives the value of P without exceeding the allowable
shearing stress in rivets, in KiloNewton:
A. 495
C. 465
B. 430
D. 450
Set 19-Timber and
Part 6-Structural Timber
Construction
& Construction
Problem 87 
Which of the following most nearly gives the uniformly distributed load the beam
can carry without exceeding the allowable shearing stress in wood, in kN/m.
A. 12.15
C. 12.96
B. 16.2
D. 16.2
Problem 88 
Which of the following most nearly give the uniform distributed load the beam csn
carry without exceeding the allowable shearing stress in glue, in kN/m.
A. 12.15
B. 10.58
C. 16.2
D. 12.96
Situation: problem 89 to 91
Two 75mm x 200mm members are bolted to a 100 x 250mm horizontal member,
as shown in Figure TD-44, with four 20-mm volts. The angle between the
members is 45 degree, the sloping member transmits a compressive force to the
horizontal member and the wood is Yakal (Special Group 1). The allowable load
in each bolt for the connection is given in Table 4-4.
Problem 89 
If failure occur in the inclined member, which of the following most nearly gives
the value of F without exceeding the allowable, in KiloNewton, load in the bolts:
A. 45.7
C. 68.9
B. 76.3
D. 89.2
Problem 90 
If failure occur in the horizontal member, which of the following most nearly gives
the value of F without exceeding the allowable, in KiloNewton, load in the bolts:
A. 62.4
C. 58.7
B. 68.9
D. 45.7
Problem 91 
Which of the following most nearly gives the safe value of F in KiloNewton:
A. 76.3
C. 58.7
B. 62.4
D. 89.2
Set 19-Timber and
Construction
Part 6-Structural Timber
& Construction
Situation: Problem 92 to 94
A concrete platform weighing 7.2kPa is to be supported during pouring by a
timber deck consisting of 25-mm planks resting on wooden joints. Allowable
stress of the plank in flexure and shear are 10.34MPa and 0.69MPa, respectively.
Allowable deflection of the plank is 1/250 of the span. It is required to determine
the spacing of the joist so that the allowable stresses and deflection will not be
exceeded. Timber plank weighs 7.86kN/m^3. E=10340MPa.
Problem 92 
Considering 1m width of plank, which of the following gives the load carried by
the plank in KiloNewton per meter.
A. 8.1
B. 6.6
C. 7.4
D. 5.2
Problem 93 
Which of the following gives the spacing of the without exceeding the allowable
shearing stress.
A. 3109mm
C. 2765mm
B. 3457mm
D. 1687mm
Problem 94 
Which of the following gives the required spacing of the joist.
A. 1680
B. 740
C. 2765
D. 820
Situation: Problem 95 to 97 (CE November 2001)
Section 3.7.3 of NSCP states the following for simple timber solid-column design.
Allowable unit stress in N square mm of cross-sectional area of square or
rectangular simple solid columns shall be determined by the following formulas,
but such unit stress shall not exceed values for compression, parallel to grain Fc
in Table 3.1.
Short columns (Ie/d of 11 or less);
F
' 
c
F
c
Intermediate columns (Ie/d greater than 11 but less than K)
K  0.671
E
F
F
c
' 
c
F
 1  I e d 

1  
c

 3  K 
Long columns (Ie/d of K or greater):
F'
c

0.30 E
I e d 
2
Where:
Fc=allowable unit stress in compression parallel to grain, MPa
'
F c = allowable unit stress in compression parallel to grain adjusted fpr Ie/d
ratio where d is the least dimension, MPa
E= modulus of elasticity, MPa
Ie= effective span length= KeL
Set 19-Timber and
Part 6-Structural Timber
Construction
& Construction
A 140mm x 140mm Apitong 3m long is used as a column. The column is hinged
at both ends with Ke=1.0. The properties of Philippine woods at 80% stress grade
is given in Table 4-3.
Problem 95 
Which of the following gives the classification of the column.
A. Long column
C. short column
B. Intermediate column
D. not approved by the code
Problem 96 
Which of the following most nearly gives the value of the allowable unit stress in
compression parallel to grain adjusted for Ie/d ratio, in MegaPascals.
A. 5.64
B. 7.54
C. 3.21
D. 4.78
Problem 97 
Which of the following most nearly gives the axial load capacity of the column in
KiloNewtons.
A. 76.5
B. 103.2
C. 93.6
D. 83.3
Set 19-Timber and
Construction
Part 6-Structural Timber
& Construction
1 to 3
From the diagram shown, the critical path is: a-b-d-f-j-k-m-n
Duration of project = 65 weeks
Earliest start of activity m = 45 weeks
4 to 6
Critical path= A-H-I-J-K-M = 19 days
Normal Cost = 1,000 + 2,000 + 1,600 + 250 + 500 + 600 + 800 + 3,000 + 400 +
1,000 + 500 + 500 + 600
Normal Cost = P12,750
Set 19-Timber and
Construction
Part 6-Structural Timber
& Construction
Crashing
Reduce Activity A to 2 days at an additional cost of P100
Reduce activity H to 3 days at an additional cost of (800/4)(3)=P600
Reduce activity D to 2 days at an additional cost of P50.
Reduce activity F to 2 days at an additional cost of P100
Reduce activity L to 1 day at an additional cost of P500.
Thus, the least possible time is 15 days and the additional cost (cost of crashing)
is P100+600+50+100+500=P1350.
Total cost = P12750+1350=P14100.
Set 19-Timber and
Construction
Part 6-Structural Timber
& Construction
7 to 9
AAE=P500,000,000.00
120% of P500,000,000=P600,000,000
60% of P500,000,000= 300,000,000
Therefore, only bidders A, C, and D shall be considered as responsive
bidders.
Average of responsive bids:
Average = (550,234,451.98+454,218,557.98+389,122,897.44)/3
Average= P464,525,302.47
AGE=1/2(AAE + Average of responsive bids)
AGE=1/2(500,000,000+464,525,302.47)=P482,262,651.23
10 to 12
AAE = P500,000,000.00
Maximum bid price= 120% of P500,000,000 = P600,000,000
Maximum bid price= 60% of P50,000,000=P300,000,000
Therefore, only bidders A,C and D shall be considered as responsive bidders.
Average of responsive bids:
Average = (550,234,451.98+454,218,557.98+389,122,897.44)/3
Average= P464,525,302.47
AGE=1/2(AAE + Average of responsive bids)
AGE=1/2(500,000,000+464,525,302.47)=P482,262,651.23
70% of AGE = P337,583,855.86
According to the rules and regulation, no award of contract shall be made to a
bidder whose bid price is higher than either the AAE or AGE, (P500,000,000) or
whose bid is lower than 70% of AGE (P337,583,855.86)
Among the responsive bidders, the award cant be made t bidder A. It may either
be awarded to bidders C or D, and among the two, Bidder D is more desirable.