Ref F2F Week 4 - Solution_unlocked.pdf

GILLESANIA Engineering
Review & Training Center
September 2022
F2F Refresher
Week 4 ( PSAD )

1. Given the following data of the solid
circular pole shown:
250 mm 0.45 kN
3 m 100 mm
3 kN
Weight of pole = 22 kg/m
Determine the maximum compressive
stress (MPa) at the base of the pole.
A. 0.954 B. 0.941
C. 0.806 D. 1.150
P
H
L
e
D

1. Given the following data of the solid
circular pole shown:
250 mm 0.45 kN
3 m 100 mm
3 kN
Weight of pole = 22 kg/m
Determine the maximum compressive
stress (MPa) at the base of the pole.
A. 0.954 B. 0.941
C. 0.806 D. 1.150
P = 3
H=0.45
L=3
e=0.1
D=250
Wp
22 9.81 3 647.46 N
0.64746 kN
D=250
y
x

b
a
3 0.64746 3.64746 kN
0.45 3 3 0.1
1.65 kN m
!
# $

%
$

'(
!
$
3647.46
)
* 250 + $
1.65 10, 250/2
)
,* 250 *
!
$.. ./ 012

2. Refer to the figure shown:
Given: = 2 m
Bar diameter = 20 mm
Concrete cover, 3 = 65 mm
Length of extension bar, 4 = 80 mm
Radius of bend = 45#
’
= 20.7MPa; (
= 414 MPa
According to 412.6 of NSCP, Development
length 78 for deformed bars in tension
terminating in a standard hook shall not be
less than:
1.
(5#
4.17
9
2. 8 5#
3. 150 mm
Determine the total length (m) of the bar.
A. 2.53 C. 2.87
B. 3.03 D. 3.36
a
L
b
Ldh
r

2. Refer to the figure shown:
Given: = 2 m
Bar diameter = 20 mm
Concrete cover, 3 = 65 mm
Length of extension bar, 4 = 80 mm
Radius of bend = 45#

9
20.7MPa; ( 414 MPa
According to 412.6 of NSCP, Development
length 78 for deformed bars in tension
terminating in a standard hook shall not be
less than:
1.
(5#
4.17
9
2. 8 5#
3. 150 mm
Determine the total length (m) of the bar.
A. 2.53 C. 2.87
B. 3.03 D. 3.36
65
L = 2
b=80
Ldh
r=4db
db = 20mm

414 20
4.17 20.7
436.4 mm
8 20 160
← 78
; 4 20 80
x1
x2 x3
= 65 ; +
+
155
+ 2000 1845
? 436.4
? 346.4
1845 346.4 2191.4
@; @;
4 80 4 80
#AB 2191.4 @ 80 2 2 80
#AB 2854 mm
$155
$ ;
C. D/E F
Q2 C 46.08 14.81 28.94 9.68
$+
+

3. A 100-mm thick one-way slab is
reinforced with 12 mm bars spaced at
200 mm on centers. Given that
9 20.7
MPa, ( 275 MPa, compute the design
moment (kN-m) of the slab. Assume 20
mm concrete cover.
A. 10.82 C. 10.58
B. 11.75 D. 9.74
4 1000
ℎ

100
20
12‡ 5
5 100 $ 20 $ 12/2 74 mm
%# )
* 12 + 113.1 mm+
%H
1000%#
I
Consider 1 m slab width

1000 113.1
200
565.5 mm+
Assuming steel yields:
3
%H(
0.85
9
4

565.5 275
0.85 20.7 1000
3 8.838 mm
3/J= 8.838/0.85
10.4 mm ?
K
5 27.75 Tension controlled
L M 5 $ 3/2 0.85
934 5 $ 3/2
0.85 20.7 8.838 1000 74 $
8.838
2
L 10.82 kN m
NL 0.9 10.82 O. PQD RS F
T
Q3 D 21.01 18.68 35.53 23.23

Situation 2 –
Dimensions:
3 4 5 0.4 m; 3.5 m
Footing width, 2 m
Effective depth = 500 mm
Uniform soil pressure
(from factored loads) = 196 kPa
Shear diagram:
156.8 kN; U 761.2 kN; -610.8 kN

9
27.5 MPa; ( 414 MPa
Main bar diameter = 20 mm
Reduction factors: Moment = 0.90
Shear = 0.75
7. Determine the required number of bars at
critical moment.
A. 10 B. 12 C. 11 D. 13
8. Determine the spacing of bars, in mm, at the
face of the column. Use 75 mm clear cover.
A. 90 B. 180 C. 95 D. 105
9. Determine the critical wide beam shear
stress, in MPa.
A. 1.02 B. 0.75 C. 0.81 D. 0.57
a b c d
W
column column
e
f
g
h

Situation 2 –
Dimensions:
3 4 5 0.4 m; 3.5 m
Footing width, 2 m
Uniform soil pressure
(from factored loads) = 196 kPa
Shear diagram:
156.8 kN; U 761.2 kN; -610.8 kN

9
27.5 MPa; ( 414 MPa
Shear = 0.75
0.4 0.4 3.5
2
column column
156.8
-610.8
761.2
h
0.4
wu
Solving for wu:
$610.8 VW 3.5 761.2
VW 392 kN/m
OR: 0 VW 0.4 156.8
VW 392 kN/m
OR: VW XW
VW 196 2 392 kN/m

Situation 2 –
Shear = 0.75
9. Determine the critical wide beam shear
stress, in MPa.
A. 1.02 B. 0.75 C. 0.81 D. 0.57
0.4 0.4 3.5
2
column column
156.8
-610.8
761.2
h
0.4
wu
VW 392 kN/m
deff=0.5
Y
W
Y
W 761.2 $ VW5
761.2 $ 392 0.5
Y
W 565.2 kN
ZW[
Y
W
4[5
4
[

2000
mm
ZW[
565,200
2000 500
5 500 mm
ZW[ ]. /^/ 012
Q9 D 30.3 24.69 30.11 13.94

Situation 2 –
0.4 0.4 3.5
2
column column
156.8
-610.8
761.2
h
0.4
wu
VW 392 kN/m
4
[

2000
mm
5 500 mm
4 2000 mm
156.8
-610.8
761.2
0.4
x

156.8

0.4
156.8 610.8
0.0817 m
0.4
A1
%= =
+
0.4 0.0817 156.8
%= 37.765 kN m
0.4+x
156.8
-610.8
0.4-x=0.3183
A2
%+
=
+
0.3183 $610.8
%+ $97.21 kN m
610.8/VW
1.558
3.5 $ 1.558
1.942
A3
A4
%?
=
+ 1.558 $610.8 $475.81 kN m
%*
=
+
1.942 761.2
%* 739.13 kN m

Situation 2 –
0.4 0.4 3.5
2
column column
0.4
VW 392 kN/m
4
[

2000
mm
5 500 mm
4 2000 mm
A1
%= 37.765 kN m
A2
%+ $97.21 kN m
A3
A4
%?
$475.81 kN m
%* 739.13 kN m
=
+
?
= %= QP. P^/ RS F
+ %= %+ %?
$/Q/. C^ RS F
? + %* C]Q. DP RS F

Situation 2 –
9
27.5 MPa; ( 414 MPa
7. Determine the required number of bars at
critical moment.
A. 10 B. 12 C. 11 D. 13
A. 90 B. 180 C. 95 D. 105
0.4 0.4 3.5
2
column column
0.4
5 500 mm
4 2000 mm
=
+
?
$/Q/. C^ RS F
C]Q. DP RS F
At critical moment:
W + 535.26 kN m
_L
W
N45+
535.26 10,
0.9 2000 500 +
1.189 MPa
c
0.85
9
(
1 $ 1 $
2_L
0.85
9
c 0.00295 cdeL 1.4/( 0.00338 ◄Use this
%H
QP. P^/ RS F
c45 0.00338 2000 500
%H 3380 mm+
f+
3380
)
* 20 + 10.75 use 11
Q7 C 15.88 42.88 28.94 11.42

Situation 2 –
9
27.5 MPa; ( 414 MPa
A. 90 B. 180 C. 95 D. 105
0.4 0.4 3.5
2
column column
0.4
5 500 mm
4 2000 mm
=
+
?
$/Q/. C^ RS F
C]Q. DP RS F
cdeL 1.4/( 0.00338
QP. P^/ RS F
%H 3380 mm+
f+
3380
)
* 20 + 10.75 use 11
75 75
20‡
2000
2000 $2 75 $20 1830 mm
gh3ijU, I
1830
11 $ 1
.DQ FF
Q8 B 21.49 18.97 28.56 30.01

Situation 3 - The bridge truss shown in
the figure is subjected to moving
loads, 8 m, I 4 m, 0 kN,
k 45°.
10. Compute the value of the maximum
ordinate of influence diagram of
member lm.
A. 1.5 C. 0.75
B. 0.5 D. 1.00
11. Compute the axial load in member lm
(kN) due to a standard wheel base 4.3
m apart, front wheel load = 72.4 kN
and rear wheel load = 19.6 kN.
A. 63.25 C C. 81.47 C
B. 63.25 T D. 81.47 T
(kN) due to a uniform moving load of
9.35 kN/m.
A. 74.8 C C. 52.6 T
B. 74.8 T D. 52.6 C
P
A
B C D
E
F G H I
J K
k k
s s s s
L

k 45°.
member lm.
A. 1.5 C. 0.75
B. 0.5 D. 1.00
A. 63.25 C C. 81.47 C
B. 63.25 TC D. 81.47 T
9.35 kN/m.
A. 74.8 C C. 52.6 T
B. 74.8 T D. 52.6 C
A
B C D
E
F G H I
J K
k 45° k
4 4 4 4
8

k 45°.
member lm.
A. 1.5 C. 0.75
B. 0.5 D. 1.00
A
B C D
E
F G H I
J K
k 45° k
4 4 4 4
8
h
FJK
a = 8 b = 8
L = 16
ℎ 4 tan 45° 4 m
q
q
34

8 8
16
4
r $
q
ℎ
$
4
4
$.
Influence line for FJK:
Q10 D 13.17 23.81 22.75 39.59

k 45°.
A. 63.25 C C. 81.47 C
B. 63.25 T D. 81.47 T
r $1
A
B C D
E
F G H I
J K
k k
4 4 4 4
L
72.4 19.6
4.3
3.7
r=
r=
3.7

$1
8
r= $0.4625
stu !
72.4 $1 19.6 $0.4625
stu !
D.. E^/ RS v
Q11 C 28.07 11.23 43.56 16.17

k 45°.
9.35 kN/m.
A. 74.8 C C. 52.6 T
B. 74.8 T D. 52.6 C
r $1
A
B C D
E
F G H I
J K
k k
4 4 4 4
L
w = 9.35 kN/m

k 45°.
9.35 kN/m.
A. 74.8 C C. 52.6 T
B. 74.8 T D. 52.6 C
r $1
A
B C D
E
F G H I
J K
k k
4 4 4 4
L
w = 9.35 kN/m
%
1
2
16 $1 $D
stu !
V% 9.35 $8
PE. D RS v
Q12 A 43.08 21.59 15.68 18.78

Situation 6 - The pile footing supports a
column, 600 mm 600 mm at the center.
The piles are precast concrete with 350
mm 350 mm dimension. Given:
Net load on the footing at Ultimate
Condition:
W 2700 kN; Ww 108 kN-m
W( 165 kN-m
Effective depth of the footing = 400 mm
Dimensions: 3 0.6 m 1.2 m
4 1 m 5 0.6 m
Strength Reduction Factor
For shear = 0.75; For moment = 0.90
19. Determine the critical punching shear
stress (MPa) of one pile due to axial load.
A. 0.54 B. 0.68 C. 0.25 D. 0.33
20. Determine the punching shear stress (MPa)
of the column due to axial load.
A. 1.5 B. 2.0 C. 2.5 D. 3.0
stress (MPa) due to axial and moment
loads of the pile cap.
A. 2.0 B. 3.0 C. 1.5 D. 2.5
x
y
a
b
b
a
d d
c
c

Condition:
W 2700 kN; Ww 108 kN-m
W( 165 kN-m
4 1 m 5 0.6 m
A. 0.54 B. 0.68 C. 0.25 D. 0.33
A. 1.5 B. 2.0 C. 2.5 D. 3.0
A. 2.0 B. 3.0 C. 1.5 D. 2.5
x
y
0.6
0.6
1
1
0.6
1.2 1.2
0.6
Pu = 2700
Part 1: due to axial load only
_
W
f

2700
9
300 kN
350+400
350+400
Y
W
ZW
Y
W
4x5

300,000
4 750 400
]. C/ 012
Q19 C 23.52 20.81 20.81 33.88

Condition:
W 2700 kN; Ww 108 kN-m
W( 165 kN-m
4 1 m 5 0.6 m
A. 0.54 B. 0.68 C. 0.25 D. 0.33
A. 1.5 B. 2.0 C. 2.5 D. 3.0
A. 2.0 B. 3.0 C. 1.5 D. 2.5
x
y
0.6
0.6
1
1
0.6
1.2 1.2
0.6
Pu = 2700
Part 2: due to axial load only
600+400
600+400
_
300 kN
ZW
Y
W
4x5

2,400,000
4 1000 400
.. / 012
Y
W W $ _ 2700 $ 300 2400 kN
Q20 A 24.1 33.3 28.36 13.36

Situation 6 –
A. 2.0 B. 3.0 C. 1.5 D. 2.5 x
y
0.6
0.6
1
1
0.6
1.2 1.2
0.6
Pu = 2700
Mux = 108
Muy = 165
A
B
C
D
E
F
G
H
I
350+400
350+400
_
W
f
y
Wwz
Σz+ y
W(
Σ+
_ |
2700
9

108 1
1 + 6

165 1.2
1.2 + 6
_ | 340.92 kN Y
W
ZW
Y
W
4x5

340,920
4 750 400
]. CDE 012
On the heavily loaded pile:

Situation 6 –
A. 2.0 B. 3.0 C. 1.5 D. 2.5 x
y
0.6
0.6
1
1
0.6
1.2 1.2
0.6
Pu = 2700
Mux = 108
Muy = 165
A
B
C
D
E
F
G
H
I
ZW
]. CDE 012
On the heavily loaded pile:
On the column:
_ }
2700
9
0 0 300 kN
600+400
600+400
ZW
Y
W
4x5

2,400,000
4 1000 400
.. / 012
Y
W W $ _ } 2700 $ 300 2400 kN
_
W
f
y
Wwz
Σz+ y
W(
Σ+
0; z 0
Q21 C 22.75 25.17 24.2 26.91

Situation 8 – For the rectangular tied column
shown in the figure.
Given:

9
27.5 MPa mL 0.80
( 414 MPa _L 0.205
Reinforcement steel ratio, c 0.05
Dimensions (4 ℎ) = 600 mm x 400 mm
Strength reduction factor = 0.65
25. How many 20 mm diameter bars are
required for the rectangular tied column?
A. 40 B. 36 C. 30 D. 28
26. What is the maximum nominal axial load
L (kN) that the column could carry?
A. 7542 B. 5354 C. 8238 D. 9236
27. If reinforcement steel ratio, c 0.03,
what is the required length (mm) of the
long side of the column?
A. 620 B. 700 C. 650 D. 670

Given:

9
27.5 MPa mL 0.80
( 414 MPa _L 0.205
25. How many 20 mm diameter bars are
required for the rectangular tied column?
A. 40 B. 36 C. 30 D. 28
% 600 400 240,000 mm+
%H c% 0.05 240,000
%H 12,000 mm+
f+
12000
)
*
20 + 38.2 Use 40
Q25 A 27.98 24.1 20.72 26.23

Given:

9
27.5 MPa mL 0.80
( 414 MPa _L 0.205
26. What is the maximum nominal axial load
L (kN) that the column could carry?
A. 7542 B. 5354 C. 8238 D. 9236
% 600 400 240,000 mm+
x 0.85
9 % $ %H (%H
x 0.85 27.5 240,000 $ 12,000
%H 12,000 mm+
414 12,000
x
L 0.8x D, CQD RS
10,297,500 N 10,297.5 kN
Q26 C 24.1 35.14 29.53 9.97

Situation 8 –
A. 620 B. 700 C. 650 D. 670
% 240,000 mm+
ℎ 400 mm c 0.03
L mL
9% 0.8 27.5 240,000
L 5280 kN
L L _L
9%ℎ
0.205 27.5 240000 400
mL 0.8 _L 0.205
L 541.2 kN m
€ ]. D
‚ ]. C]/
€ ]. PC
%
L
mL
9
5,280,000
0.72 27.5
266,667 mm+
% 4ℎ 4 400 266,667 mm+
4 ^^^. P FF Q27 D 19.17 25.75 31.46 22.17

Situation 8 –
A. 620 B. 700 C. 650 D. 670
ℎ 400 mm
c 0.03
€ ]. D
‚ ]. C]/
€ ]. PC
%
266,667 mm+
4 ^^^. P FF
‚

].
.D/
L 5280 kN L 541.2 kN m
L L _L
9
%ℎ
L 0.185 27.5 266,667 400
L /EC. P RS F

Situation 4 – The beam ƒM shown is
supported by a quarter-circular arc %M
having a radius of 4 m. The frame is pin-
supported at %, ƒ and M. The pins at % and
ƒ are in single shear while the pin at M is
in double shear. Allowable shear stress of
the pin is 112 MPa. V 3 kN/m.
13. Determine the minimum diameter (mm)
of the pin at ƒ.
A. 10 B. 12 C. 8 D. 6
of the pin at M.
A. 8 B. 6 C. 12 D. 10
15. Determine the resultant reaction (kN) at
%.
A. 12 B. 8.49 C. 9.36 D. 6
A
B
C
w
R

A
B
C
w = 3 kN/m
4 m
4
m
Av
Ah
Bh
Bv
Σ„ 0
3 4 4/2 $ %8 4 0
%8 6 kN
Σs8 0
ƒ8 %8 6 kN
6
6

of the pin at ƒ.
A. 10 B. 12 C. 8 D. 6
of the pin at M.
A. 8 B. 6 C. 12 D. 10
%.
A. 12 B. 8.49 C. 9.36 D. 6
A
B C
w = 3 kN/m
4 m
Av
Ah
Bh
Bv
6
6
Cv
Ch
C
Cv
Ch
In beam BC:
3 4 4/2
Σ„ 0 $ M 4 0
M 6 kN
Σs8 0 M8 ƒ8 6 kN
6
6
ƒ
6
6
6
6
_„ 6+ 6+
_„ 8.48 kN
_| _q _„ 8.48 kN
Diameter of pin @ B (single shear):

_„
% eL
† 112 MPa
% eLdeL

8480
112
75.71 mm+
)
*
5 eL
+
75.71
5 eLdeL
9.82 mm
use 10 mm
Q13 A 40.17 21.88 26.23 11.13
Q15 B 17.23 39.79 17.23 25.07

of the pin at M.
A. 8 B. 6 C. 12 D. 10
%.
A. 12 B. 8.49 C. 9.36 D. 6
A
B C
w = 3 kN/m
4 m
Av
Ah
Bh
Bv
6
6
Cv
Ch
C
Cv
Ch
6
6
6
6
6
6
_„ 6+ 6+
_„ 8.48 kN
_| _q _„ 8.48 kN
Diameter of pin @ C (double shear):

_
2% eL
† 112 MPa
% eLdeL

8480
112 2
37.9 mm+
)
*
5 eL
+
37.9 5 eLdeL
6.9 mm use 8 mm
Q14 A 29.82 31.56 18.59 19.55

Situation 7 – A flag pole is supported by tension
wires %ƒ, %M and % to resist an uplift force
of 185 N acting on the axis of the pole. In
this problem, 3 3 m, 4 4 m and ℎ 12
m.
22. Calculate the tension (kN) in wire %ƒ.
A. 82.46 C. 47.43
B. 65.00 D. 58.42
23. Calculate the tension (kN) in wire %M.
A. 58.42 C. 65.00
B. 47.43 D. 82.46
24. Calculate the tension (kN) in wire %.
A. 47.43 C. 58.42
B. 82.46 D. 65.00
z

‡
%
ƒ

M
3
4
3
4

m.
A. 82.46 C. 47.43
B. 65.00 D. 58.42
A. 58.42 C. 65.00
B. 47.43 D. 82.46
A. 47.43 C. 58.42
B. 82.46 D. 65.00
z

‡
%
ƒ

M
3
4
3
4
12

m.
z

‡
%
ƒ

M
3
4
3
4
185 f
ˆ„
ˆq
ˆ‰
%ƒ
Using matrix:
3i 0Š 12
$ 12‹ %ƒ 153
%M $4i 3Š $ 12‹ %M 13
% 0i $ 4Š $ 12‹ % 160
3Œ%
?
=?

=?
Ž=+
=?
Ž*
=?
?
=?
Ž=+
=?

=,
Ž*
=,
Ž=+
=,
3Œƒ
0
0
185
$Trn Inv MatA MatB
ˆ„ 82.462 kN
ˆq 65 kN
ˆ‰ 47.43 kN

m.
A. 82.46 C. 47.43
B. 65.00 D. 58.42
A. 58.42 C. 65.00
B. 47.43 D. 82.46
A. 47.43 C. 58.42
B. 82.46 D. 65.00
z

‡
%
ƒ

M
3
4
3
4
185 f
ˆ„
ˆq
ˆ‰
12
$Trn Inv MatA MatB
ˆ„ 82.462 kN
ˆq 65 kN
ˆq 47.43 kN
Q22 A 30.69 25.27 26.72 16.46
Q23 C 23.62 24.88 33.01 17.62
Q24 A 34.17 19.75 24.98 20.33

Situation 9 – Refer to the Figure shown.
Given:
ℎ= 125 mm; ℎ+ 475 mm; 60 mm
Clear concrete cover = 40 mm
Tension bars, %H 8 – 28 mm diameter
Compression bars, %H
9
4 – 28 mm diameter
Stirrups = 12 mm diameter
Concrete 28-day compressive strength = 28 MPa
Steel strength, (8 275 MPa
Reduction factor for shear = 0.75
28. What is minimum beam width “4” (mm) based on
spacing and concrete cover according to NSCP.
A. 300 B. 320 C. 280 D. 340
29. Given: Factored shear, Y
W 480 kN
Spacing of 4 legs of stirrups = 160 mm
How much is the minimum beam width “4” (mm)?
A. 350 B. 600 C. 550 D. 500
W 480 kN
A. 400 B. 450 C. 375 D. 425
h1
h2
c
b

Given:
ℎ= 125 mm; ℎ+ 475 mm; 60 mm
Clear concrete cover = 40 mm
Tension bars, %H 8 – 28 mm diameter
Compression bars, %H
9
4 – 28 mm diameter
28. What is minimum beam width “4” (mm) based on
spacing and concrete cover according to NSCP.
A. 300 B. 320 C. 280 D. 340
125
475
60
bw
40
40
40
40
x
largest of 5#, 25 mm and *
?5A
28 mm
4[ œ
40 2 12 2
←12‡
5# 28 mm
28 4 28 3
4[ œ
Q]] FF
Q28 A 45.5 23.52 18.97 11.23

W 480 kN
A. 350 B. 600 C. 550 D. 500
125
475
60
bw
40
40
40
40
←12‡
d
5 125 475 $ 40 12 28/2 60/2
5 504 mm
% 4 )
* 12 + 452.4 mm+
Y
H
% (5
I

452.4 275 504
160
391.89 kN
Y
L Y
W/N 480/0.75 640 kN
Y
Y
L $ Y
H 640 $ 391.89 248.11 kN
Y
0.17
9
4[5
4[deL

ž 248.11 kN
248,110
0.17 28 504
4[deL
/EP. Q FF
Q29 C 19.36 15.68 46.37 17.81

W 480 kN
A. 400 B. 450 C. 375 D. 425
125
475
60
bw
40
40
40
40
←12‡
d
5 504 mm
% 2 )
*
12 + 226.2 mm+
Y
H
% (5
I

226.2 275 504
70
447.88 kN
Y
L Y
W/N 480/0.75 640 kN
Y
Y
L $ Y
H 640 $ 447.88 192.12 kN
Y
0.17
9
4[5
4[deL

ž 192.12 kN
192,120
0.17 28 504
4[deL
ECE FF
Q30 D 13.94 15.3 27.98 41.63

Situation 1 – Refer to the truss shown:
Given: I = 3 m; ℎ = 6 m; = 2 kN
4. Determine the axial force (kN) in
member '.
A. 4 T C. 6 T
B. 4 C D. 6 C
member lŸ.
A. 3.35 T C. 7.83 C
B. 3.35 C D. 7.83 T
member 'Ÿ.
A. 12.3 C C. 1.12 T
B. 12.3 T D. 1.12 C
P P P P P
s s s s s s
h
A
B C D E F
G
H I J

Given: I = 3 m; ℎ = 6 m; = 2 kN
member '.
A. 4 T C. 6 T
B. 4 C D. 6 C
member lŸ.
A. 3.35 T C. 7.83 C
B. 3.35 C D. 7.83 T
member 'Ÿ.
A. 12.3 C C. 1.12 T
B. 12.3 T D. 1.12 C
P P P P P
s s s s s s
h
A
B C D E F
G
H I J
R=2.5P R=2.5P

Given: I = 3 m; ℎ = 6 m; = 2 kN
member '.
A. 4 T C. 6 T
B. 4 C D. 6 C
member lŸ.
A. 3.35 T C. 7.83 C
B. 3.35 C D. 7.83 T
member 'Ÿ.
A. 12.3 C C. 1.12 T
B. 12.3 T D. 1.12 C
P=2
s s=3 s
h=6
D E F
G
I J
R=2.5P=5
FIJ = FHI
FJE
FEF
k
k
tan k 6/3 k 63.43°
Σs 0
st} sin 63.43° 2 $ 5 0
st} Q. Q/ RS T
Σ} 0
s¡t 6 2 3 $ 5 6 0
s¡t E RS C s¡¢
Q4 B 13.07 60.41 13.17 13.17
Q5 A 55.37 19.46 11.71 13.26

Given: I = 3 m; ℎ = 6 m; = 2 kN
member 'Ÿ.
A. 12.3 C C. 1.12 T
B. 12.3 T D. 1.12 C
P P P P P=2
s s s s s s
h
A
B C D E F
G
H I J
R=2.5P R=5
2 2
s s=3
6
D E F
G
I J
R=5
k
FIE
Σs 0
s¡} sin 63.43°
k 63.43°
k
2 2 $ 5 0
s¡} .. ..D RS C
Q6 D 10.45 16.46 27.59 45.11

Situation 5 - A man weighing 1.2 kN walks
through a log of wood 250 mm in
diameter and simply supported at a
length of 3 m.
16. What is the maximum shear (kN)?
A. 0.6 B. 0.8 C. 1.2 D. 1.8
17. What is the maximum moment (kN-m)?
A. 1.5 B. 0.8 C. 1.2 D. 0.9
18. If the log is replaced with 300 mm wide
by 100 mm thick section of wood, what
is the maximum weight of a man can
walk through the log if the allowable
flexural stress, s# 4.2 MPa?
A. 4.1 B. 2.4 C. 2.8 D. 3.2
P = 1.2 kN
L = 3
D=250
A B
Maximum shear

length of 3 m.
A. 0.6 B. 0.8 C. 1.2 D. 1.8
A. 1.5 B. 0.8 C. 1.2 D. 0.9
A. 4.1 B. 2.4 C. 2.8 D. 3.2
P = 1.2 kN
L = 3
D=250
A B
Maximum shear
Y
dAw .. C RS
Maximum moment:
L/2 L/2
dAw

4

]. O RS F
Q16 C 28.75 6.97 50.24 13.75
Q17 D 10.45 9.78 14.62 64.76

length of 3 m.
A. 0.6 B. 0.8 C. 1.2 D. 1.8
A. 1.5 B. 0.8 C. 1.2 D. 0.9
A. 4.1 B. 2.4 C. 2.8 D. 3.2
L = 3
D=250
A B
Maximum shear
Y
dAw .. C RS
Maximum moment:
P
L/2 L/2
dAw

4

]. O RS F
Part 3:
4 300
5 100
#
6
45+ † 4.2
dAw
4.2 300 100 +
6
2,100,000 N mm
dAw 2.1 kN m
de7

4
† 2.1 kN m
dAw
4 2.1
3
C. D RS
Q18 C 14.71 21.3 44.05 19.07

Situation 10 - Refer to the figure
shown.
The stresses in a bar subjected to
uniaxial stress is plotted as shown.
Given, 3 40 MPa.
31. Determine the maximum normal
stress (MPa).
A. 50 C. 80
B. 60 D. 40
32. Determine the maximum shear
stress (MPa).
A. 40 C. 60
B. 20 D. 80
33. What angle does the maximum
shear stress occur from the origin.
A. 90° C. 60°
B. 30° D. 45°
£
¤
a a

z

z
40
40
Maximum normal stress:
£dAw 80 MPa
Maximum shear stress:
¤dAw
¤dAw 40 MPa
£dAw
90°
45°
z
z
Q31 C 2.03 3.39 73.86 20.23
Q32 A 77.06 3.1 6.49 12.97
Q33 D 33.98 3.48 4.74 57.5

Ref F2F Week 4 - Solution_unlocked.pdf

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Ref F2F Week 4 - Solution_unlocked.pdf