ARC - 4109

CE 423
STEEL
&
TIMBE
R
DESIG
N
Tuesday I 10:00 AM - 1:00 PM

Engr. Dativa D. Hernandez

Instructor
ADIGUE, KIMBERLIE V.

Problem 1.
A 40-mm diameter bar is used as a simply supported beam 2 m long. Determine the largest
uniformly distributed load which can be applied over the right half of the beam if the flexural
stress is limited to 60 MN /m².

∑𝑀 = 0
Solution :
𝑐
2𝑅 = 𝑊 (1)(0. 5)
1
𝑅 = 0. 25 𝑊 𝑁
1
𝑅 = 𝑊 (1) − 0. 25 𝑊
2
𝑅 = 0.75W
2

∑𝑀 = 0
𝐵
WX = 0.25 W
X= 0.25 m

𝑀 = 0.25W (1) = 0.25 W N. m

𝐵
𝑀 = 0.25 W + 1
(0. 25 𝑤)(0. 25)
𝐷 2
𝑀 = 0.25125W N. m
𝐷

𝑀
𝑆
f =
𝑀 𝑀
f = 𝝅
3
=π 𝑑3
𝑑
4

32(0.28125 𝑊)(1000)
π (40)
60 = 3

W = 1340 N/m
ALVAREZ, JANELLA M.

Problem 2.
A beam 10 m long simply supported 1 m from each end. It is made of two C 380 x 50
channels riveted back to back and used with the webs vertical. Determine the total uniformly
distributed load that can be carried along its entire length without exceeding a flexural stress of
120 MPs.

𝑅 = 𝑅= 5 𝑊
Solution:
1 2
𝑀 = 1
(1)(− 𝑊) =− 0. 5 𝑊 𝑁. 𝑚
𝐵 2
𝑀 = − 0. 5 𝑊 + 1
(4𝑊)(4)
𝐸 2
𝑀 = 7. 5 𝑊 𝑁. 𝑚
𝐸
𝑀 = − 0. 5𝑊
𝐶
𝑚𝑎𝑥 𝑀 = 7. 5 𝑊
1

𝑓 =
𝑀
�
�

𝑓𝑟𝑜𝑚 𝑡𝑎𝑏𝑙𝑒,
𝑓𝑜𝑟 𝑜𝑛𝑒 (380 𝑥 50,
3 3
𝑆 = 687 𝑥 10 𝑚𝑚
𝑋
120 = (7. 5 𝑊)(1000)
3
2 (687 𝑥 10 )
W = 21,948 N/m

Conclusion:
The calculation for the 21,948 N/m load is based on the maximum allowable flexural stress of
the beam. By analyzing the beam, which consists of two C 380 x 50 channels and is simply

modulus 𝑆
supported, the support reactions and maximum bending moment are determined. The section
3 3 𝑀
𝑋 = 687 𝑥 10 𝑚𝑚 for each channel is used in the flexural stress formula 𝑓 � =
where 𝑓 is the allowable stress (120 MPa), 𝑀 is the maximum moment, and 𝑆 is the section �

modulus. Solving this equation gives the maximum uniformly distributed load the beam can
support, which is 21,948 N/m, ensuring the stress remains within the safe limit of 120 MPa.
BARCENAS, ANNA LOREINE H.
PROBLEM 3.

A wooden beam 150mm wide by 300 mm deep is loaded as shown . If the maximum flexural
stress is 8MN/ 2, find the maximum values of W and P that can be applied simultaneously.
𝑚

Solution:

∑𝑀 = 0
Equilibrium of Moments Point A:
𝐴
6𝑅 = 3𝑃 + 𝑊 (2)(7)
2

𝑅3𝑃=+14𝑊 62
𝑅 =P+2W−𝑅
1 2
𝑅 =P+2W−6 3𝑃 +14𝑊

1 6

𝑅1 =
3𝑃 +14𝑊
6

Bending Moment at Point B:

𝑀𝐵 =
3𝑃 +14𝑊 3𝑃 +2𝑊
6 (3)= 6 N.m

𝑀 = 1 (2W)(2)=−2WN.m
Bending Moment at Point C:
𝐶 2

Flexural Stress Calculation:

6𝑀2
f= 𝑏ℎ

6 (2𝑊)(1000)
2
8= 150 (300)

W = 9000N/m

Solving for P:

6 (3𝑃 − 2𝑊)(1/2)(1000)
2
8= 1500 (300)
3P − 2W=36,000
3P = 36,000+2(9000)

P = 18,000N
BOQUIREN, JOSHUA FRANCISCO P.

Problem 4.
A rectangular beam, 120mm wide by 400mm deep, is loaded as shown. If w = 3KN/m, find P to
cause a maximum bending stress of 10MPa.

Solution:

∑𝑀 = 0
𝐴
5𝑅 = 3(5) (2.5) + 4 P
2
𝑅 = 7.5 + 0.8
2
𝑅
1
𝑅
= 3(5) + P -(7.5 + 0.8)

1
= 7.5 + 0.2

7.5 + 0.2 = 3X
7.5 + 0.2𝑃
3
𝑀 =
X =
1 7.5 + 0.2𝑃

𝐷 2 3
(7.5 + 0.2P)
2

𝑀𝐷 =
(7.5 + 0.2𝑃)
6 KN.m

𝑀 =
� 1
2
𝑀 = (6 + 0.8P) KN.m
�
(1)(4.5 + 0.8P + 7.5 + 0.8P)

𝐵
F = 6M / bh² 2
6 (7.5 + 0.8𝑃) (1/6) (1000)²
(120)(400)²
10 =

(7.5 + 0.2P)2 = 192

7.5 + 0.2 = 13.856

P = 31.8 KN
CATAPANG, AEDYN KEITH T.

Problem 5.

Three planks 100 mm by 150 mm, arranged as shown and supported by bolts 0.4 m apart,
are used to support a concentrated load P at the center of a simply supported span 6m long. If P
causes a maximum flexural stress of 12 MPa, determine the bolt diameters, assuming the shear
between the planks is transmitted by friction only. The bolts are tightened to a tension of 1.40
MPa, and the coefficient of friction between the plans is 0.40.

Solution:

max V = P/2 KN

max M = 1.5 P KN. m

f=6M/bh²

12 = 6 (1.5P) (1000)² / 150 (300)²

P = 18 KN

max V= P/ 2

=9 KN

Frictional Resistance

F = Ve Ｑ

Ｑ = 100 (150)(100) = 1500 x 10³ mm³

I = 150(300)³ = 337.5 x 10⁶ mm⁴

12

Calculate the Frictional Force: The bolts provide pre-tension that generates a normal force
between the planks.
F = (9000) (400) (1500 x 10³)

337.5 x 10⁶

F = uN

16,000 = 0.40 (N)

N = 40,000 N

Determine Bolt Diameter: Using the shear force, the shear stress τ\tauτ on each bolt can be
related to the bolt area:

S=P

140 = 40,000

d² 4

d = 19.07mm
CLOSA, JOSEF MARL M.

Problem 6.
A timber beam with a square cross section supports the load shown. Determine the cross
sectional dimension if the allowable flexural stress is 10MPa. What is the maximum shearing
stress developed in the beam?

Solution: 3 2
𝑏ℎ 𝑏ℎ
I= 12 iS= 6
But b = h
3
ℎ
S=
6
f = 10 MPa
2 2

−𝑊𝐿 𝑃𝐿 𝑃𝑎 𝑏
a.) Max. M = M
-8
12 𝐿
A
= -2
2 2

𝑤𝐿 𝑃𝐿 𝑃𝑎 𝑏
12
-8
𝐿
MA = - - 2

3 2 3 3 2

−30 𝑥 30 (4) 4𝑥10 (4) 2𝑥10 (3) (1)

12 8 (4)
2
MA = - -
MA = - 7125 N - m

● Formula for flexural stress is:

f= 𝑀
𝑆
𝑀 6𝑀
f= ℎ
3
; h3 = 𝑓
6
3
ℎ
6
● Where S is the section modulus, and for a square cross-section, S =
● Given that the allowable flexural stress =10MPa = 10 x 10 N/m , we can solve for the 6 2

length of h:

7125 (6)

10 𝑥 10
h3 = 6
= 0.163 m
h = 163 mm

3𝑉
2
b.) Max. J =
𝐴 2 2

𝑤𝐿 𝑃𝐿 𝑃𝑎𝑏
MB = - 12 - 8 -
�
2
3 2
� 3 3 2

−3 𝑥 10 (4) 4𝑥10 (4) 2𝑥10 (3)(1)

- 8
12 (4)
2
MB = -
MB = - 6375 N - m
MB = - 6375 = -7125 - 2 x 103(3) - 4 x 103(2) - 3 x 103(4)
VA = 9687.5 N
 Jmax = 3(9687.5)
2
2(0.163)
Jmax = 546.92 kPa

Therefore, the cross-sectional dimensions for the support beam is 163 mm x 163 mm,
and the maximum shearing stress is 546.92 kPa.
DEL MUNO, IRRA D.

Problem 7.
In the overhanging beam shown, determine P so that the moment over each support equals the
moment at mid span.

Solution:
𝑅 = 𝑃 + 5(4) = 𝑃+ 20
1

Where:
𝑀 =𝑀
𝐵 𝐸

𝑀 = 𝑃(1) + 5(1)(0. 5) = 𝑃 + 2. 5
𝐵

Thus:
𝑀 = (𝑃 + 20)(3) − 𝑃(4) − 5(4)(2)
𝐸

𝑀 = 3𝑃 + 60 − 4𝑃 − 40
𝐸

𝑀 = 20 − 𝑃
𝐸

𝑃 + 2. 5 = 20 − 𝑃

P = 8.75 KN

Therefore, the value of the load P that ensures the moment at the supports equals the
moment at the mid-span is 𝑃 = 8. 75 𝑘𝑁
EXPLANATION:
1
1. Reaction at Support B (𝑅 )

● Use equilibrium conditions to solve for the reactions at the supports, 𝑅 and 𝑅 . We
1 2

assume that the beam is symmetrically loaded, and the uniform load spans the section
between the supports (𝑅 = 𝑅 ).
1 2

● From equilibrium (sum of vertical forces), we calculate the reaction at support B, R1:

𝑅 = 𝑃 + 5 (4)
1

The term 5×4 represents the total force due to the uniform load of 5 kN/m, acting
over a distance of 4 m (the length between B and C).

So, 𝑅 = 𝑃 + 20
1

2. Moment at Support B (𝑀 )
𝐵

● Calculate the moment at support B using the formula for moments:

𝑀 = 𝑃(1) + 5 (1)(0. 5)
𝐵

This represents the moment at B due to the point load P at A and the uniform load
over a 1 meter distance between A and B.

● P(1) is the moment due to the point load P at a distance of 1 m B.

● 5(1)(0.5) is the moment due to the uniform load of 5 kN/m over a 1-meter
distance between B and C, with its center of gravity at 0.5 m.

Simplifying: Moment at support B

𝑀 = 𝑃 + 2. 5
𝐵

𝐵
3. Moment at Mid-Span E (𝑀 )
 ● Calculate the moment at the mid-span E. Using the reaction at B (𝑅 = 𝑃 + 20), the
1

distance between B and E is 3 m, and the moment arm from P at point D is 4 m. The
moment at E is calculated as:

𝑀 = 𝑅 (3) − 𝑃(4) − 5(4)(2)

𝐸 1

Wherein:

● (R1)(3) is the moment due to the reaction force R1 at a distance of 3 m.

● P(4) is the moment due to the point load P at a distance of 4 m from the mid-span.
● 5(4)(2) is the moment due to the uniform load of 5 kN/m acting over 4 m, with its
center of gravity at 2 m.

Substitute 𝑅 = 𝑃 +
20 into the equation:
1

𝑀 = (𝑃 + 20) (3) − 𝑃(4) − 5(4)(2)

𝐸

Simplify the terms: Moment at the midspan E.

𝑀 = 3𝑃 + 60 − 4𝑃 − 40
𝐸

𝑀 = 20 − 𝑃
𝐸

4. Equating Moments at B and E

● The problem states that the moment at support B equals the moment at mid-span E:

𝑀 =𝑀
𝐵 𝐸

● Substitute the expressions for 𝑀 and 𝑀 :

𝐵
𝐸

𝑃 + 2. 5 = 20 − 𝑃

Solve for P:

𝑃 + 𝑃 = 20 − 2. 5
2𝑃 = 17. 5

𝑃 = 17.5
2
= 8. 75 𝑘𝑁

Therefore, the value of the load P that ensures the moment at the supports equals the
moment at the mid-span is 𝑃 = 8. 75 𝑘𝑁
DUMAS, JONALYN O.

Problem 11.
Determine the minimum width b of the beam shown if the flexural stress is not to exceed 10
MPa.

∑𝑀 = 0
𝐷
3𝑅 = 500(1) + 2000(4)(2)
1
𝑅 = 7000𝑁
1
𝑅 = 5000 + 2000(4) − 7000
2
𝑅 = 6000𝑁
2
𝑀 = 1
(1)(1 − 2000) = − 1000𝑁. 𝑚
𝐵 2
𝑀 =− 1000 + 2
(5000 + 1000
𝐶 2
𝑀 = 5000𝑁. 𝑚
𝐶
𝑚𝑎𝑥. 𝑀 = 500𝑂𝑁. 𝑚

𝑓 = 𝑏ℎ2
6𝑀

10 = (𝑏)(200)2
6(5000)(1000)

𝑏 = 75𝑚𝑚

Therefore, the minimum width b of beam is 75mm if the flexural stress is not to exceed to
10 MPa.
ESCAREZ, HARVIE C.

Problem 12.
Draw the shear and moment diagram

● Summing moments about point A to solve for the

reaction at D:

∑MA = 0
6R2 = 10(60)(5)

This is a moment equation where:

● R2 is the reaction at D.
● 10(6)(5) is the total moment caused by the UDL,
considering it acts at its centroid (which is at 3
M from B).

Solving this:

R2 = 50 KN

Reaction at A:

R1 = 6(10) - 50
R1 = 10 KN

VAB = 10
MAB = 10X
VBC = 10 - 10 (X - 2)
MBC = 10X - 10 (X - 2)( 𝑥 − 2
2
)
2
MBC = 10X - 5 (X - 2)
VCD = 10 (8 - X)

= 10(8 - X)( 8 − 𝑥
MCD
2
)
MCD = -5(8 - X)2

Therefore,the shear diagram shows constant shear in region AB, decreasing linearly in
region BC, and decreasing further in region CD; the moment diagram starts at zero at A,
increases linearly to B, then forms a quadratic curve between B and D, peaking near B.
GENETA, SEAN HARVEY F.

Problem 14.
A concentrated load of 0 KN is applied at the center of a simply supported, beam 8m long. Select the
lightest suitable W shape section using an allowable stress of 120 MN/m^2.

Solution.

max, 𝑀 = 4𝑃𝐿 (at midspan)

𝑀 =
(8)
90
4
= 180 kN.

𝑓 = �𝑀
�
2

120 =
180 (1000)
4

S = 1500 x 10
3 3

𝑚𝑚

From table,

for W 530 x 74, 𝑆 = 1550 x 10 𝑚𝑚

3 3

𝑥
weight of beam = (74.7 kg/m)(9.81) = 732.8 N/m
2
𝑊
max. M (due to weight of beam 𝐿 (at midspan)
8
𝑀 = 5.8624 kN.m
=

𝐵
𝑀 = 180 + 5.8624 = 185.8642 kN.m
𝑇
Actual stress, 2

𝑓 =
(185.8624) (1000)
3
1550 𝑥 10

𝑓 = 119. 9 𝑀𝑃𝑎 < 120 𝑀𝑃𝑎 (safe)

Use W 530 x 74.

Explanation.

Calculate the Maximum Moment (M):

1. The maximum moment for a simply supported beam with a center load is
𝑀 =4
𝑃𝐿

Substitute the value: M=

𝑀 =
(8)
90
4
= 180 kN.

(S): Formula: 𝑓� = 𝑀
2. Determine required section modulus

� 2
120 =
180 (1000)
4

S = 1500 x 10
3 3

𝑚𝑚

W 530 x 74, 𝑆
3. Select a W shape section:
3 3
𝑥 = 1550 x 10 𝑚𝑚

4. Calculate moment due to beam’s weight

weight of beam = (74.7 kg/m)(9.81) = 732.8 N/m
2
𝑊𝐿
8
𝑀 = 5.8624 kN.m
max. M (due to weight of beam = (at midspan)

𝑀 = 5.8624 kN.m
5. Total maximum moment and check the actual stress:
𝐵
𝑀 = 180 + 5.8624 = 185.8642 kN.m
𝑇
2

𝑓 =
(185.8624) 3
(1000)
1550 𝑥 10

𝑓 = 119. 9 𝑀𝑃𝑎 < 120 𝑀𝑃𝑎 (safe)

Use W 530 x 74.

HUBILLA, MISYLEE A.

Problem 15.
A 10-m beam simply supported at the ends carries a uniformly distributed load of 16 kN/m over
its entire length. What is the lightest W shape beam that will not exceed a flexural stress of 120
MPa? What is the actual maximum stress in the beam selected?

Solution:

𝑀
f =
𝑆
2

𝑀 =
𝑊𝐿
8 2

𝑀 =
(16)(10)
8
𝑀 = 200 𝐾𝑁.2 𝑚

120 =
(200)(1000)
�
�3
S= 1,667 x 10
3

𝑚𝑚
From table,
for WG10 x 82, Sx = 1870 x 10³ mm³
weight of the beam = (81.90 kg/m) (9.81) = 803N/m

𝑊 = 16 + 0. 803 = 16. 803 𝑘𝑁/𝑚

𝑇
2

𝑀 𝑊 𝐿
=
𝑇
8
𝑇 2

𝑀𝑇 =
(16.803)(10)
8 = 210.04 kN.m
𝑀
𝑆
f =
2

(210.04)(1000)
3
1870 𝑥 10
f =
f = 112.3 MPa (actual answer)
LAT, AXL SEBASTIAN B.

Problem 18

A cantilever beam of 1.50 m long is to support a uniform load of 4.39 kn/m and a
concentrated load of 1.78 kn at the end. Design the size of a circular beam using yacal with
flexural stress of 12.41 MPa and a unit shearing stress perpendicular to grain of 1.40 MPa.

Given:

· Length of the beam L= 1.50

· Uniform load w= 4.39 kN/m

· Concentrated load P= 1.78 kN

· Allowable flexural stress = 12.41 MPa

· Allowable shearing stress= 1.40 MPa

Solution:
2
4.39 (1.5) (100)
2
M = 1.78 (1.50)(100) +

M = 760.875 cm-KN

760.875
1.241
S.M. =

𝑙
�
S.M. =
�

𝑑
2
S.M. =

4
π𝑑 (2)
613.114 =
64 𝑑

3 32(613.114)
𝑑 = π

d = 18.415 cm
Conclusion:
The required diameter of the cantilever beam to satisfy both flexural and shear stress
limits is approximately 18.415 cm.

CHECKING FOR SHEAR STRESS:

4𝑉
v=
3𝐴

V = 1.78 + 4.39 (1.50)

V = 8.365

4(8.365)
0.14 = 3
𝐴

A = 80.0 𝑐
2

𝑚
2
π𝑑
4
=
80
d = 10.09 cm

d = 20 cm
LIMBO, ARABELA A.

Problem No. 19.

A wooden beam 150mm × 250mm × 3.6 meters is subjected to a uniform load through and its
entire length. How far from the end may the cross section be reduced by boring one vertical hole,
one inch in diameter without weakening the beam. Use rough dimensions.

𝑀 =
𝑤𝐿
8

𝑅 =
𝑤𝐿
2

=
3.60
𝑊 2
= 1. 8 𝑤 𝐾𝑁
2

𝑀 =
𝑤(3.6)
(100) 8
= 162 𝑤 𝑐𝑚 𝐾𝑁

𝑓 = 𝐵𝐷2
6𝑀

𝑓 =
6(1622𝑤)
15 (25)
2
𝑓 = 0. 10368 𝑤 𝐾𝑁/𝑐𝑚
x= distance from left support to the hole

𝑀𝑥 = 𝑅𝑥 − 𝑊𝑥
2
𝑥

𝑀𝑥 = 1. 8 𝑤𝑥
2

𝑀𝑥 = 1000(1. 8 𝑤𝑥 )𝑐𝑚 − 𝐾𝑁
2
𝑥 𝑤
2
2
𝑀𝑥 = 50 𝑤 (3. 6 − 𝑥 )𝑐𝑚 − 𝐾𝑁
𝑀𝑥 = 15𝑐𝑚 − 2. 5 = 12. 5

𝑓 =
6𝑀
2
𝑏𝑑
0. 10368𝑤
2
6(50𝑤)(3.6𝑥−𝑥 )

=
(12.5)(25)
2
1080𝑥 − 300𝑥 = 810
2
𝑥 − 3. 6 ++ 2. 72 = 0
−(−3.6)

𝑥 (−3.6𝑥) −4(1)(2.7)

=
−
2

𝑥 = 3. 266𝑚
By Shear:
3𝑣

=
2𝑏𝑑
f𝑣

𝑓𝑣 2 (15)(25)
3(1.8𝑤)

=
𝑓𝑣 = 25
1.8𝑤
0
𝑣 = 𝑅 − 𝑤𝑥
𝑣 = 1. 8𝑤 − 𝑤𝑥
𝑣 = 𝑤(1. 8 − 𝑥)
𝑣
3(𝑤)(1.8−𝑥)

=
2(15)(25)

=𝑣 =2 (12.5)(25)
1.8𝑤 3𝑤 (1.8−𝑥)
250
1125𝑤 = 1350𝑤 − 750𝑥
750x=225
x=0.3m from the left the support

Therefore, to calculate the shear force and left support reaction for a beam with a hole, first
determine the reactions and shear force distribution using static equilibrium equations based on
the beam’s loading and support conditions. The hole reduces the cross-sectional area, which
increases shear stress at that section, but the overall shear force remains the same as calculated
from the external loads.
LUISTRO, AIYHESA G.

Problem No. 20.

A wooden 6 m long is simply supported at the left end and another one at 1.20 m from the right
end the section of the beam is 75 mm by 200 mm. if the allowable stress for bending is 105 kg/cm³ and 12
kg/cm² for sheer, compute the maximum uniform unit load per meter (including its own weight) the beam
can carry

∑𝑀𝑅2 = 0
R1(4.8) = W(6)(1.8)

∑𝑀𝑅1 = 0
R1. = 2.25 W kg

R2(4.8) = 6W(3)
R2. = 3.75W kg

By Ratio & Proportion:

x/2.25 = (4.8 - x) / 2.55
2.55 x = 19.8 - 2.25x
4.8 x = 10.8
4.8 x. = 2.55

Max. Moment = 2.53 (kg.m)

Max. Shear. = 2.55 (kg)

Due to Bending

fb = MC/ I
I = bd^3 / 12 = (7.5)(20)^3 / 12
I = 5000 cm^4
c = 19 cm
M = 2.53(kg. m) x 100 cm / m
M = 253 kg.m
fb = 105 kg/m^2
105= 253W (10) / 5000
W= 207.5 kg/m
Due to Shear

V = 3V/ 2bd
12 = 3(2.55W) / 2(7.5)(20)
W = 470.58 kg/m

Safe Uniform load including its own

Wt = W = 207.5 kg/m

The problem involves a simply supported wooden beam subjected to a uniformly distributed load.
The maximum allowable load the beam can carry is determined by calculating the maximum bending and
shear stresses and comparing them to the given allowable limits. The maximum bending stress occurs at
the mid-span, while the maximum shear stress occurs at the supports. By analysing the stress distributions
and comparing them to the allowable values, the safe working load for the beam can be determined.
LUNA, LEAH CHENG T.

Problem 21.
A simply supported wooden beam 30mm in depth with a span of 7m carries a uniformly
distributed load of 7.6 kN/m including its own weight. Determine the width of the beam when
f=12.5MPc. Use nominal dimensions.

Given data:
 Depth of the beam, d = 300 mm
 Span of the beam, L = 7 m
 Uniformly distributed load, w = 7.6 kN/m
 Allowable bending stress, f = 12.5 MPa = 12.5 N/mm² = 12.5 x 106 N/m²

Required:
 Width of the beam (b) ?

Solution:
 Since the length and load is given, we can get the width of the beam by using the formula
of Moment:
M = wL2
8

 Substitute the given values:

M = 7.6 (7)2 KN.m
8

 The maximum bending moment for a simply supported beam with a uniformly
distributed load is:
M = 46.55 x 106 N.mm
  Since the value of Moment is found (M = 46.55 x 106 N.mm), and the depth of the beam
and the allowable bending stress is given, we can use the formula of the bending stress in
a rectangular section:
F = 6M
bd2

 Derive it to get the width of the

beam: b = 6M
fd2

 Substitute the given values:

b = 6 (46.55 x 106 N.mm)
12.5MPa (300mm)2

 The width of the beam:

b = 248.3mm, or let’s say 250mm for approximates

Thus, use 250 x 300 mm

MAMIS, DENVER GIL V.

Problem 22
Determine the safe concentrated load P at the center of the trapezoidal beam having a
simple span of 6.00 m if the allowable bending stress is 10.34 MPa. Neglect the weight of
beam.

Get the area of the rectangle. Then the areas of the two triangles. Then get the total area
of the trapezoid.

A1 = 7.5(12.5) = 93.75 cm2

A2 = 2(1/2)(11.5)(12.5) = 143.75 cm2

A = 237.50 cm2

Get the centroid of the rectangle and two triangles to get the Y1 and the Y2

Y1 = 12.5/2 = 6.25 cm

Y2 = 12.5/3 =4.17 cm

Ay = A1Y1 + A2Y2

237.50y = 93.75(6.25) + 143.75(4.17)

Y = 4.99 cm say 5.0 cm

Solve for I = I1 + I2

I1 = 7.5(12.5)3 /12+ 93.75(6.25 - 5)2 = 1367.20 cm4

I2 = 1/2(2)[(11.5)(12.5)3/36] + 143.75(5 - 4.17)2 = 722.94

cm4 I = 1367.20 + 722.94 = 2090.14cm4

From flexure formula, f = Mc/I

 10.34/10 = M(7.5)/2090.14

M = 288.16 KN.m capacity of section

Actual moment = PL/4 = P(6)/4 = 1.5P

KN.m Hence Beam: 1.5 P(100) + 288.16

P = 1.92 KN
MARQUEZ, MHACEE EUNICE

Problem 23.

A simply supported wooden beam carrying a uniform load has a span of 9.15 m. The beam has
adequate lateral supports. Allowable stresses are: for bending = 12.41 MPa, for deflection of
1/360 of span. E = 13.790 MPa. Determine the depth of the beam so that when the allowable
stress of 12.61 MPa is reached, the deflection of the beam is 1/360 of the span.
If the width of the beam is 300 mm, how much total uniform load can it safely support?

Solution:
Calculate moment due to uniform load, w.

𝑀 =
𝑤 ( 9.15)(100)
8

𝑀 = 114. 375 𝐾𝑁 − 𝑐𝑚

From:
𝑓 =
62𝑀
𝑏𝑑
2
𝑏𝑑
6𝑀

= 𝑓
6 (114.375 𝑤)
2
𝑏𝑑 1.241
=
2
𝑏𝑑 = 552. 98 𝑤 (1)

𝑦
Calculate allowable:
9.15 (100)

=
360
𝑦 = 2. 54 𝑐𝑚

𝑦 =
3
5𝑤𝐿
384 𝐸𝐼

𝐼=
3

𝑏𝑑
12

but y = 2.54; equating

2. 54 =
3
5𝑤𝐿
384 𝐸 1 𝑏𝑑
3

3 2
𝑏𝑑 3

=
5𝑤𝐿
12 384 𝐸 (2.54)
 3 3

=
𝑏𝑑 5 (𝑤)(9.15)(100)
12 384 (1379)(2.54)
3
𝑏𝑑
1
= 2847. 77 𝑤
2
bd³ = 34173.17 w

Divide
3
(2) by (1)
=
𝑏𝑑 34173.17 𝑤
552.98 𝑤
2
𝑏𝑑

d = 61.80 cm say 65 cm

b) Total Uniform Load

𝑓𝑏 =
62 𝑀
𝑏𝑑
2

𝑀 =
𝑓
𝑏𝑑 6
2

𝑀 =
1.241 (30)(61.80)
6
𝑀 = 23698. 38 𝐾𝑁 − 𝑐𝑚

𝑀 =
2
𝑤𝐿
8
2

23698
= 𝑤(9.15)
8
.38
100

w = 22.64 Kn/m
MONTEALTO, KAREN A.

Problem 24.
24. A cantilever beam 60 mm wide by 200 mm high and 6m long carries a load that varies
uniformly from zero at the free end to 1000 N/m at the wall.
a. Compute the magnitude and location of the maximum flexural stress
b. Determine the type and magnitude of the stress in fiber 40 mm from the top of the beam at a
section 3m from the free end.

Solution:

𝑀 = 2 1 (1000)(6)(2)
a. max, M is at fixed end

𝑀 = 6000 𝑁. 𝑚
𝑓 = 𝑏ℎ 2
6𝑀

𝑓 =
6(6000)(1000)
2
(60)(200)

max, f = 15 mPa (at fixed end)

𝑀 =2 1 (500)(3)(1) = 750 𝑁. 𝑚
b. At 3 m from the free end,

𝐼 =
𝑏ℎ
12
3

𝐼 =
60(200)
1
2 6 4
𝐼 = 40𝑋10 𝑚𝑚
𝑓 =
𝑀𝑦
�

𝑓 =
(750)(1000)(60)
�
6
40×10

f = 1.125 Mpa (tensile)

Therefore, at a section 3 m from the free end, the stress in the fiber 40 mm from the top is 1.125
MPa. This stress is proportional to the distance from the neutral axis and is less than the
maximum stress at the fixed end. To find the magnitude and location of maximum flexural stress
in a beam, determine the maximum bending moment, which occurs where the moment is greatest
due to applied loads, and then calculate the stress at that point by considering the distance from
the neutral axis to the outermost fiber of the beam.
ONCE, MARY AUDREY D.

Problem 26.
From the figure below. Draw the Shear and Moment Diagram
Show the Computation on how to get the maximum moment

Load Diagram:

Shear Diagram:

Moment Diagram:

TO SOLVE FOR MAXIMUM MOMENT

Step 1: Calculate the Reactions 𝑅 and 𝑅

1
2

Use symmetry to determine the reaction forces at bothe supports:

𝑅 = 𝑅 = 1/2 (𝑊)( 𝐿 ) = 𝑊𝐿

1 2 2 4

Step 2: Establish the Relation Between 𝑦 and 𝑥

The load intensity 𝑦 varies linearly with 𝑥 based on the triangular load
distribution:

=
𝑦 𝑊
𝐿

𝑥 2
Solve for 𝑦:

𝑦 = �
2𝑊𝑋 �

Step 3: Compute the Shear Force 𝑉

The shear force at any point along the beam is given by:
 𝑉 =𝑅
− 𝑋
1 1
2 𝑦

Substituting 𝑦 = 2𝑊𝑋
�
into the equation:
�

𝑉 = −2 𝑋𝐿( )
𝑊𝐿 1 2𝑊𝑋
4

Simplify the expression:

𝑉 = −
2
𝑊𝐿 𝑊𝑋
4 𝐿

Step 4: Compute the Bending Moment 𝑀

The bending moment at any point along the beam is calculated as:

𝑀 =𝑅 𝑋 −
𝑋
𝑦( 3 )
1 𝑋
1 2
2𝑊
Substituting 𝑦 = :
𝑋

�
�
𝑀 = −𝑋 )
2
(
𝑊𝐿𝑋 1 2𝑊𝑋
4 6 𝐿
Simplify the expression:

𝑀 = −
3
𝑊𝐿𝑋 𝑊𝑋
4 3𝐿

The maximum moment occurs at the midpoint 𝑥 = . Substitute 𝑥 =

Step 5: Find the Maximum Moment
𝐿 𝐿
2 2
into the
moment equation:

𝑊 3 𝐿
𝑀 = ( ) −
𝑊𝐿 𝐿
4 2 ( )
3𝐿 2
Simplify:2
2

𝑀 = −24
𝑊𝐿 𝑊𝐿
8
Final result for the maximum moment:
2

𝑀
𝑊𝐿 = 12

Therefore the maximum moment of the figure is 𝑀 = 1𝑊𝐿

2

2
ORILLA, NEIL BRYAN S.
PROBLEM 27.

From the figure below:

a. Draw the Shear and Moment Diagram

b. Show the computation on how to get the maximum moment

ΣMa = 0

R (L) = ( 𝑊𝐿 ) ( 2 L)
2 2 3

𝑊𝐿
2
R2 =

𝑊𝐿
R1 = 2 𝑊
𝐿
- 3

𝑊𝐿
6
R1 =

𝑊
𝐿
y=

𝑊𝑋
𝐿
y=

− 1
v = R1
2
xy

𝑊𝐿 1 𝑊𝑋
v= 6 -2 X (𝐿 )

2
𝑊𝐿 𝑊𝑋
v= 6 - 2𝐿

1
M=R X- xy ( 𝑋 )
1 2 3

𝑊𝐿𝑋 1 2 𝑊𝑋
M= 6 -6 x 𝐿
( )

−
𝑊𝑋
M= 6 6𝐿
𝑊𝐿𝑋

for max. M, V = 0
 3
𝑊𝐿 𝑊𝑋
0= 6 - 2𝐿

2
𝐿 𝑋
0= 3 - 𝐿

𝐿
√3
x=

2
𝑊𝐿 𝐿 𝑊 𝐿 3
6 ( √3 ) - 6𝐿 (√3 )
M=

2 2
𝑊𝐿 𝑊𝐿
M= 6√3 - 18√3

2
𝑊𝐿
9√3
M=
PUNTO, JULIANE PATRICE A.

Problem 28

A cast-iron beam 10 m long and supported as shown in Fig. P-557 carries a uniformly distributed
load of intensity wo (including its own weight). The allowable stresses are fbt ≤ 20 MPa and fbc ≤
80 MPa. Determine the maximum safe value of wo if x = 1.0 m.

By symmetry:
R1 = R2 = 5 W KN
Mb = Wx² / 2 KN m

Me = 5W(5-x) - 5W(2.5)
Me = 5W(2.5-x) KN m

At Section B:
ft = My (t) / I

20 = (Wx ²/²) (1000)² (180) / 36 * 10⁶

(1) Wx² = 8 ; x = 1m
W = 8 KN/m
fc = My (c) / I
80 = (Wx ²/²) (1000)² (50) / 36 * 10⁶
Wx² = 115.2 ; x = 1m
W = 115.2 KN/m
Tension Controls
At Section E :
ft = My (t) / I

20 = 5W (2.5-x) (1000)² (180) / 36 * 10⁶

(2) 5W(2.5-x) = 14.4 ; x = 1m
5W (2.5-1) = 14.4
W = 1.92 KN/m

80 = 5W (2.5-x) (1000)² (180) / 36 * 10⁶

5W(2.5-x) = 16
5W (2.5-1) = 16
W = 2.13 KN/m

For safe load wo, use W = 1.92 KN/m

The moments at various places along the beam are obtained using the formula Fb=My/I. These
moments aid in identifying the associated distributed load values that generate the fiber stresses
at the top and bottom of the beam in both compression and tension. The safe load is ultimately
identified as the lowest value of that guarantees the beam stays within acceptable stress ranges.
KYLA D. SANMOCTE

Problem 29.

Timbers 300x300 mm, spaced 0.90m apart on centers, are driven into the ground and act as
cantilever beams to back up the sheet piling of a cofferdam. What is the maximum safe height of
water behind the dam if the density of water is 1000 kg/m 3 and the bending stress is limited to
MN/m2?

Solution:

P = WhA

P = 1000 (9.81) (ℎ)(0.9h)

2

P = 4414.5 h2 N

M = P (ℎ)
3

M = 4414.5 h2 (ℎ) N.m

2

6𝑀
f=( )
𝑏𝑑⬚
2

6 (1471.5 ℎ )(1000)
3
8= 300/300

h = 2.90 m
SULIT, JOSE EMMANUEL M.

Problem 30.

A beam simply supported on a 12 m span carries a uniformly distributed load of 30 KN/m. Over
the middle 6m. Using an allowable stress of 140 MPa. determine the lightest suitable W shape
beam. What is the actual maximum stress in the selected beam?

Solution:

R1 = R2 = 30 (3) = 90 KN

MB = 90 (3) = 270 KN . m

ME = 270 + ½ (90)(3)

ME = 405 KN . m

𝑓 =
𝑀
�
�

2
405 (1000)
140 = 𝑆

S = 2893 x 103 mm3

From table,
For W 610 x 125, Sx = 3220 x 102 mm2
WB = (125)(9.81) = 1226 n/m = 1.226 KN/m
For uniform load on entire span
2
𝑊𝐿
Max, M = ( at
midspan) 8

2
(1226)(12)
8
MB = = 22.068 KN .
m
MT = 405 + 22.068 = 427.068 KN . m

Actual Stress

2
427.068 (1000)

3220 𝑥 10
f= 3
= 132.6 MPa < 140 MPa