ENGINEERING
SOLVED PROBLEMS
Structural Engineering and
Construction
JOHN MARK A. GUIMBA
Civil Engineer
Engineer, Department of Public Works and Highways
BSCE (2019), University of the Philippines Diliman -
Magna Cum Laude
5th Place, November 2019 CE Board Exam
Champion, Sy^2 + Associates Structural Engineering Quiz,
2019
3rd Place, National MTAP Finals Individual Category, 2014
, Solved Problems in Structural Engineering and Construction
PART I
ENGINEERING MECHANICS
PROBLEM MECH-1. For the hanging bridge shown, weight 𝜔𝜔 = 900 N/m, s1 = 15 m, s2 = 25 m,
and h1 = 3m.
1. What is the weight of the cable?
2. What is the value of h2?
3. What is the maximum tension in the cable in kN?
4. What is the value of x2 in m?
Solution:
𝑘𝑘𝑘𝑘
𝑊𝑊𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡 = 0.9 ∙ (15 𝑚𝑚 + 25𝑚𝑚)
𝑚𝑚
𝑊𝑊𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡 = 36 kN Ans. #1
Recall: For catenary curves, we set a coordinate system in 2D plane such that the lowest point C
is at an elevation 𝑎𝑎 from the origin. The curve is then defined by the equation:
𝑦𝑦 2 = 𝑠𝑠 2 + 𝑎𝑎2
𝑥𝑥 𝑥𝑥
where 𝑦𝑦 = 𝑎𝑎 cosh and 𝑠𝑠 = 𝑎𝑎 sinh .
𝑎𝑎 𝑎𝑎
At point A, 𝑦𝑦 = 𝑎𝑎 + ℎ1 = 𝑎𝑎 + 3 and 𝑠𝑠1 = 15. Then,
(𝑎𝑎 + 3)2 = 152 + 𝑎𝑎2
6𝑎𝑎 = 216
𝑎𝑎 = 36
Now, at point B, 𝑦𝑦 = 𝑎𝑎 + ℎ2 = 36 + ℎ2 and 𝑠𝑠2 = 25. Then,
(ℎ2 + 36)2 = 252 + 362
ℎ2 + 36 = 43.829
ℎ2 = 7.829 m Ans. #2
Now, 𝑇𝑇𝑚𝑚𝑚𝑚𝑚𝑚 occurs at support B. Hence,
Problems Solved by Engr. John Mark A. Guimba, CE |1
, Solved Problems in Structural Engineering and Construction
𝑇𝑇𝑚𝑚𝑚𝑚𝑚𝑚 = 𝜔𝜔𝑦𝑦𝑚𝑚𝑚𝑚𝑚𝑚
𝑇𝑇𝑚𝑚𝑚𝑚𝑚𝑚 = 0.9(43.829)
𝑇𝑇𝑚𝑚𝑚𝑚𝑚𝑚 = 39.446 kN Ans. #3
Finally,
𝑥𝑥2
𝑦𝑦2 = 𝑎𝑎 cosh
𝑎𝑎
𝑥𝑥2
43.829 = 36 cosh
36
43.829
𝑥𝑥2 = 36 cosh−1
36
𝑥𝑥2 = 23.332 m Ans. #4
PROBLEM MECH-2. A precast concrete beam of length L is to be lifted from the casting bed and
transported so that the maximum bending moment is as small as possible. The beam is lifted by
two strings placed symmetrically.
1. How far from each end should the strings be attached?
2. What is the maximum moment if the beam weighs 𝑤𝑤 (N/m)?
Solution:
By conducting a static analysis and drawing the bending moment diagram,
The magnitudes of the maximum positive and maximum negative moments must be equal. Hence,
𝐿𝐿2 𝑥𝑥𝑥𝑥 𝑤𝑤𝑥𝑥 2
𝑤𝑤 � − �=
8 2 2
Problems Solved by Engr. John Mark A. Guimba, CE |2