MYR (RM)
RM
Malaysia Ringgit
$
United States Dollar
$
Brunei Darussalam Dollar

CONTOH QUESTION

Lesson 4/4 | Study Time: 10 Min

Question 1 (Topic: Trusses – Method of Joints)

Diagram: A statically determinate truss with vertical loads at points C and E.

Question:

The figure below shows a statically determinate truss supported by a pinned support at point A and a roller support at point D. A vertical load of 20 kN acts at point C and 30 kN at point E.

i. Determine the support reactions at A and D.

ii. Using the method of joints, determine the forces in members AB, BC, and CD.


Answer Summary:

(i) Use equilibrium equations:






MA=0:RD(6)30(4)20(2)=0RD=26.67kN\sum M_A = 0: R_D(6) - 30(4) - 20(2) = 0 \Rightarrow R_D = 26.67\,kN
Fy=0:RA+RD=50RA=23.33kN\sum F_y = 0: R_A + R_D = 50 \Rightarrow R_A = 23.33\,kN

(ii) Use joints C and B to solve for member forces:



  • Member AB = Tension 15.56 kN




  • Member BC = Compression 17.78 kN




  • Member CD = Compression 20.89 kN




Question 2 (Topic: Cable Structures)

Diagram: A cable supported at A and D with vertical loads at B (25 kN) and C (50 kN).

Question:

Neglecting the self-weight of the cable, solve the following:

i. Determine the support reactions at A and D.

ii. Calculate the tension in segments AC and CD of the cable.


Answer Summary:

(i) Use moment and force equilibrium:









MA=0:RD(7)=25(2)+50(4.5)=325RD=46.43kN\sum M_A = 0: R_D(7) = 25(2) + 50(4.5) = 325 \Rightarrow R_D = 46.43\,kN
Fy=0:RA=7546.43=28.57kN\sum F_y = 0: R_A = 75 - 46.43 = 28.57\,kN

(ii) Use geometry and vector components (assuming geometry is available):



  • Tension in AC32.14 kN




  • Tension in CD53.57 kN




Question 3 (Topic: Influence Lines)

Diagram: A simply supported beam AB, 12 m long, with point C located 4.5 m from A.

Question:

The figure shows a simply supported beam subjected to a moving unit load.

i. Draw the influence line diagram for reactions at supports A and B.

ii. Draw the influence line diagrams for bending moment and shear force at point C.


Answer Summary:

(i) Influence line for support reactions:




  • Reaction at A: Linearly decreases from 1 (at A) to 0 (at B).




  • Reaction at B: Linearly increases from 0 (at A) to 1 (at B).



(ii) At Point C (4.5 m from A):












  • Bending Moment Influence Line:




    • Triangle shape peaking at C.




    • Maximum value when load is directly at point C.






  • Shear Force Influence Line:




    • Jump at C:




      • Left of C = +1




      • Right of C = –1