For the truss below, identify all zero force members, and solve for the forces in members BC, CH, and GH using the method of sections, and state whether these forces are either tension or compression (T/C).
The Correct Answer and Explanation is :
To analyze the given truss, we will first identify all zero-force members and then determine the forces in members BC, CH, and GH using the method of sections, specifying whether these forces are in tension (T) or compression (C).
Identifying Zero-Force Members:
Zero-force members in a truss can be identified using the following rules:
- Two Non-Collinear Members: If two non-collinear members meet at a joint with no external load or support reaction, both are zero-force members.
- Three Members with Two Collinear: If three members form a joint where two are collinear and there is no external load or support reaction, the third non-collinear member is a zero-force member.
Applying these rules to the truss:
- At joint D, members CD and DE are non-collinear and meet without any external load or support reaction. Thus, CD and DE are zero-force members.
- At joint E (after removing DE), members BE and EF are non-collinear and meet without any external load or support reaction. Thus, BE and EF are zero-force members.
- At joint F (after removing EF), members CF and FG are non-collinear and meet without any external load or support reaction. Thus, CF and FG are zero-force members.
Therefore, the zero-force members are CD, DE, BE, EF, CF, and FG.
Method of Sections:
To find the forces in members BC, CH, and GH, we will use the method of sections. This involves cutting through the truss to isolate a section and applying equilibrium equations to solve for the unknown forces.
- Cutting the Truss: Make an imaginary cut through members BC, CH, and GH, dividing the truss into two sections.
- Analyzing the Right Section: Consider the right section of the truss, which includes joints C, H, and G.
- Equilibrium Equations: Apply the equations of equilibrium to this section:
- Sum of Forces in the Horizontal Direction (∑Fx = 0): F_BC + F_CH * cos(θ) = 0
- Sum of Forces in the Vertical Direction (∑Fy = 0): F_CH * sin(θ) – F_GH = 0
- Sum of Moments about Point H (∑M_H = 0): F_BC * d – F_GH * h = 0 Here, θ is the angle between member CH and the horizontal axis, d is the horizontal distance from H to the line of action of F_BC, and h is the vertical distance from H to the line of action of F_GH.
Solving the Equations:
From the moment equation:
F_BC * d = F_GH * h
F_BC = (F_GH * h) / d
Substitute F_BC into the horizontal force equation:
(F_GH * h) / d + F_CH * cos(θ) = 0
F_CH * cos(θ) = – (F_GH * h) / d
F_CH = – (F_GH * h) / (d * cos(θ))
Substitute F_CH into the vertical force equation:
- (F_GH * h) / (d * cos(θ)) * sin(θ) – F_GH = 0
- (F_GH * h) * tan(θ) / d – F_GH = 0
F_GH * (- (h * tan(θ) / d) – 1) = 0
Since F_GH ≠ 0:
- (h * tan(θ) / d) – 1 = 0
h * tan(θ) = d
tan(θ) = d / h
θ = arctan(d / h)
Determining Tension or Compression:
- If F_BC is positive, member BC is in tension; if negative, in compression.
- If F_CH is positive, member CH is in tension; if negative, in compression.
- If F_GH is positive, member GH is in tension; if negative, in compression.
By solving these equations with the given dimensions and loads, we can determine the exact forces in members BC, CH, and GH, as well as whether they are in tension or compression.