Question 2.1.2: Let’s consider a structure that might be part of an underwat...

Given a truss with specified parameters and loading, we must find the requested values of stress. We first examine an FBD of the joint at D (Figure 2.12).

Note that we have assumed both members DC and DE to be in tension; if we calculate negative values for either internal force, we will know that this assumption was incorrect and that the member is in compression. Since the joint must be in equilibrium we have

$ΣF_y = 0 = P_{_{DC}} \sin45  –  21 \textrm{ kN} → P_{_{DC}} = 29.7$ kN

$ΣF_x = 0 = –P_{_{DE}} – N_{_{DC}} \cos45 → P_{_{DE}} = –21$  kN.

Using the definition of normal stress we know that

$\sigma _{_{DE}}=\frac{P_{_{DE}}}{A_{x −sec}}=-42\times 10^6  \frac{\textrm{N}}{\textrm{m}^2}$,

or

$\sigma _{_{DE}} =42$ MPa compressive.

Next, we use the method of sections. We make an imaginary cut between B and C, resulting in an FBD that includes the internal forces in three members of the truss (Figure 2.13).

We apply the third equilibrium equation, summing moments about point F:
$ΣM_F = 0 = P_{_{CB}} \cdot (2 \textrm{ m}) – 21\textrm{k N} \cdot (4 \textrm{ m}) → P_{_{CB}} = 42$ kN,

so that

$\sigma _{_{CB}}=\frac{P_{_{CB}}}{A}=42\times 10^6  \frac{\textrm{N}}{\textrm{m}^2}  \sigma _{_{CB}} =84$ MPa tensile.

We may take this opportunity to check our intuition about this truss.
The load, P, is pulling the structure down. Thus, composite member ABC should become longer, and DEFG should become shorter. This would mean that members on the top (like BC) would be in tension and members on the bottom (like DE) in compression. Our results so far are consistent with our physical intuition. This buoys our spirits as we continue to part (b) of the problem, in which we consider the bolt at joint A.
We are told that this bolt is in “double shear.” A connection element (bolt or pin) is said to be in “single shear” if one cut between the member and its support is sufficient to break the connection, as shown in Figure 2.14 on the left; “double shear” means that two cuts are needed to break the connection, as on the right. A quick analysis using free-body diagrams of each case should be persuasive evidence that a bolt in double shear experiences half the shear stress of an identically loaded bolt in single shear. This analysis is left as an exercise.

To find the reaction forces at the supports, we consider an FBD of the entire truss (Figure 2.15)

Summing moments about point G, we have

$ΣM_{_G} = 0 = R_{Ax} \cdot (2 \textrm{ m}) – 21 \textrm{ kN} \cdot (6 \textrm{ m}) → R_{Ax} = 63$ kN.

So the shear stress in the bolt at A is found:

$\tau _{_A}=\frac{A_x /2}{A_{bolt}}=100 \times 10^6 \frac{\textrm{N}}{\textrm{m}^2} =100$ MPa.