Problem 436 - Howe Truss With Counter Braces | Engineering Mechanics Review at MATHalino

Problem 436
In Figure P-420, assume that counter diagonals act from B to E and from E to F in addition to the counter diagonals CD and DG shown in the figure. Assuming that these counter diagonals can support tension only, determine which diagonals are acting and the force in each.

Solution 436

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Sections through the 2nd and 3rd panels

From the Free Body Diagram of Section Through M-M

Without the diagonals:
$\Sigma F_v = 190 - 80 = 110 ~ \text{kN upward}$

This 110 kN force must be countered by downward components of the diagonals. Member CD will be in compression. Since diagonals can support only tension, the force in member CD must be zero. The acting diagonal therefore is member BE.

$\Sigma F_v = 0$

$\frac{3}{5}F_{BE} + 80 = 190$

$F_{BE} = 183.33 ~ \text{kN}$ answer

From the Free Body Diagram of Section Through N-N

Without the diagonals:
$\Sigma F_v = 190 - 80 - 200 = -90 ~ \text{kN} = 90 ~ \text{kN downward}$

Since ΣF_v is downward, the counter diagonals must act upward. DG is in compression and therefore redundant, the acting member is the diagonal EF.

$\Sigma F_v = 0$

$\frac{3}{5}F_{EF} + 190 = 80 + 200$

$F_{EF} = 150 ~ \text{kN}$ answer

Problem 436 - Howe Truss With Counter Braces

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