# Structural Analyses of Trusses, method of joints

Symmetrical truss is given:

I have issue with the sign convention for joint H.

See below the solution for joint I. As you can see, when calculating equilibrium in y (vertical) direction, forces that act upwards are written with "+" sign, and forces acting downwards with "-" negative sign.

Yet for joint H below, forces acting downwards are written as positive, which is why $F_{HJ}$ turns out to be a negative force (compression force). But it's not consistent with the solution for previous joint.

How would you go about finding force in member HJ?

The answer presented is not inconsistent, but it is badly laid out.

They have not visibly defined "upwards as positive" (they should have...). The equation that's confusing you is:

$$\sum F_y = 0; F_{HJ}\cos(45) + 100\sin(75) + 273.2\cos(45) = 0$$

In this instance, every single force on the Free Body Diagram in the 'y' direction is pointing downwards, so, to follow the "upwards is potitive" convention, we could write:

$$-F_{HJ}\cos(45) - 100\sin(75) - 273.2\cos(45) = 0$$

Which, I'm sure you'll agree, is equivalent to the originally presented formula.

When answering questions like this, you should have always define the positive directions by an annotation with the FBD, and show any extra steps (such as multiplying an equation through by '-1'), to prevent confusion for casual readers.

Note further, that a 'trailing minus' at the start of an equation can get lost, and is often the root of a sign-error; what they did made sense, but they should have made it clearer.

• another confusing point is do you really write signs based on the direction of arrows? Or whether force is compression or tension? We know that $F_{IH}$ is a tension force. So when doing calculations for both joints I and H, $F_{IH}$ should be written as a positive sign, since it's a tensile force? – Jack Jan 10 '18 at 5:32
• Yes, you do really write signs of Forces that way. Tension is a scalar (like speed), and force is a vector (like velocity). Imagine a rope with a weight hanging on it. There is a force at each end of the rope, one down (due to the mass), and one up (due to the reaction at the support). Clearly these forces are equal and opposite, so cannot have the same sign. Which one is positive depends on your chosen sign convention (is the direction of gravity, or "upwards" positive?), but they must always be opposite since the system is in equilibrium. – Jonathan R Swift Jan 10 '18 at 12:15
• well in that case, how come the 200lb force on joint I was written as negative, because it's assumed it's given as a compressive force? yet for joint H similar kind of force with 100lb isn't assumed to be compressive (not written with negative sign). At joint H, 100lb should be with negative sign, and $F_{IH}$ with positive sign, since it's a tensile force. Yet in the equation given they're both of the same sign... – Jack Jan 12 '18 at 20:45
• The force on a joint is neither compressive nor tensile - those classifications only make sense related to the members between joints. The 200lb force on Joint I is both downwards, and to the left, i.e. it is negative in both the X and Y directions. That is why it is written as negative. not "because it's given as a compressive force". In Joint H, it appears positive because they have multiplied the whole equation through by minus 1, in order to avoid leading with a minus sign, as explained in my answer. – Jonathan R Swift Jan 13 '18 at 23:02
• yeah right..., so unknown forces are assumed to be tensile and thus written as positive. But you also need to account for x and y axis directions? Well in that case, for joint H, you know $F_{IH}$ is compressive force as found from previous equation, so it'll be pointing downwards and to the right from joint H. So it's in positive x axis and negative y direction. Is this how you reason here? Also, why is it that in soln 6, for $\sum F_x = 0; F_{HJ}$ is double minus? It is already known that $F_{HJ}$ is compressive, so it's pointing in negative x and y directions. Why double minus when in x? – Jack Jan 14 '18 at 4:41