2: Forces in Statically Determinate Beams and Plane Frames

2.7: External Reactions in Frames (Part 1)

2.7.1: Loads on Inclined Members, Examples

Basic Structure

Figure 2.7-1: Inclined Beam

We will show the effects of different kinds of loads on the structure shown in Fig. 2.7-1.

Snow Loads

Figure 2.7-2: Snow Loads

Snow loads act vertically on the structure and are computed by determining the weight of snow plus rain resting on a horizontal surface (i.e., the ‘ground’ snow load). These are often converted to distributed loads acting on frame members by multiplying by the frame spacing (member loaded width, really). When expressed as a uniform load such as 10kN/m, the distance is measured horizontally.

\[\begin{split} \sum M_a &= -10\times4\times2 + V_b\times4 = 0\\ V_b &= 20\\ \sum F_y &= -10\times4 + V_b + V_a = 0\\ V_a &= 20\\ \sum F_x &= H_a = 0\\ H_a &= 0\\ \end{split}\]

Dead Loads

Figure 2.7-3: Dead Loads

Dead loads act vertically on the structure and are computed by determining the self weight of the structure. This usually means estimating the weight of a unit area of the inclined surface, and multiplying that by the loaded width to get a distributed load. When expressed as a uniform load such as 10kN/m, the distance is measured along the incline.

\[\begin{split} \sum M_a &= -10\times5\times2 + V_b\times4 = 0\\ V_b &= 25\\ \sum F_y &= -10\times5 + V_b + V_a = 0\\ V_a &= 25\\ \sum F_x &= H_a = 0\\ H_a &= 0 \end{split}\]

Wind Loads

Figure 2.7-4: Wind Loads

Wind loads act perpendicular to the surface and are computed by determining wind pressure acting on that surface and multiplying by the loaded width to get a distributed load acting on the member. When expressed as a uniform load such as 10kN/m, the distance is measured along the incline.

\[\begin{split} \sum M_a &= -10\times5\times2.5 + V_b\times4 = 0\\ V_b &= 31.25\\ \sum F_y &= -10\times5\times{4\over5} + V_b + V_a = 0\\ V_a &= 8.75\\ \sum F_x &= 10\times5\times{3\over5} - H_a = 0\\ H_a &= 30 \end{split}\]

2.7.2: Loads on Inclined Members, General Relationships

Here are some general relationships, expressed in terms of the geometry of the basic structure. For each load case are shown the moment about point a, $M_a$, the total vertical component, $F_y$, and the total horizontal component, $F_x$. Also shown are the distributed components normal to the member, $w_n$ and parallel to it, $w_p$. These distributed components, $w_n$ and $w_p$, are particularly important when it comes time to construct the NVM diagrams for sloped members, but can also be useful for local member equilibrium.

Basic Structure

Figure 2.7-5: Inclined Beam

The following relationships hold for this structure:

\[\begin{split} h &= L\cos\theta\\ L &= h/\cos\theta \end{split}\]

Snow Load

Figure 2.7-6: Snow Load and Distributed Components

For the distributed snow load, w:

\[\begin{split} F_x &= 0\\ F_y &= w h\\ M_a &= w h {h\over2}\\ w_n &= { {w h \cos\theta}\over L}\\ w_p &= { {w h \sin\theta}\over L} \end{split}\]

Dead Load

Figure 2.7-7: Dead Load

For the distributed dead load, w:

\[\begin{split} F_x &= 0\\ F_y &= w L\\ M_a &= w L {h\over2}\\ w_n &= { {w L \cos\theta}\over L} = w \cos\theta\\ w_p &= { {w L \sin\theta}\over L} = w \sin\theta \end{split}\]

Wind Load

Figure 2.7-8: Wind Load

For the distributed wind load, w:

\[\begin{split} F_x &= w L \sin\theta\\ F_y &= w L \cos\theta\\ M_a &= w L {L\over2}\\ w_n &= w\\ w_p &= 0 \end{split}\]

2.7.3: Example 2.7.3-1

Determine the reactions for the gable frame subjected to the unbalance snow load shown in Fig. 2.7-9. Also, calculate the member end forces, draw an FBD for each individual member, and check the equilibrium of the two hip joints.

Reactions

Figure 2.7-9: Gable Frame

The frame is a three-hinged arch, and so we know it is both statically determinate and stable. Nevertheless, we can count $m=4$, $r=4$, $j=5$, and $c=1$. Thus $3m+r=13$ and $3j+c=13$ and therefore statically determinate if stable. It is stable.

Figure 2.7-10: Complete Frame Free Body Diagram

Figure 2.7-10 shows the free body diagram of the complete frame. There are four unknowns, but because the supports a and e are at the same elevation, we can write some equilibrium equations that only involve only one unknown and are thus solvable:

\[\begin{split} \sum M_a &= 0~~~~~\text{+CCW}\\ &= -24\times6\times{6\over2}+V_e\times12\\ V_e &= 36~~~~~(\therefore\uparrow)\\ \sum F_y &= 0~~~~~\text+\uparrow\\ &= V_a - 24\times6 + V_e\\ &= V_a - 144 + 36\\ V_a &= 108~~~~~(\therefore\uparrow) \end{split}\]

Figure 2.7-11: Partial Frame Free Body Diagram

Fig. 2.7-11 shows a free body diagram of the right portion of the structure, resulting from cutting the structure at the condition at c.

\[\begin{split} \sum M_c &= 0~~~~~\text{+CCW}\\ &= V_e\times6 - H_e\times6\\ H_e &= V_e\\ H_e &= 36~~~~~(\therefore\leftarrow) \end{split}\]

We could have done this a little more simply by recognizing the portion c-d-e is a two-force member, subjected only to forces at the pin on either end. That means that the resultant of $V_e$ and $H_e$ (and also of $V_c$ and $H_c$ must lie along the line joining the ends, c-e. As that line has a 1:1 slope in this problem, the components $V_e$ and $H_e$ must be equal.

Figure 2.7-12: Summary Free Body Diagram

Going back to the first free body diagram, we can now write:

\[\begin{split} \sum F_x &= 0~~~~~\text+\rightarrow\\ &= H_a - H_e\\ &= H_a - 36\\ H_a &= 36~~~~~(\therefore\rightarrow) \end{split}\]

Fig. 2.7-12 shows the results of the reaction calculation on a summary freebody diagram.

Member End Forces

Figure 2.7-13: Column a-b

To determine the forces (normal, shear and moment) at each of the member ends, we draw a FBD for each. Fig. 2.7-13 shows this for column a-b. The left portion of the figure shows a FBD with the unknowns labelled; the right figure shows the summary results.

As this is a particularly simple FBD, we will simply write the equilibrium expressions.

\[\begin{split} &N_{ba} = 108~\text{kN}\\ &V_{ba} = 36~\text{kN}\\ &M_{ba} = 36\times4 = 144~\text{kNm} \end{split}\]

Figure 2.7-14: Column d-e

Similarly, column d-e is shown in Fig. 2.7-14.

\[\begin{split} &N_{de} = 36~\text{kN}\\ &V_{de} = 36~\text{kN}\\ &M_{de} = 36\times4 = 144~\text{kNm} \end{split}\]

Figure 2.7-15: Roof Beam c-d

Fig. 2.7-15 shows roof beam c-d. It has not loads applied directly to it. From our previous work above, we know the horizontal and vertical components applied to the end coming from the pin c, so we show that in the top figure of the FBD. To get the shears and normal forces at end d, it is probably easiest to construct the x’-y’ axis system, with x’ parallel to the beam, as shown in the figure.

Having done that, we can then write equilibrium equations involving only one unknown each.

\[\begin{split} &\theta = \tan^{-1}{2\over6} = 18.44^\circ\\ &\sum F_{y'} = 0:\\ &V_{dc} = 36\times\cos\theta - 36\times\sin\theta\\ &V_{dc} = 22.77\\ &\sum F_{x'} = 0:\\ &N_{dc} = 36\times\cos\theta + 36\times\sin\theta\\ &N_{dc} = 45.53\\ &\sum M_b = 0:\\ &M_{dc} = 36\times6 - 36\times2\\ &M_{dc} = 144\\ \\ &N_{cd} = N_{dc} = 45.53\\ &V_{cd} = V_{dc} = 22.77\\ \end{split}\]

Figure 2.7-16: Roof Beam b-c

Fig. 2.7-16 shows the last remaining roof beam, b-c. It is slightly more complicated than the previous one due to the fact that it has a distributed load applied. We show that in a form thats most convenient for our calculations. We also show the pin forces at c being equal and opposite to the forces shown on the adjacent beam c-d.

\[\begin{split} &\sum F_{y'} = 0:\\ &V_{bc} = 24\times6\times\cos\theta - 36\times\cos\theta - 36\times\sin\theta\\ &V_{bc} = 91.08\\ &\sum F_{x'} = 0:\\ &N_{bc} = 24\times6\times\sin\theta + 36\times\cos\theta - 36\times\sin\theta\\ &N_{bc} = 68.30\\ &\sum M_b = 0:\\ &M_{bc} = 24\times6\times3 - 36\times6 - 36\times2\\ &M_{bc} = 144\\ \\ &V_{cb} = 24\times6\times\cos\theta - 91.08\\ &V_{cb} = 45.53\\ &N_{cb} = 68.30 - 24\times6\times\cos\theta\\ &N_{cb} = 22.77 \end{split}\]

Joint Equilibrium

Transferring the equal and opposites of the member end forces to the joints allows us to check joint equilibrium.

Figure 2.7-17: Joint b

Fig. 2.7-17 is a free body diagram of joint b, the one that connects members a-b and b-c. The joint has no dimensions – it is of zero size. Checking equilibrium gives us:

\[\begin{split} &\sum M = 144 - 144 = 0~~~~~\text{O.K.}\\ &\sum F_x = 36 + 91.08\sin\theta - 68.30\cos\theta = -0.002~~~~~\text{O.K.}\\ &\sum F_y = 108 - 91.08\cos\theta - 68.03\sin\theta = -0.001~~~~~\text{O.K.} \end{split}\]

Therefore, equilibrium is satisfied at joint b.

Figure 2.7-18: Joint d

Fig. 2.7-18 is a free body diagram of joint d, the one that connects members c-d and d-e. Checking equilibrium gives us:

\[\begin{split} &\sum M = 144 - 144 = 0~~~~~\text{O.K.}\\ &\sum F_x = -36 - 22.77\sin\theta + 45.53\cos\theta = -0.004~~~~~\text{O.K.}\\ &\sum F_y = 36 - 22.77\cos\theta - 45.53\sin\theta = 0.004~~~~~\text{O.K.} \end{split}\]

Therefore, equilibrium is satisfied at joint d.

For joint c, we can examine end c of member c-d in Fig. 2.7-15, and end c of b-c in Fig. 2.7-16, we notice that the moments are zero, and that the shears and normal forces are equal and opposite. Therefore, by inspection, we see that equilibrium is also statisfied at joint c.

The fact that joint equilibrium checks is very good evidence that you have not made errors in calculation. Unfortunately, it is not proof. It is necessary, but not sufficient to prove correctness.