# Question 3.8: A load P is hung from the end of a rigid beam AD, which is s......

A load P is hung from the end of a rigid beam AD, which is supported by a pin at end A and by two uniform hanger rods (see Fig. 1). How much load, to the nearest 100 pounds, can be applied without exceeding an allowable tensile stress of $σ_{\text{allow}}$ = 30 ksi in either of the rods? Assume small-angle

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Plan the Solution This problem is not quite as straightforward as the previous one that served to introduce statically indeterminate structures.
Now the load is unknown, but the allowable axial stresses are known. Since the axial stress in an element is just the (internal) axial force in the element divided by its area, it is clear that we must eventually get equations that relate $F_1$ and $F_2$, the axial forces in the two rods, to the load P.
Let us begin by writing down the three fundamental types of equations.
Equilibrium: We must ask ourselves the question: What free-body diagram(s) would lead to an equilibrium equation (or equations) that relate the external load P to the internal element forces $F_1$ and $F_2$? Clearly, a free-body diagram of the rigid beam AD should be used.
Since we are not specifically asked to determine the reactions $A_x$ and $A_y$, we do not need to sum forces. A moment equation for the free-body diagram in Fig. 2 will directly relate $F_1$ and $F_2$ to P, so we write

$+\circlearrowleft\left(\sum{M} \right) _A = 0 : aF_1 + bF_2$ – cP = 0      Equilibrium (1)

Since there are two unknown forces and only one equilibrium equation, the structure is statically indeterminate. Since the beam AD could be supported by just one hanger rod, one of the rods is redundant.
Element Force-Deformation Behavior: It will be convenient to use the form given in Eq. 3.14.

$e = fF, \text{where} f ≡ \frac{L}{AE}$        (3.14)

$e_1 = f_1F_1, e_2 = f_2F_2$    Element Force-Deformation Behavior      (2)

where the values of the flexibility coefficients for the hanger rods, labeled (1) and (2) in Fig. 1, are

$f_1 = \left(\frac{L}{AE}\right)_1 = \frac{80 in.}{(1.0 in^2)(10 \times 10^3 ksi)} = 8.00 (10^{-3})$ in./kip

$f_2 = \left(\frac{L}{AE}\right)_2 = \frac{50 in.}{(0.5 in^2)(10 \times 10^3 ksi)} = 10.00 (10^{-3})$ in./kip

Geometry of Deformation: Because AD is assumed to be rigid, we can sketch a deformation diagram that relates the elongations of the hanger rods to the rotation of the beam AD about A (Fig. 3). Since the rotation angle θ is assumed to be small, we can assume that points B and C move vertically downward (instead of along arcs of circles with centers at A), and we can therefore use the properties of similar triangles to get the compatibility equation

$e_2 = \left(\frac{b}{a}\right)e_1$    Deformation Geometry  (3)

Therefore, the rigid beam constrains the two hanger-rod elongations to satisfy this compatibility equation.
Solution of the Equations: To solve for the internal forces, we can eliminate the element elongations, $e_1$ and $e_2$, by substituting the force-deformation equations into the compatibility equation to get the following compatibility equation in terms of the unknown element forces.

$f_2F_2 = \left(\frac{b}{a}\right)f_1F_1$    Compatibility in Terms of Forces  (4)

In Eqs. (1) and (4) we have two equations in terms of the two unknown internal forces. These can be $\underline{\text{solved simultaneously}}$ to give expressions that relate the forces $F_1$ and $F_2$ to the external load P.

$F_1 = \left( \frac{acf_2} {a^2f_2 + b^2f_1}\right) P$      (5a,b)

$F_2 = \left( \frac{acf_1} {a^2_2f_2 + b^2f_1}\right) P$

Allowable Load: Let us now substitute numerical values from Fig. 1 and numerical values for the flexibility coefficients $f_1$ and $f_2$ into Eqs. (5). From Eqs. (5),

$F_1 = \left(\frac{15}{21}\right)P, F_2 = \left(\frac{24}{21}\right)P$,  (6a,b)
Now we can relate the external load P to the allowable stress. With $F_1$ and $F_2$ given by Eqs. (6), we get the following expressions for the stresses in the rods:

$σ_1 = \frac{F_1}{A_1} = \frac{(15/21)(P kips)}{(1.0 in^2)}$    (7a,b)

$σ_2 = \frac{F_2}{A_2} = \frac{(24/21)(P kips)}{(0.5 in^2)} > σ_1$

Since rod (2) is more highly stressed than rod (1), we set $σ_2$ equal to the allowable stress and get

$P_{\text{allow}} = \frac{21} {24} (0.5 in^2)σ_{\text{allow}}$ = 13.1 kips

Review the Solution One way to verify the results is to check compatibility.

Is      $f_2F_2 = \left(\frac{a}{b}\right) f_1F_1?$         Yes

$[10.00(10^{-3})$ in./kip] $\left(\frac{24}{21}\right)$ P(kips) =

$\left(\frac{100 in.} {50 in.}\right) [8.00(10^{-3})$ in./kip] $\left(\frac{15} {21}\right)$ P(kips)

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