Rigid Bar on Two Inclined Support Bars (Statically Indeterminate): 04-BS-6, December 2018, Q1

Question 1: Rigid Bar on Two Inclined Support Bars (Statically Indeterminate) (20 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. Rigid bar $ABCD$, pinned at $A$ (1-in diameter pin, double shear). Point $E$ is on the wall, directly above $A$ by $d=220\text{ in}$. Bar (1) runs from $E$ to $B$ ($x_B=a=220\text{ in}$ from $A$); bar (2) runs from $E$ to $C$ ($x_C=a+b=360\text{ in}$ from $A$); load $P=6\text{ kips}$ acts down at $D$ ($x_D=a+b+c=460\text{ in}$). Both bars: $A_{bar}=0.45\text{ in}^2$, $E=30{,}000\text{ ksi}$, $\sigma_Y=110\text{ ksi}$.

Find. Part (a): normal stress $\sigma_1,\sigma_2$ in bars (1),(2). Part (b): downward deflection $\delta_D$. Part (c): shear stress $\tau_{pin}$ at A.

Rigid bar ABCD pinned at A; inclined bars (1) E-B and (2) E-C prop the bar against the tip load P at D.

Approach. The rigid bar is pinned only at A, so its only support against rotation is the two inclined bars — one equilibrium equation (moment about A) with two unknown bar forces, closed by the compatibility of the bar rotating as a rigid body about A: each bar behaves as a linear vertical spring whose stiffness follows from its own elongation-displacement geometry.

Compatibility — effective spring stiffness of each bar. Let the rigid bar rotate by a small angle $\theta$ about $A$ (down at $D$ positive); the vertical displacement at a station a distance $x$ from $A$ is $\delta(x)=\theta x$. Since $E$ is fixed, each bar's elongation is the component of that displacement along the bar's own axis: $\Delta_i=\delta(x_i)\,d/L_i$, where $L_i=\sqrt{x_i^2+d^2}$. The bar force is $F_i=(A E/L_i)\Delta_i$, so its VERTICAL (supporting) component is $F_i\,d/L_i=\theta\left(AEd^2/L_i^3\right)x_i=k_i\,\delta(x_i)$, i.e. each bar acts as a vertical spring of stiffness $k_i=AEd^2/L_i^3$ at its attachment point. $L_1=\sqrt{220^2+220^2}=311.13\text{ in}$, $L_2=\sqrt{360^2+220^2}=421.90\text{ in}$.
Moment equilibrium about A — solve for the rotation. Only the VERTICAL bar components (and P) create moment about A (both B and C sit at the same height as A, so horizontal bar components have zero moment arm): $k_1\theta x_B\cdot x_B+k_2\theta x_C\cdot x_C=P\,x_D$. Substituting the numbers gives $\theta=P x_D\big/\left[AEd^2\left(x_B^2/L_1^3+x_C^2/L_2^3\right)\right]=\boxed{1.2674\times10^{-3}\text{ rad}}$.
Part (a) — bar forces and stresses. $F_1=\theta\,AE\,x_B d/L_1^2=8.555\text{ kips}$, so $\sigma_1=F_1/A=8.555/0.45=\boxed{19.01\text{ ksi}}$. $F_2=\theta\,AE\,x_C d/L_2^2=7.613\text{ kips}$, so $\sigma_2=F_2/A=7.613/0.45=\boxed{16.92\text{ ksi}}$. Both are far below $\sigma_Y=110\text{ ksi}$, confirming the small-rotation elastic analysis is valid.
Part (b) — deflection at D. $\delta_D=\theta\,x_D=1.2674\times10^{-3}\times460=\boxed{0.583\text{ in (downward)}}$.
Part (c) — pin reaction and shear stress. Resolve each bar force into horizontal/vertical components ($F_{1x}=F_1x_B/L_1=6.049\text{ kips}$, $F_{1y}=F_1d/L_1=6.049\text{ kips}$ — equal, since bar (1) happens to sit at exactly $45^{\circ}$ because $x_B=d$; $F_{2x}=F_2x_C/L_2=6.496\text{ kips}$, $F_{2y}=F_2d/L_2=3.970\text{ kips}$). Equilibrium of the rigid bar: $A_x=F_{1x}+F_{2x}=12.55\text{ kips}$; $A_y=P-F_{1y}-F_{2y}=6-6.049-3.970=-4.019\text{ kips}$ (the two bars alone over-support the vertical load, so the pin must pull DOWN by $4.02\text{ kips}$ to close equilibrium). Resultant $R_A=\sqrt{A_x^2+A_y^2}=\sqrt{12.55^2+4.019^2}=\boxed{13.17\text{ kips}}$. Double shear across the 1-in pin: $\tau_{pin}=R_A/(2A_{pin})=13.17/(2\times0.7854)=\boxed{8.39\text{ ksi}}$.

Final results
Quantity	Value
σ₁ (bar 1)	19.01 ksi
σ₂ (bar 2)	16.92 ksi
δ_D	0.583 in (downward)
Pin reaction R_A	13.17 kips
τ_pin (double shear)	8.39 ksi

Topic: Statically indeterminate rigid bar on spring-like inclined bars

Key relations: Rigid-body rotation compatibility $\delta(x)=\theta x$; bar elongation $\Delta=\delta\,d/L$; effective vertical stiffness $k=AEd^2/L^3$; moment equilibrium $\sum k_i\theta x_i^2=P x_D$.

Why this works: A rigid bar pinned at only one point and propped by two inclined two-force members is one equation short (moment equilibrium alone has two unknown bar forces); each inclined bar behaves exactly like a linear spring at its own attachment point once the small-rotation geometry is used to relate elongation to the bar's vertical displacement, closing the system with one compatibility unknown ($\theta$) instead of two force unknowns.

Common pitfall: Treating the two bars as if they shared the same lever arm or the same stiffness; each bar's vertical stiffness scales with $x_i^2/L_i^3$, so the farther/steeper bar does NOT automatically carry the larger force. Also easy to forget that horizontal bar forces do not enter the moment equation about A because B and C are level with A.

Source: Hibbeler, Mechanics of Materials, 10th ed., §4.4 Statically Indeterminate Axially Loaded Members

Question 1 of 8: Rigid Bar on Two Inclined Support Bars (Statically Indeterminate)

Question 1: Rigid Bar on Two Inclined Support Bars (Statically Indeterminate) (20 marks)