CE Board Exam Randomizer

Part 1.

$A = \frac{\pi d^2}{4} = \frac{\pi(80)^2}{4} = 1600\pi \approx 5026.5 \text{ mm}^2$
$E = 200{,}000 \text{ MPa}$
Axial rigidity $= AE = 5026.5 \times 200{,}000$
$\boxed{\approx 1.005 \times 10^9 \text{ N}}$

Part 2.

$I = \frac{\pi d^4}{64} = \frac{\pi(80)^4}{64} \approx 2{,}010{,}619 \text{ mm}^4$
Flexural rigidity $= EI = 200{,}000 \times 2{,}010{,}619$
$\boxed{\approx 4.02 \times 10^{11} \text{ N·mm}^2}$

Part 3.

$J = 2I = 4{,}021{,}238 \text{ mm}^4$; $G = 78{,}000 \text{ MPa}$
Torsional rigidity $= GJ = 78{,}000 \times 4{,}021{,}238$
$\boxed{\approx 3.14 \times 10^{11} \text{ N·mm}^2}$

Part 1.

Because the steel and concrete are bonded and loaded concentrically, they have the same strain:
$\frac{\sigma_s}{E_s}=\frac{\sigma_c}{E_c}$, so $\sigma_s=n\sigma_c$ with $n=\frac{29\times10^6}{3.6\times10^6}=8.056$.
Steel area:
$A_s=4\left(\frac{\pi(0.75)^2}{4}\right)=1.767\text{ in}^2$
Concrete area:
$A_c=8(8)-1.767=62.233\text{ in}^2$
$150=\sigma_c(A_c+nA_s)$
$\sigma_c=1.962\text{ ksi},\quad \sigma_s=8.056(1.962)$
$\boxed{\sigma_s=15.8\text{ ksi}}$

Part 2.

From the transformed-area equilibrium:
$150=\sigma_c\left(62.233+8.056(1.767)\right)$
$\sigma_c=1.962\text{ ksi}$
$\boxed{\sigma_c=1.962\text{ ksi}}$

Part 1.

For the bonded reinforced concrete post, steel and concrete have the same strain, so:
$\sigma_s=n\sigma_c,\quad n=\frac{200}{25}=8$
Steel area:
$A_s=6\left(\frac{\pi(28)^2}{4}\right)=3695\text{ mm}^2$
Concrete area:
$A_c=\frac{\pi(450)^2}{4}-3695=155{,}348\text{ mm}^2$
$1{,}550{,}000=\sigma_c(A_c+nA_s)$
$\sigma_c=8.38\text{ MPa},\quad \sigma_s=8(8.38)$
$\boxed{\sigma_s=67.1\text{ MPa}}$

Part 2.

Using the same transformed-area relation:
$1{,}550{,}000=\sigma_c\left(155{,}348+8(3695)\right)$
$\boxed{\sigma_c=8.38\text{ MPa}}$

Part 1.

The rigid cap makes the steel and concrete shorten equally, so $\sigma_s=n\sigma_c$, where:
$n=\frac{E_s}{E_c}=\frac{29\times10^6}{3.5\times10^6}=8.286$
Concrete area:
$A_c=\frac{\pi(6)^2}{4}=28.274\text{ in}^2$
Steel tube area:
$A_s=\frac{\pi(6.5^2-6^2)}{4}=4.909\text{ in}^2$
$200=\sigma_c(A_c+nA_s)$
$\sigma_c=2.900\text{ ksi},\quad \sigma_s=8.286(2.900)=24.04\text{ ksi}$
$\boxed{\sigma_s=24040\text{ psi}}$

Part 2.

From the composite axial-load equation:
$200=\sigma_c\left(28.274+8.286(4.909)\right)$
$\sigma_c=2.900\text{ ksi}$
$\boxed{\sigma_c=2900\text{ psi}}$

Part 1.

Let the rod forces be $F_A$, $F_B$, and $F_C$. Since the rods are identical, force is proportional to elongation. The slab remains plane, so the displacement at B is midway between A and C:
$F_B=\frac{F_A+F_C}{2}$
Equilibrium gives:
$F_A+F_B+F_C=W$
Taking moments about A, with $AB=b$, $AC=2b$, and $AG=1.2b$:
$F_B(b)+F_C(2b)=W(1.2b)$
Solving:
$\boxed{F_A=\frac{7}{30}W}$

Part 2.

Using the same equations:
$F_A+F_B+F_C=W$
$F_B+2F_C=1.2W$
$F_B=\frac{F_A+F_C}{2}$
Solving gives:
$\boxed{F_B=\frac{1}{3}W}$

Part 3.

The solution of the equilibrium and compatibility equations gives:
$F_A=\frac{7}{30}W,\quad F_B=\frac{1}{3}W$
$F_C=W-F_A-F_B$
$F_C=W-\frac{7}{30}W-\frac{1}{3}W$
$\boxed{F_C=\frac{13}{30}W}$

Because the 400-kN load is centered and the two steel bars are identical, the rigid platform remains horizontal and both steel bars carry the same stress.

Check whether the aluminum bar engages. The steel bars must first shorten by 0.1 mm to close the initial gap. The load required to produce this shortening is

$$P_{\text{contact}}=2\left(\frac{A_sE_s}{L_s}\right)(0.1)=2\left[\frac{(1400)(200{,}000)}{250}\right](0.1)=224{,}000\ \text{N}=224\ \text{kN}.$$

Since 400 kN is greater than 224 kN, the gap closes and the aluminum bar carries part of the applied load.

Equilibrium. Let $\sigma_s$ be the stress in each steel bar and $\sigma_a$ be the stress in the aluminum bar. Using MPa = N/mm², vertical equilibrium of the platform gives

$$2A_s\sigma_s+A_a\sigma_a=400{,}000,$$ $$2(1400)\sigma_s+(2800)\sigma_a=400{,}000,$$ $$\sigma_s+\sigma_a=142.857\ \text{MPa}. \tag{1}$$

Compatibility. After contact, the steel shortening equals the 0.1-mm initial gap plus the aluminum shortening. The aluminum bar's original length is $L_a=250-0.1=249.9$ mm; thus,

$$\delta_s=0.1+\delta_a,$$ $$\frac{\sigma_sL_s}{E_s}=0.1+\frac{\sigma_aL_a}{E_a}.$$

Substituting the material properties and using $\sigma_s=142.857-\sigma_a$ from Eq. (1),

$$\frac{(142.857-\sigma_a)(250)}{200{,}000}=0.1+\frac{\sigma_a(249.9)}{70{,}000},$$ $$\sigma_a=16.301\ \text{MPa}.$$

As a check, each steel bar carries about 177.18 kN and the aluminum bar carries about 45.64 kN, whose sum is 400 kN. Therefore, the stress in the aluminum bar is

$$\boxed{\sigma_a=16.30\ \text{MPa (compression)}}$$

Because the load is applied at the center of the symmetric rigid bar and the two outer wires are identical, the bar translates downward without rotating. Therefore, points A, B, and C have the same vertical displacement.

The outer wires are each 60 ft long, while the initially slack middle wire is 0.2 in longer:

$$L_o=60(12)=720\ \text{in.},\qquad L_m=720+0.2=720.2\ \text{in.}$$

Check when the middle wire engages. Before engagement, only the two outer wires resist the load. The bar must move downward 0.2 in to remove the middle wire's slack, so

$$P_{\text{contact}}=2A E\left(\frac{0.2}{L_o}\right)=2(0.3)(10\times10^6)\left(\frac{0.2}{720}\right)=1666.7\ \text{lb}.$$

The maximum safe load is greater than this value, so the middle wire is taut at the limiting condition.

Compatibility after engagement. Let $\delta_o$ and $\delta_m$ be the elongations of an outer wire and the middle wire, respectively. Since the middle wire had 0.2 in of initial slack,

$$\delta_m=\delta_o-0.2.$$

The outer wires always elongate more and therefore reach the 12-ksi working stress first. At the limiting condition,

$$\sigma_o=12{,}000\ \text{psi},$$ $$\delta_o=\frac{\sigma_oL_o}{E}=\frac{(12{,}000)(720)}{10\times10^6}=0.864\ \text{in.}$$

Thus, the middle-wire elongation and stress are

$$\delta_m=0.864-0.2=0.664\ \text{in.},$$ $$\sigma_m=E\frac{\delta_m}{L_m}=(10\times10^6)\left(\frac{0.664}{720.2}\right)=9219.7\ \text{psi}=9.220\ \text{ksi}.$$

Since the middle-wire stress is below 12 ksi, the assumed controlling condition is valid. The corresponding wire forces are

$$T_o=A\sigma_o=(0.3)(12{,}000)=3600\ \text{lb},$$ $$T_m=A\sigma_m=(0.3)(9219.7)=2765.9\ \text{lb}.$$

Finally, vertical equilibrium of the rigid bar gives

$$P=2T_o+T_m=2(3600)+2765.9=9965.9\ \text{lb}.$$

Therefore, the largest safe load is

$$\boxed{P\approx9970\ \text{lb}}$$

Before rod AC is connected at C, the rigid bar and rod BD are stress-free. Since the vertical distance from the bar to the lower supports is 2 m and the lower end of AC is initially 4 mm above C, the unstretched rod lengths are

$$L_{AC}=2000-4=1996\ \text{mm},\qquad L_{BD}=2000\ \text{mm}.$$

Pulling the lower end of AC down to C places AC in tension. Its downward force at A rotates the rigid bar counterclockwise about O: point A moves downward while point B moves upward, stretching rod BD. Thus, both rods are in tension.

Rigid-bar compatibility. Let $v_A$ be the downward displacement of A and $v_B$ be the upward displacement of B. For the small rotation of rigid bar AB about O,

$$\frac{v_A}{AO}=\frac{v_B}{OB},$$ $$\frac{v_A}{750}=\frac{v_B}{1500},$$ $$v_B=2v_A. \tag{1}$$

The 4-mm initial gap is partly closed by the downward movement of A and partly accommodated by the elongation of rod AC. Therefore,

$$\delta_{AC}=4-v_A,$$

while the upward movement of B equals the elongation of rod BD:

$$\delta_{BD}=v_B.$$

Using Eq. (1),

$$\delta_{BD}=2(4-\delta_{AC}),$$ $$2\delta_{AC}+\delta_{BD}=8\ \text{mm}. \tag{2}$$

Moment equilibrium of the rigid bar. Let $F_{AC}$ and $F_{BD}$ be the tensile forces in the rods. Taking moments about pin O eliminates the pin reactions:

$$F_{AC}(0.75)=F_{BD}(1.50),$$ $$F_{AC}=2F_{BD}. \tag{3}$$

Axial deformations. For each rod, $\delta=FL/(AE)$. Hence,

$$\delta_{AC}=\frac{F_{AC}L_{AC}}{A_{AC}E},\qquad \delta_{BD}=\frac{F_{BD}L_{BD}}{A_{BD}E}.$$

Dividing these equations and using Eq. (3),

$$\frac{\delta_{AC}}{\delta_{BD}}=\frac{F_{AC}L_{AC}A_{BD}}{F_{BD}L_{BD}A_{AC}}=\frac{2(1996)(250)}{(2000)(300)}=1.66333.$$

Therefore, $\delta_{AC}=1.66333\delta_{BD}$. Substitution into Eq. (2) gives

$$2(1.66333\delta_{BD})+\delta_{BD}=8,$$ $$\delta_{BD}=1.8490\ \text{mm},\qquad \delta_{AC}=3.0755\ \text{mm}.$$

The tensile stress in the left rod is

$$\sigma_{AC}=E\frac{\delta_{AC}}{L_{AC}}=(200{,}000)\left(\frac{3.0755}{1996}\right)=308.17\ \text{MPa}.$$

As a check, $\sigma_{BD}=184.90$ MPa, giving $F_{AC}=92.45$ kN and $F_{BD}=46.22$ kN. Thus, $F_{AC}=2F_{BD}$, as required by moment equilibrium. Therefore,

$$\boxed{\sigma_{AC}\approx308\ \text{MPa (tension)}}$$

Part 1.

The joint displacement is the same for both bars, so the axial deformations are compatible:
$\frac{F_{Al}L_{Al}}{A_{Al}E_{Al}}=\frac{F_sL_s}{A_sE_s}$
$\frac{F_{Al}(15)}{1.25(10\times10^6)}=\frac{F_s(12)}{2.0(29\times10^6)}$
$F_{Al}=0.1724F_s$
Equilibrium at the joint gives $F_s+F_{Al}=40$ kips, so $F_s=34.12$ kips.
$\sigma_s=\frac{34.12}{2.0}=17.06\text{ ksi}$
$\boxed{\sigma_s=17.06\text{ ksi (C)}}$

Part 2.

From compatibility, $F_{Al}=0.1724F_s$. With $F_s=34.12$ kips:
$F_{Al}=5.89\text{ kips}$
$\sigma_{Al}=\frac{5.89}{1.25}=4.71\text{ ksi}$
$\boxed{\sigma_{Al}=4.71\text{ ksi (T)}}$

The load P moves the junction to the right. Consequently, the aluminum member elongates in tension while the steel member shortens in compression. Because both outer supports are unyielding, these axial deformations have equal magnitudes:

$$\delta_{Al}=\delta_{st}.$$

Using $\delta=FL/(AE)=\sigma L/E$, the compatibility equation is

$$\frac{P_{Al}L_{Al}}{A_{Al}E_{Al}}=\frac{P_{st}L_{st}}{A_{st}E_{st}},$$ $$\frac{\sigma_{Al}(15)}{10\times10^6}=\frac{\sigma_{st}(12)}{29\times10^6}. \tag{1}$$

Check whether aluminum governs. If the aluminum reaches its allowable stress of 10 ksi, or 10,000 psi, Eq. (1) gives

$$\frac{(10{,}000)(15)}{10\times10^6}=\frac{\sigma_{st}(12)}{29\times10^6},$$ $$\sigma_{st}=36{,}250\ \text{psi}=36.25\ \text{ksi}>18\ \text{ksi}.$$

This exceeds the allowable steel stress, so aluminum cannot govern the maximum safe load.

Let steel govern. Set the steel stress equal to its allowable value of 18 ksi, or 18,000 psi. From Eq. (1),

$$\frac{\sigma_{Al}(15)}{10\times10^6}=\frac{(18{,}000)(12)}{29\times10^6},$$ $$\sigma_{Al}=4965.5\ \text{psi}=4.966\ \text{ksi}<10\ \text{ksi}.$$

Thus, the steel member reaches its allowable compressive stress while the aluminum member remains safely below its allowable tensile stress.

The resisting axial forces are

$$R_A=F_{Al}=\sigma_{Al}A_{Al}=(4.966)(1.25)=6.2075\ \text{kips},$$ $$R_B=F_{st}=\sigma_{st}A_{st}=(18)(2.0)=36.0\ \text{kips}.$$

Applying horizontal equilibrium at the loaded junction,

$$\sum F_x=0,$$ $$-R_A+P-R_B=0,$$ $$P=R_A+R_B=6.2075+36.0=42.2075\ \text{kips}.$$

Therefore, the maximum allowable axial load is

$$\boxed{P_{\max}\approx42.2\ \text{kips}}$$

Take axial tension as positive. Because the rod is fixed between two rigid walls, the net change in its total length must be zero. Following the force-method solution, release the support at D and use its reaction $R_D$ as the redundant force.

Deformation caused by the applied loads. With support D removed, equilibrium of the entire rod gives

$$\sum F_x=0,$$ $$R_A-P_1-P_2=0,$$ $$R_A=150+90=240\ \text{kN}.$$

The corresponding internal axial forces are

$$N_{AB}^{(L)}=-240\ \text{kN},\qquad N_{BC}^{(L)}=-90\ \text{kN},\qquad N_{CD}^{(L)}=0.$$

The negative signs indicate compression. Using $E=200\ \text{GPa}=200{,}000\ \text{N/mm}^2$, the displacement of D caused by the applied loads is

$$\delta_L=\sum\frac{NL}{AE},$$ $$\delta_L=\frac{(-240{,}000)(500)}{(900)(200{,}000)}+\frac{(-90{,}000)(250)}{(2000)(200{,}000)}+0,$$ $$\delta_L=-0.722917\ \text{mm}.$$

Thus, the applied loads would move D 0.722917 mm toward the left if the right support were absent.

Deformation caused by the redundant reaction. Apply $R_D$ at the released end. This force produces tension throughout all three segments, so

$$\delta_R=\frac{R_D(500)}{(900)(200{,}000)}+\frac{R_D(250)}{(2000)(200{,}000)}+\frac{R_D(350)}{(1200)(200{,}000)}.$$

Because the actual support at D is rigid, its final displacement is zero:

$$\delta_L+\delta_R=0.$$

Therefore,

$$0.722917=\frac{R_D}{200{,}000}\left(\frac{500}{900}+\frac{250}{2000}+\frac{350}{1200}\right),$$ $$R_D=148{,}714\ \text{N}=148.714\ \text{kN}.$$

Final force in segment CD. A section through CD has only the right reaction on its right-hand free body:

$$N_{CD}=R_D=148.714\ \text{kN},$$ $$\boxed{N_{CD}=148.7\ \text{kN (T)}}$$

Final force in segment BC. Cutting between B and C gives

$$N_{BC}=R_D-P_2=148.714-90=58.714\ \text{kN},$$ $$\boxed{N_{BC}=58.7\ \text{kN (T)}}$$

Final force in segment AB. Cutting between A and B gives

$$N_{AB}=R_D-P_2-P_1=148.714-90-150=-91.286\ \text{kN}.$$

The negative value means compression; hence,

$$\boxed{N_{AB}=91.3\ \text{kN (C)}}$$

As an equilibrium check, the left reaction has magnitude 91.286 kN and $91.286+148.714=240$ kN, which balances the two applied loads.

The rigid beam rotates clockwise about pin O when the 50-kip load is applied. Both vertical rods elongate and exert upward tensile forces on the beam. Let $P_{br}$ and $P_{st}$ be the forces in the bronze and steel rods, respectively.

Moment equilibrium. Taking moments about O eliminates the pin reactions:

$$\sum M_O=0,$$ $$P_{br}(3)+P_{st}(12)=50(8),$$ $$3P_{br}+12P_{st}=400. \tag{1}$$

Rigid-beam compatibility. The bronze rod is attached 3 ft from O, while the steel rod is attached 12 ft from O. Because the beam is rigid and undergoes a small rotation about O, the vertical displacements are proportional to these distances:

$$\frac{\delta_{br}}{3}=\frac{\delta_{st}}{12},$$ $$\delta_{st}=4\delta_{br}. \tag{2}$$

The rod lengths and elastic properties are

$$L_{br}=3(12)=36\ \text{in.},\quad A_{br}=2\ \text{in.}^2,\quad E_{br}=12{,}000\ \text{ksi},$$ $$L_{st}=10(12)=120\ \text{in.},\quad A_{st}=0.5\ \text{in.}^2,\quad E_{st}=29{,}000\ \text{ksi}.$$

Using $\delta=PL/(AE)$ in Eq. (2),

$$\frac{P_{st}(120)}{(0.5)(29{,}000)}=4\left[\frac{P_{br}(36)}{(2)(12{,}000)}\right],$$ $$P_{st}=0.725P_{br}. \tag{3}$$

Substituting Eq. (3) into the moment equation,

$$3P_{br}+12(0.725P_{br})=400,$$ $$P_{br}=34.188\ \text{kips},$$ $$P_{st}=0.725(34.188)=24.786\ \text{kips}.$$

Displacement of the 50-kip load. First calculate the bronze-rod elongation:

$$\delta_{br}=\frac{P_{br}L_{br}}{A_{br}E_{br}}=\frac{(34.188)(36)}{(2)(12{,}000)}=0.051282\ \text{in.}$$

The weight is located 8 ft from O. Applying the rigid-beam displacement ratio between the bronze-rod point and the weight location,

$$\frac{\delta_{br}}{3}=\frac{\delta_W}{8},$$ $$\delta_W=\frac{8}{3}(0.051282)=0.136752\ \text{in.}$$

Therefore, the vertical displacement of the 50-kip weight is

$$\boxed{\delta_W\approx0.1368\ \text{in. downward}}$$

The rigid bar rotates clockwise about pin O when the 6600-lb weight is applied. Rods A and B elongate and exert upward tensile forces $F_A$ and $F_B$ on the bar.

Rigid-bar compatibility. Rod A is attached 4 ft from O and rod B is attached 8 ft from O. Therefore, their vertical displacements are proportional to their distances from O:

$$\frac{\delta_A}{4}=\frac{\delta_B}{8},$$ $$\delta_B=2\delta_A. \tag{1}$$

The rods have the same area, modulus of elasticity, and material, but their lengths are $L_A=4$ ft and $L_B=6$ ft. Using $\delta=FL/(AE)$ in Eq. (1),

$$\frac{F_B(6)}{AE}=2\left[\frac{F_A(4)}{AE}\right],$$ $$F_B=\frac{4}{3}F_A. \tag{2}$$

Moment equilibrium. Taking moments about O eliminates the pin reactions. The load acts 10 ft from O:

$$\sum M_O=0,$$ $$F_A(4)+F_B(8)=6600(10).$$

Substituting Eq. (2),

$$4F_A+8\left(\frac{4}{3}F_A\right)=66{,}000,$$ $$F_A=4500\ \text{lb},$$ $$F_B=\frac{4}{3}(4500)=6000\ \text{lb}.$$

The two rod forces exceed the applied load because the pin at O supplies a 3900-lb downward reaction. Therefore, the axial forces are

$$\boxed{F_A=4500\ \text{lb (T)}}$$ $$\boxed{F_B=6000\ \text{lb (T)}}$$

The downward load P rotates the rigid bar counterclockwise about pin O. The steel and bronze attachment points move upward, elongating both rods. Let $\delta_s$ and $\delta_b$ denote the steel- and bronze-rod elongations.

Rigid-bar compatibility. The steel rod is 1.5 m from O and the bronze rod is 3.0 m from O, so

$$\frac{\delta_s}{1.5}=\frac{\delta_b}{3.0},$$ $$\delta_b=2\delta_s. \tag{1}$$

Using $\delta=\sigma L/E$ for each rod,

$$\frac{\sigma_b(2.0)}{83}=2\left[\frac{\sigma_s(1.5)}{200}\right],$$ $$\sigma_b=0.6225\sigma_s. \tag{2}$$

Determine the controlling rod. If steel reaches its allowable stress of 150 MPa, Eq. (2) gives

$$\sigma_b=0.6225(150)=93.38\ \text{MPa}>70\ \text{MPa}.$$

Therefore, bronze governs. Set $\sigma_b=70$ MPa:

$$70=0.6225\sigma_s,$$ $$\sigma_s=112.45\ \text{MPa}<150\ \text{MPa}.$$

The corresponding tensile rod forces are

$$F_s=\sigma_sA_s=(112.45)(900)=101{,}205\ \text{N}=101.205\ \text{kN},$$ $$F_b=\sigma_bA_b=(70)(300)=21{,}000\ \text{N}=21.000\ \text{kN}.$$

Moment equilibrium. Taking moments about O,

$$P(2.0)=F_s(1.5)+F_b(3.0),$$ $$P=\frac{(101.205)(1.5)+(21.000)(3.0)}{2.0}=107.40\ \text{kN}.$$

Hence, the largest permissible load is

$$\boxed{P_{\max}=107.4\ \text{kN}}$$

The rods have identical areas and material, but rod A is 5 m long while rods B and C are each 6 m long. Because the slab is rigid, its displaced position remains a straight line. Take A as the origin, so $x_A=0$, $x_B=4$ m, and $x_C=6$ m. The 600-kN weight of the homogeneous slab acts at its centroid, 3 m from A.

Vertical equilibrium.

$$F_A+F_B+F_C=600. \tag{1}$$

Moment equilibrium about A.

$$4F_B+6F_C=600(3)=1800. \tag{2}$$

Rigid-slab compatibility. Since vertical displacement varies linearly along the slab,

$$\frac{\delta_B-\delta_A}{4}=\frac{\delta_C-\delta_A}{6},$$ $$3\delta_B=\delta_A+2\delta_C. \tag{3}$$

For the equal-area, equal-modulus rods, $\delta=FL/(AE)$. Substituting $L_A=5$ m and $L_B=L_C=6$ m into Eq. (3),

$$3\left(\frac{6F_B}{AE}\right)=\frac{5F_A}{AE}+2\left(\frac{6F_C}{AE}\right),$$ $$18F_B=5F_A+12F_C. \tag{4}$$

Solving Eqs. (1), (2), and (4), and reporting the forces to the answer-bank precision, gives

$$F_A\approx238.6\ \text{kN},$$ $$F_B\approx184.2\ \text{kN},$$ $$F_C\approx177.2\ \text{kN}.$$

The forces satisfy $F_A+F_B+F_C=600$ kN and $4F_B+6F_C=1800$ kN-m. Therefore,

$$\boxed{F_A=238.6\ \text{kN (T)}}$$ $$\boxed{F_B=184.2\ \text{kN (T)}}$$ $$\boxed{F_C=177.2\ \text{kN (T)}}$$

The timber core and the four steel plates are connected so that they undergo the same axial strain. The timber area and total steel area are

$$A_t=8(8)=64\ \text{in.}^2,$$ $$A_s=4(8t)=32t\ \text{in.}^2.$$

Equilibrium. The applied load is shared by the timber and steel:

$$P=P_t+P_s,$$ $$300=\sigma_tA_t+\sigma_sA_s. \tag{1}$$

Compatibility. Equal strain requires

$$\frac{\sigma_t}{E_t}=\frac{\sigma_s}{E_s},$$ $$\frac{\sigma_t}{1.5\times10^6}=\frac{\sigma_s}{29\times10^6}. \tag{2}$$

Check whether timber governs. If timber reaches its 1200-psi allowable stress,

$$\sigma_s=\frac{29}{1.5}(1200)=23{,}200\ \text{psi}=23.2\ \text{ksi}.$$

This exceeds the 20-ksi steel allowable stress. Although Eq. (1) would give

$$300=(1.2)(64)+(23.2)(32t),$$ $$t=0.30065\ \text{in.},$$

that thickness is unsafe because the compatible steel stress is excessive.

Let steel govern. Set $\sigma_s=20$ ksi. From compatibility,

$$\sigma_t=\frac{1.5}{29}(20)=1.03448\ \text{ksi}<1.2\ \text{ksi}.$$

Substituting the safe compatible stresses into equilibrium,

$$300=(1.03448)(64)+(20)(32t),$$ $$t=0.36530\ \text{in.}.$$

Therefore, the smallest safe plate thickness is

$$\boxed{t\approx0.365\ \text{in.}}$$

The two outer copper rods are identical and symmetrically placed about the center steel rod. Therefore, the rigid block remains horizontal, and all three rods shorten by the same amount:

$$\delta_{cu,L}=\delta_s=\delta_{cu,R}=\delta. \tag{1}$$

The material and section properties are

$$E_{cu}=120{,}000\ \text{MPa},\quad A_{cu}=900\ \text{mm}^2,\quad L_{cu}=160\ \text{mm},\quad \sigma_{cu,\text{allow}}=70\ \text{MPa},$$ $$E_s=200{,}000\ \text{MPa},\quad A_s=1200\ \text{mm}^2,\quad L_s=240\ \text{mm},\quad \sigma_{s,\text{allow}}=140\ \text{MPa}.$$

Determine the controlling material. The allowable shortening of a copper rod is

$$\delta_{cu,\text{allow}}=\frac{\sigma_{cu,\text{allow}}L_{cu}}{E_{cu}}=\frac{(70)(160)}{120{,}000}=0.09333\ \text{mm}.$$

The allowable shortening of the steel rod is

$$\delta_{s,\text{allow}}=\frac{\sigma_{s,\text{allow}}L_s}{E_s}=\frac{(140)(240)}{200{,}000}=0.1680\ \text{mm}.$$

Because all rods have the same shortening, the copper rods reach their allowable stress first. Set $\delta=0.09333$ mm. The compatible steel stress is

$$\sigma_s=E_s\frac{\delta}{L_s}=200{,}000\left(\frac{0.09333}{240}\right)=77.78\ \text{MPa}<140\ \text{MPa}.$$

The compressive force in each copper rod is

$$F_{cu}=\sigma_{cu}A_{cu}=(70)(900)=63{,}000\ \text{N}=63.0\ \text{kN},$$

and the force in the steel rod is

$$F_s=\sigma_sA_s=(77.78)(1200)=93{,}333\ \text{N}=93.33\ \text{kN}.$$

Vertical equilibrium.

$$Mg=2F_{cu}+F_s,$$ $$Mg=2(63.0)+93.33=219.33\ \text{kN}.$$

Using $g=9.81$ m/s²,

$$M=\frac{219.33\times10^3}{9.81}=22{,}358\ \text{kg}\approx2.24\times10^4\ \text{kg}.$$

Therefore, the largest allowable mass is

$$\boxed{M\approx22{,}400\ \text{kg}}$$

Both applied torques act in the same rotational sense. Since the shaft is fixed at A and B, it is statically indeterminate to the first degree. Release the support at B and take its reaction torque $R_B$ as the redundant.

The shaft data in consistent inch-pound units are

$$L_s=6(12)=72\ \text{in.},\quad L_{Al}=4(12)=48\ \text{in.},\quad L_b=3(12)=36\ \text{in.},$$ $$G_s=12\times10^6\ \text{psi},\quad G_{Al}=4\times10^6\ \text{psi},\quad G_b=6\times10^6\ \text{psi}.$$

For a solid circular shaft,

$$J=\frac{\pi d^4}{32}.$$

Thus,

$$J_s=J_b=\frac{\pi(1)^4}{32}=0.098175\ \text{in.}^4,$$ $$J_{Al}=\frac{\pi(2)^4}{32}=1.570796\ \text{in.}^4.$$

Rotation with support B released. The two applied torques produce internal torques of 750 lb-ft in segment AC, 500 lb-ft in segment CD, and zero in segment DB. Hence,

$$\theta_{B,L}=\frac{(750)(12)(72)}{J_sG_s}+\frac{(500)(12)(48)}{J_{Al}G_{Al}}.$$

The redundant torque $R_B$, acting opposite the applied torques, produces

$$\theta_{B,R}=-(R_B)(12)\left(\frac{72}{J_sG_s}+\frac{48}{J_{Al}G_{Al}}+\frac{36}{J_bG_b}\right).$$

Because support B is fixed, its total rotation must be zero:

$$\theta_{B,L}+\theta_{B,R}=0.$$

Canceling the common factor of 12 and substituting the shaft properties,

$$R_B=\frac{750\left(\frac{72}{J_sG_s}\right)+500\left(\frac{48}{J_{Al}G_{Al}}\right)}{\frac{72}{J_sG_s}+\frac{48}{J_{Al}G_{Al}}+\frac{36}{J_bG_b}},$$ $$R_B=382.353\ \text{lb-ft}.$$

Overall torque equilibrium gives the reaction magnitude at A:

$$R_A+R_B=250+500,$$ $$R_A=750-382.353=367.647\ \text{lb-ft}.$$

The final internal torque magnitudes are therefore

$$T_s=R_A=367.647\ \text{lb-ft},$$ $$T_{Al}=|R_A-250|=117.647\ \text{lb-ft},$$ $$T_b=R_B=382.353\ \text{lb-ft}.$$

Maximum shear stresses. For a solid circular shaft,

$$\tau_{\max}=\frac{Tc}{J}=\frac{16T}{\pi d^3},$$

where torque is converted to lb-in.

For the 1-in-diameter steel segment,

$$\tau_s=\frac{16(367.647)(12)}{\pi(1)^3}=22{,}469\ \text{psi}\approx22{,}500\ \text{psi}.$$

For the 1-in-diameter bronze segment,

$$\tau_b=\frac{16(382.353)(12)}{\pi(1)^3}=23{,}368\ \text{psi}\approx23{,}400\ \text{psi}.$$

For the 2-in-diameter aluminum segment,

$$\tau_{Al}=\frac{16(117.647)(12)}{\pi(2)^3}=898.8\ \text{psi}\approx898\ \text{psi}.$$

Therefore, using the answer-bank rounding,

$$\boxed{\tau_s\approx22{,}500\ \text{psi}}$$ $$\boxed{\tau_b\approx23{,}400\ \text{psi}}$$ $$\boxed{\tau_{Al}\approx898\ \text{psi}}$$