MOS-I (Lecture 1-2) Engr. Khalid Yousaf

MECHANICS OF SOLIDS-I Simple Stress and Strain: Kinds of stresses and strains, Difference between stress and pressure, Load Extension Diagrams for different Materials, Hook’s Law, Moduli of elasticity, Lateral strain, Volumetric Strain, Poisson’s Ratio, Temperature stresses and Compound bars. Analysis of Beams: Shear force and bending moment diagrams of beams under different loading conditions, Theory of simple bending, Moment of resistance and section modulus, Applications of flexure formula, Shear Stresses in Beams, Shear Centre, Shear Flow. Column and Struts: A short and long axially loaded columns, their modes of failure, and conditions, equivalent length, Euler’s formula, and Empirical formula like Rankine Gordon Formula etc., Slenderness Ratio. Circular Shafts: Theory of Torsion for solid and hollow circular shafts. Springs: Open coil springs, closed coil springs, leaf springs. Strain Energy: Strain Energy due to direct loads, shear force, bending moments, torque and impact loads. Recommended Books: 1. Pytel, A. & F. L.Singer, Strength of Material, Harper & Row Publishers, New York. 2. Hibbler, R. C., Mechanics of Materials, Prentice Hall, 6th Edition, 2004.

3. Warnock, F. V., Benham, P. P., Mechanics of Solids and Strength of Materials, Pitman Publishing, 1970. 4. James M. Gere & Barry. J. Goodno, Mechanics of Materials, 7th Edition, 2008, CL Engineering 5. James M. Gere & Stephen P. Timoshenko, Mechanics of Materials, 4th Edition, 1997, PWS Pub Co.

Normal Stresses:

Stress is defined as the strength of a material per unit area or unit strength. It is the force on a member divided by area, which carries the force, formerly express in psi, now in N/mm2 or MPa. Mathematically 𝛔 = 𝐴𝑃 where P is the applied normal load in Newton and A is the area in mm2. The maximum stress in tension or compression occurs over a section normal to the load. Normal stress is either tensile stress or compressive stress. Members subject to pure tension (or tensile force) is under tensile stress, while compression members (members subject to compressive force) are under compressive stress. Compressive force will tend to shorten the member. Tension force on the other hand will tend to lengthen the member.

Problem#101: The bar ABCD in Fig. (a) consists of three cylindrical steel segments with different lengths and cross-sectional areas. Axial loads are applied as shown. Calculate the normal stress in each segment. Solution: We begin by using equilibrium analysis to compute the axial force in each segment of the bar. Step#1:Considering section 1 and summing up the forces acting on it. 𝑃𝐴𝐵 = 4000 lb 𝐴𝐴𝐵 =1.2 in2 𝛔 = 𝐴𝑃

𝛔𝐴𝐵 = 𝛔𝐴𝐵 =

𝑃𝐴𝐵 𝐴 𝑃𝐴𝐵 𝐴 4000 1.2

𝛔𝐴𝐵 = 𝛔𝐴𝐵 = 3333.3 𝑃𝑠𝑖 (𝑇) Step#2: Adding up forces in section 2, we get 𝑃𝐵𝐶 = - 5000 lb 𝐴𝐵𝐶 =1.8 in2 𝛔𝐵𝐶 =

𝑃𝐵𝐶 𝐴 5000 − 1.8

𝛔𝐵𝐶 = 𝛔𝐵𝐶 = −2777.8 𝑃𝑠𝑖 =2777.7 Psi (C)

Step#3: Taking section 3 under consideration. 𝑃𝐶𝐷 = 7000 lb 𝐴𝐶𝐷 =1.6 in2 𝛔𝐶𝐷 =

𝑃𝐶𝐷 𝐴 7000 1.6

𝛔𝐶𝐷 = 𝛔𝐶𝐷 = 4375 𝑃𝑠𝑖 (𝑇)

Problem 102: For the truss shown in the shown in the figure. Determine the stress in the member AC and BD. The cross sectional area of the each member of the truss is 900 mm2. Solution: The three assumptions taken in the elementary analysis of the truss are , 1- Weights of the members are neglected. 2- All connections are smooth pins. 3- All external loads are applied directly to the pins. Step#1: Calculate the reactions at each support ∑ Fy =0 Ay + Hy = 50 kN ∑ MA =0 30 × 4 + 70 × 12 − Hy × 16 = 0 Hy = 60 kN And Ay = 40 kN Step#2: Considering Section 1 ∑ Fy =0 Ay + 𝟑𝟓FAB = 0 FAB = − 𝟓𝟑Ay FAB = − 𝟓𝟑 × 40 = − 66.7 kN

1

∑ Fy =0 FAC + 𝟒𝟓FAB = 0 FAC = − 𝟒𝟓FAB FAC = − 𝟒𝟓 × (−66.7) C FAC = 53.4 𝑘𝑁 Step#3: ∑ ME =0 Ay × 8 + FBD × 3 − 30 × 4 = 0 40 × 8 + FBD × 3 − 30 × 4 = 0 3FBD = −200 FBD = −66.7 kN Step#4: The area of each member 𝐴 = 900 mm2 =900 × 10−6 m2 The stresses in members AC and BD are, 𝛔 = 𝐴𝑃 53.4×103 𝛔𝐴𝐶 = 900×10−6 𝛔𝐴𝐶 = 59.3 × 106 𝛔𝐴𝐶 = 59.3MPa (T) −66.7×103 𝛔𝐵𝐷 = 900×10−6 𝛔𝐵𝐷 = −74.1 × 106 𝛔𝐵𝐷 = −74.1MPa =74.1MPa (C)

Problem 103: Figure (a) shows a two-member truss supporting a block of weight W. The cross-sectional areas of the members are 800 mm2 for AB and 400 mm for AC . Determine the maximum safe value of W if the working stresses are 110 MPa for AB and 120 MPa for AC. Solution. Being members of a truss, AB and AC can be considered to be axially loaded bars . The forces in the bars can be obtained by analyzing the FBD of pin A in Fig. (b). The equilibrium equations are, ∑ Fx =0 PAC 𝐶𝑜𝑠60𝑜 - PAB 𝐶𝑜𝑠40𝑜 = 0 PAC 𝑆𝑖𝑛60𝑜 + PAB 𝑆𝑖𝑛40𝑜 - W= 0 Solving above equations simultaneously, PAC = 0.7779𝑊 PAB = 0.5077𝑊 Area of each member A = 800 mm2 = 800 × 10−6 m2

Stress in each member: 𝛔= 𝑃𝐴 𝜎𝐴𝐶 = 𝑃𝐴𝐴𝐶

120 × 106 = 0.7779W 800 × 10−6 W = 123409 N = 123 kN

𝜎𝐴𝐵 = 𝑃𝐴𝐵𝐶

110 × 106 = 0.5077W 800 × 10−6 W = 173330 N = 173 kN

For safe loading, use W= 123 kN

Alternatively: Make a polygon with the force vectors as shown in the figure. Applying Law of Sine to this polygon , we have For Member AC: 𝑃𝐴𝐶 W o = sin 50 sin 100 𝑃𝐴𝐶 =0.7779W 𝛔𝐴𝐶 𝐴=0.7779W 120(800 × 10−6)=0.7779W W = 123409 N =123 kN For wire AB: 𝑃𝐴𝐵 W o = sin 30 sin 100 𝑃𝐴𝐵 =0.5077W 𝛔𝐴𝐵 𝐴=0.5077W 110 × 106(800 × 10−6)=0.5077W W = 173330 N = 173 kN For safe load W use W = 123 kN o

o

Problem 104 A hollow steel tube with an inside diameter of 100 mm must carry a tensile load of 400 kN. Determine the outside diameter of the tube if the stress is limited to 120 MN/m2. Solution: Given that Inside diameter d = 100 mm Tensile load P = 400 kN = 400000 N Stress 𝛔 = 120 MN/m2 Outer diameter D=? Since

or Area

𝛔 = 𝐴𝑃 𝑃 = 𝛔A 𝐴 = π4(D2 − d2) 400000 = 120 π4(D2 − d2) 400000 = 30π (D2 − 1002) 400000+300000π

𝐷2 = 30π 𝐷 = 119.35 𝑚𝑚

Problem 105: A homogeneous 800 kg bar AB is supported at either end by a cable as shown in Fig. P-105. Calculate the smallest area of each cable if the stress is not to exceed 90 MPa in bronze and 120 MPa in steel. Solutions: Given that Mass of the bar M = 800 kg W = 7848 N Stress in bronze Stress in steel

σbr = 90 Mpa σst = 90 Mpa

By symmetry: Pbr =Pst =12(7848) Pbr=3924 N Pst=3924 N For bronze cable:

Pbr = σbrAbr 3924 = 90Abr Abr = 43.6 mm2 For steel cable: Pst = σstAst 3924 = 120Ast Ast = 32.7 mm2

Problem106: The homogeneous bar shown in Fig. P-106 is supported by a smooth pin at C and a cable that runs from A to B around the smooth peg at D. Find the stress in the cable if its diameter is 0.6 inch and the bar weighs 6000 lb. Solution: Given that Diameter of the cable d = 0.6 in Weight of the bar w = 600 lb ΣMC = 0 5T+10( 𝟑𝟑𝟒T) = 5(6000) T=2957.13 lb T = σA 1 2957.13 = σ[4π(0. 62)] 𝛔 = 10458.72 psi

Problem 107: A rod is composed of an aluminum section rigidly attached between steel and bronze sections, as shown in Fig. P-107. Axial loads are applied at the positions indicated. If P = 3000 lb and the cross sectional area of the rod is 0.5 in2, determine the stress in each section. Solution: Given that Applied load P = 3000 lb = 3 kips cross-Sectional Area A = 0.5 in2 For steel: σstAst = Pst σst(0.5) = 4P σst(0.5) =4 ⨉ 3 σst = 24 ksi For aluminum: σalAal = Pal σal(0.5) = 4P σal(0.5) = 4 ⨉ 3 σal = 24 ksi For bronze: σbrAbr = Pbr σbr(0.5) = 9 σbr = 18 ksi

Problem 108: An aluminum rod is rigidly attached between a steel rod and a bronze rod as shown in Fig. P-108. Axial loads are applied at the positions indicated. Find the maximum value of P that will not exceed a stress in steel of 140 MPa, in aluminum of 90 MPa, or in bronze of 100 MPa. Solutions: Given that Stress in steel 𝛔𝑠𝑡 = 140 Mpa Steel area 𝐴𝑠𝑡 = 500 mm2 Stress in aluminum 𝛔𝑎𝑙 = 90 Mpa Aluminum area 𝐴𝑎𝑙 = 400 mm2 Stress in bronze 𝛔𝑏𝑟 = 100 Mpa Bronze area 𝐴𝑏𝑟 = 200 mm2 Max. applicable load P=? For bronze: σbrAbr = Pbr 1000(200) = 2P P = 10000 N For aluminum: σalAal = Pal 90(400) = P P = 36000 N For steel: σstAst = Pst 100(500) = 5P P = 10000 N For safe value of P , use P = 10000 N = 10 kN

Problem 109: Determine the largest weight W that can be supported by two wires shown in Fig. P-109. The stress in either wire is not to exceed 30 ksi. The cross-sectional areas of wires AB and AC are 0.4 in2 and 0.5 in2, respectively. Solution: Given that Maximum applicable stress 𝛔 = 30 ksi Cross-sectional area of AB 𝐴𝐴𝐵 = 0.4 in2 Cross-sectional area of AB 𝐴𝐵𝐶 = 0.5 in2 Largest supportable weight W=? The free body diagram of the joint A is given by For wire AB: By sine law (from the force polygon): 𝑇𝐴𝐵 W o = sin 40 sin 80o 𝑇𝐴𝐵 =0.6527W 𝛔𝐴𝐵 𝐴𝐴𝐵 =0.6527W 30(0.4)=0.6527W W = 18.4 kips For wire AC: 𝑇𝐴𝐶 W o = sin 60 sin 80o 𝑇𝐴𝐶 =0.8794W 𝛔𝐴𝐶 𝐴𝐴𝐶 =0.8794W 30(0.5)=0.6527W W = 17.1 kips For safe load W use W = 17.1 kips

Problem 110: A 12-inches square steel bearing plate lies between an 8-inches diameter wooden post and a concrete footing as shown in Fig. P-110. Determine the maximum value of the load P if the stress in wood is limited to 1800 psi and that in concrete to 650 psi. Solution: Given that Area of steel bearing plate 𝐴𝑠𝑡 = 12 in2 Diameter of wooden post 𝑑𝑤𝑑 = 8 in Stress in wood 𝛔𝑤𝑑 = 1800 Psi Stress in concrete 𝛔𝑐𝑜𝑛 = 650 Psi For Wood: Pw = σwAw Pw = σw π4 d2 Pw = 1800[π4 82] Pw = 90477.9 lb For concrete: PC = 𝛔CAC PC = 650 ⨉ 122 PC = 93600 lb. For safe load ‘P’ use P = Pw = 90477.9 lb

Problem 111: For the truss shown in Fig. P-111, calculate the stresses in members CE, DE, and DF. The cross sectional area of each member is 1.8 in2. Indicate tension (T) or compression (C). Solution: Given that Cross Sectional Area of each member A = 1.8 in2

σ σ σ

CE

=?

DE

=?

=? Σ 𝐹𝑦 = 0 RA + RF =30 (1) Σ 𝑀𝐴 = 0 24 RF =16(30) RF =20k put in (1) RA = 10k At joint F: Σ 𝐹𝑦 = 0 - 35DF = 20 DF = -33.33k = 33.33k (C) At joint D: By symmetry DF

BD = DF = 33.33k (C) Σ 𝐹𝑦 = 0 𝐷𝐸 = 35𝐵𝐷 + 35DF

𝐷𝐸 =

3 5

33.33 +

3 5

33.33

𝐷𝐸 = 40k (T) At joint E: Σ 𝐹𝑦 = 0 3 5

𝐶𝐸 + 30 = 40

CE = 10.67k (T) Stresses: (Stress = Force/ Area)

σCE = 10.67 1.8 𝜎𝐶𝐸 = 5.93 𝑘𝑠𝑖 (T) 40 𝜎𝐷𝐸 = 1.8

𝜎𝐷𝐸 = 22.22 𝑘𝑠𝑖 (𝑇)

𝜎𝐷𝐹 = 33.33 1.8

𝜎𝐷𝐹 = 18.52 𝑘𝑠𝑖 (𝐶)

Problem#112: Determine the cross-sectional areas of members AG, BC, and CE for the truss shown in Fig. P-112. The stresses are not to exceed 20 ksi in tension and 14 ksi in compression. A reduced stress in compression is specified to reduce the danger of buckling. Soln. ∑𝐹𝑦 = 0 𝐴𝑦 = 40 + 25 = 65 𝑘 ∑𝑀𝐴 = 0 18 × 𝐷𝑥 = 40 × 4 + 25 × 8 𝐷𝑥 = 20𝑘 ∑𝐹𝑥 = 0 𝐴𝑥 = 𝐷𝑥 = 20 𝑘 Check: ∑𝑀𝐷 = 0 12 × 𝐴𝑦 = 18 × 𝐴𝑥 + 40 × 8 + 25 × 4 12 × 65 = 18 × 20 + 40 × 8 + 25 × 4 780 𝑘𝑖𝑝. 𝑓𝑡 = 780 𝑘𝑖𝑝. 𝑓𝑡 (ok!) For member AG (At joint A): ∑𝐹𝑦 = 0 3 𝐴𝐵 13

= 65

AB = 78.12 k

∑𝐹𝑥 = 0 𝐴𝐺 + 20 =

2 𝐴𝐵 13 2 × 78.12 13

𝐴𝐺 + 20 = 𝐴𝐺 = 20.33 𝑘 since 𝑃 𝜎= 𝐴

𝜎𝑡𝑒𝑛𝑠𝑖𝑜𝑛 = 20.33

(𝑇)

𝑃 𝐴

20= 𝐴 𝐴𝐺 𝐴𝐴𝐺 = 1.17 𝑖𝑛2 For member BC (At section through MN): ∑𝑀𝐹 = 0 2 12 × 𝐷𝑥 = −6 × 13 𝐵𝐶 2

12 × 20 = −6 × 13 𝐵𝐶 B𝐶 = −72.11 𝑘 = 72.11 𝑘 𝑃 𝜎𝑐𝑜𝑚𝑝𝑟𝑒𝑠𝑖𝑜𝑛 = 𝐴 72.11 𝐴𝐵𝐶

14 = 𝐴𝐵𝐶 = 5.15 𝑖𝑛2

𝐵𝐶

(𝐶)

For member CE (At joint D): ∑𝐹𝑥 = 0 2 20 = 13 𝐶𝐷 CD = 36.06 k ∑𝐹𝑦 = 0 3 𝐶𝐷 13

𝐷𝐸 =

3 13

× 36.06 𝐷𝐸 = 30 𝑘 𝐷𝐸 =

At joint E: ∑𝐹𝑦 = 0 30 =

3 𝐸𝐹 13

𝐸𝐹 = 36.06 𝑘 ∑𝐹𝑥 = 0 2 C𝐸 = − 13 𝐸𝐹 C𝐸 = −

2 × 13

36.06

𝐶𝐸 = −20 𝑘 = 20 𝑘 𝑃 𝜎𝑐𝑜𝑚𝑝𝑟𝑒𝑠𝑖𝑜𝑛 = 14 =

20 𝐴𝐴𝐺

𝐴

𝐴𝐴𝐺 = 1.43 𝑖𝑛2

(𝐶)

Problem#113:Find the stresses in members BC, BD, and CF for the truss shown in Fig. P-113. Indicate the tension or compression. The cross sectional area of each member is 1600 mm2. Soln. For member BD: (See FBD 01) ∑𝑀𝐶 = 0 4 3 × 𝐵𝐷 = 3 × 60 5 𝐵𝐷 = 75 𝑘 (T) 𝑃 𝜎= 𝐴

75(1000)

𝜎𝐵𝐷 = 1600 𝜎𝐵𝐷 = 46.875 Mpa (T) For member CF: (See FBD 01) ∑𝑀𝐷 = 0 1 4 × 𝐶𝐹 = 4 × 900 + 7 × 60 2

𝐶𝐹 = 275.77 𝑘𝑁 𝑃 𝜎= 𝐴

(𝐶)

275.77(1000)

𝜎𝐶𝐹 = 1600 𝜎𝐶𝐹 = 172.375𝑀𝑃𝑎

(𝐶)

For member BC: (See FBD 02) ∑𝑀𝐷 = 0 4 × 𝐵𝐶 = 7 × 60 𝐵𝐶 = 105 𝑘𝑁 (𝐶) 𝑃 𝜎= 𝐴

105(1000)

𝜎𝐵𝐶 = 1600 𝜎𝐵𝐶 = 62.625𝑀𝑃𝑎

(𝐶)

Problem#114:The homogeneous bar ABCD shown in Fig. P-114 is supported by a cable that runs from A to B around the smooth peg at E, a vertical cable at C, and a smooth inclined surface at D. Determine the mass of the heaviest bar that can be supported if the stress in each cable is limited to 100 MPa. The area of the cable AB is 250 mm2 and that of the cable at C is 300 mm2. Soln. ∑𝐹𝑥 = 0 𝑇𝐴𝐵 𝐶𝑜𝑠300 = 𝑅𝐷 Sin300 𝑅𝐷 = 1.1305𝑇𝐴𝐵 ∑𝐹𝑦 = 0 𝑇𝐴𝐵 + 𝑇𝐶 + 6 × 𝑇𝐴𝐵 Sin300 + 𝑅𝐷 Cos500 = 𝑊 𝑇𝐴𝐵 + 𝑇𝐶 + 𝑇𝐴𝐵 Sin300 + 1.1305𝑇𝐴𝐵 Cos500 = 𝑊 𝑇𝐶 + 2.226𝑇𝐴𝐵 = 𝑊 𝑇𝐶 = 𝑊 − 2.226𝑇𝐴𝐵 ∑𝑀𝐷 = 0 4 × 𝑇𝐴𝐵 + 2 × 𝑇𝐶 + 6 × 𝑇𝐴𝐵 Sin300 = 3𝑊 7 × 𝑇𝐴𝐵 + 2 × (𝑊 − 2.226𝑇𝐴𝐵 ) = 3𝑊 2.5466𝑇𝐴𝐵 = 𝑊 𝑇𝐴𝐵 = 0.3927𝑊 𝑇𝐶 = 𝑊 − 2.226𝑇𝐴𝐵 𝑇𝐶 = 𝑊 − 2.2267(0.3927𝑊) 𝑇𝐶 = 0.1256𝑊

Based on cable AB: 𝑇 𝜎𝐴𝐵 = 𝐴𝐵 𝐴𝐴𝐵 0.3967𝑊 = 250

100 𝑊 = 63661.83 𝑁 Based on cable at C: 𝑇 𝜎𝐶 = 𝐶 𝐴𝐶 0.1256𝑊 = 300

100 𝑊 = 238853.5 𝑁 Safe value of W 𝑊 = 63661.83 𝑁 𝑊 = 𝑚𝑔 63661.83 = 𝑚 (9.81) 𝑚 = 6490 𝑘𝑔 𝑚 = 6.49 𝑀𝑔

Assignment.

Q#1: Axial loads are applied to the compound rod that is composed of an aluminum segment rigidly connected between steel and bronze segments. What is the stress in each material given that P = 10 kN? Q#2: Axial loads are applied to the compound rod that is composed of an aluminum segment rigidly connected between steel and bronze segments. Find the largest safe value of P if the working stresses are 120 MPa for steel, 68MPa for aluminum, and110 MPa for bronze.

Q#3: The wood pole is supported by two cables of 0.25 in. diameter. The turnbuckles in the cables are tightened until the stress in the cables reaches 60000 psi. If the working compressive stress for wood is 200 psi, determine the smallest permissible diameter of the pole.

Q#4:Find the maximum allowable value of P for the column. The cross-sectional areas and working stresses (𝜎𝑤 ) are shown in the figure. Q#5:Determine the smallest safe cross-sectional areas of members CD, GD, and GF for the truss shown. The working stresses are 140 MPa in tension and 100 MPa in compression.(The working stress in compression is smaller to reduce the danger of buckling.) Q#6.Find the stresses in members BC, BD, and CF for the truss shown.Indicate tension or compression.The crosssectional area of each member is 1400 mm2. Q#7:Determine the smallest allowable cross-sectional areas of members BD, BE, and CE of the truss shown. The working stresses are 20 000 psi in tension and 12000 psi in compression. (A reduced stress in compression is specified to reduce the danger of buckling.)

Engr. Khalid Yousaf BS Civil Engineering(The University of Lahore)

Q#8: The uniform 300-lb bar AB carries a 500-lb vertical force at A. The bar is supported by a pin at B and the 0.5 in. diameter cable CD. Find the stress in the cable. Q#9: Determine the smallest allowable cross-sectional areas of members CE, BE, and EF for the truss shown. The working stresses are 20 ksi in tension and 14 ksi in compression. (The working stress in compression is smaller to reduce the danger of buckling.) Q#10:For the Pratt bridge truss and loading shown, determine the average normal stress in member BE, knowing that the cross-sectional area of that member is 5.87 in2.

Q#11:A solid brass rod AB and a solid aluminum rod BC are connected together by a coupler at B, as shown in Figure Q#11.The diameters of the two segments are 𝑑1 = 60𝑚𝑚 and 𝑑2 = 50𝑚𝑚, respectively. Determine the axial stresses 𝜎1 (in rod AB) and 𝜎2 (in rod BC).

Engr. Khalid Yousaf

BS Civil Engineering

Q#12:A hollow circular nylon pipe (see Fig 1-23) supports a load 𝑃𝐴 = 7800𝑁,which is uniformly distributed around a cap plate at the top of the lower pipe. A second load 𝑃𝐵 is applied at the bottom. The inner and outer diameters of the upper and lower parts of the pipe are 𝑑1 = 51𝑚𝑚, 𝑑2 = 60𝑚𝑚,𝑑3 = 57𝑚𝑚, and 𝑑4 = 63𝑚𝑚, respectively. The upper pipe has a length 𝐿1 = 350𝑚𝑚; the lower pipe length is 𝐿2 = 400𝑚𝑚.Neglect the self weight of the pipes. (a) Find 𝑃𝐵 so that the tensile stress in upper part is 14.5 MPa. What is the resulting stress in the lower part? (b) If 𝑃𝐴 remains unchanged, find the new value of 𝑃𝐵 so that upper and lower parts have same tensile stress.

(c) Find the tensile strains in the upper and lower pipe segments for the loads in part (b) if the elongation of the upper pipe segment is known to be 3.56 mm and the downward displacement of the bottom of the pipe is 7.63 mm.

The University of Lahore

Q#13: A 1- in. diameter solid bar (1), a square solid bar (2), and a circular tubular member with 0.2 in. wall thickness (3), each supports an axial tensile load of 5 kips. (a) Determine the axial stress in bar (1). (b) If the axial stress in each of the other bars is 6 ksi, what is the dimension, b, of the square bar, and what is the outer diameter, c, of the tubular member? Q#14:The Washington Monument (Figure Q#14.) stands 555 ft. high and weighs 181,700 kips (i.e., approximately 182 million pounds). The monument was made from over 36,000 blocks of marble and granite. As shown in Fig. 1b, the base of the monument is a square that is 665.5 in. long on each side, and the stone walls at the base are 180 in. thick.Determine the compressive stress that the foundation exerts over the cross section at the base of the monument, assuming that this normal stress is uniform. Q#15:The pin-jointed planar truss in is subjected to a single downward force P at joint A. All members have a cross-sectional area of 500 mm2. The allowable stress in tension is (𝜎𝑇 )allow 300 MPa, while the allowable stress (magnitude) in compression is (𝜎𝐶 )allow 200 MPa. Determine the allowable load, 𝑃allow .

Q#16:A column in a two-story building is fabricated from square structural tubing having the cross-sectional dimensions shown in Fig b. Axial loads 𝑃𝐴 = 200 𝑘𝑁 and 𝑃𝐵 = 200 𝑘𝑁 are applied to the column at levels A and B, as shown in Fig a. Determine the axial stress 𝜎1 in segment AB of the column and the axial stress 𝜎2 in segment BC of the column. Neglect the weight of the column itself. Q#17:Each member of the truss in Fig. is a solid circular rod with diameter 𝑑 = 10 𝑚𝑚. Determine the axial stress 𝜎1 in the truss member (1) and the axial stress 𝜎2 in the truss member (6). Q#18:The three-part axially loaded member in Fig. consists of a tubular segment (1) with outer diameter (𝑑𝑜 )1 = 1.00 𝑖𝑛. and inner diameter (𝑑𝑖 )1 = 0.75 𝑖𝑛. a solid circular rod segment (2) with diameter 𝑑2 = 1.00 𝑖𝑛 ., and another solid circular rod segment (3) with diameter 𝑑2 = 0.75 𝑖𝑛. The line of action of each of the three applied loads is along the centroidal axis of the member. Determine the axial stresses 𝜎1 , 𝜎2 , and 𝜎3 in each of the three respective segments.