EBK MECHANICS OF MATERIALS
7th Edition
ISBN: 9780100257061
Author: BEER
Publisher: YUZU
expand_more
expand_more
format_list_bulleted
Concept explainers
Videos
Textbook Question
Chapter 5.3, Problem 88P
Solve Prob. 5.87, assuming that the T-shaped beam is inverted.
5.87 Determine the largest permissible value of P for the beam and loading shown, knowing that the allowable normal stress is +8 ksi in tension and –18 ksi in compression.
Fig. P5.87
Expert Solution & Answer
Want to see the full answer?
Check out a sample textbook solution
Students have asked these similar questions
!
Required information
Air at 25°C (cp=1006 J/kg.K) is to be heated to 58°C by hot oil at 80°C (cp = 2150 J/kg.K) in a cross-flow heat exchanger
with air mixed and oil unmixed. The product of heat transfer surface area and the overall heat transfer coefficient is 750
W/K and the mass flow rate of air is twice that of oil.
NOTE: This is a multi-part question. Once an answer is submitted, you will be unable to return to this part.
Air
Oil
80°C
Determine the effectiveness of the heat exchanger.
In an industrial facility, a counter-flow double-pipe heat exchanger uses superheated steam at a temperature of 155°C to
heat feed water at 30°C. The superheated steam experiences a temperature drop of 70°C as it exits the heat exchanger.
The water to be heated flows through the heat exchanger tube of negligible thickness at a constant rate of 3.47 kg/s. The
convective heat transfer coefficient on the superheated steam and water side is 850 W/m²K and 1250 W/m²K,
respectively. To account for the fouling due to chemical impurities that might be present in the feed water, assume
a fouling factor of 0.00015 m²-K/W for the water side.
The specific heat of water is determined at an average temperature of (30 +70)°C/2 = 50°C and is taken to be
J/kg.K.
Cp=
4181
Water
Steam
What would be the required heat exchanger area in case of parallel-flow arrangement?
The required heat exchanger area in case of parallel-flow arrangement is
1m².
A single-pass crossflow heat exchanger is used to cool jacket water (cp = 1.0 Btu/lbm.°F) of a diesel engine from 190°F to 140°F, using
air (Cp = 0.245 Btu/lbm.°F) at inlet temperature of 90°F. Both air flow and water flow are unmixed. If the water and air mass flow rates
are 85500 lbm/h and 400,000 lbm/h, respectively, determine the log mean temperature difference for this heat exchanger. Assume
the correction factor F to be 0.92.
Air flow
(unmixed)
Water flow
(unmixed)
The log mean temperature difference of the heat exchanger is
°F.
Chapter 5 Solutions
EBK MECHANICS OF MATERIALS
Ch. 5.1 - 5.1 through 5.6 For the beam and loading shown,...Ch. 5.1 - 5.1 through 5.6 For the beam and loading shown,...Ch. 5.1 - 5.1 through 5.6 For the beam and loading shown,...Ch. 5.1 - 5.1 through 5.6 For the beam and loading shown,...Ch. 5.1 - 5.1 through 5.6 For the beam and loading shown,...Ch. 5.1 - 5.1 through 5.6 For the beam and loading shown,...Ch. 5.1 - 5.7 and 5.8 Draw the shear and bending-moment...Ch. 5.1 - 5.7 and 5.8 Draw the shear and bending-moment...Ch. 5.1 - 5.9 and 5.10 Draw the shear and bending-moment...Ch. 5.1 - 5.9 and 5.10 Draw the shear and bending-moment...
Ch. 5.1 - 5.11 and 5.12 Draw the shear and bending-moment...Ch. 5.1 - 5.11 and 5.12 Draw the shear and bending-moment...Ch. 5.1 - 5.13 and 5.14 Assuming that the reaction of the...Ch. 5.1 - 5.13 and 5.14 Assuming that the reaction of the...Ch. 5.1 - 5.15 and 5.16 For the beam and loading shown,...Ch. 5.1 - 5.15 and 5.16 For the beam and loading shown,...Ch. 5.1 - For the beam and loading shown, determine the...Ch. 5.1 - For the beam and loading shown, determine the...Ch. 5.1 - 5.19 and 5.20 For the beam and loading shown,...Ch. 5.1 - 5.19 and 5.20 For the beam and loading shown,...Ch. 5.1 - Draw the shear and bending-moment diagrams for the...Ch. 5.1 - 5.22 and 5.23 Draw the shear and bending-moment...Ch. 5.1 - 5.22 and 5.23 Draw the shear and bending-moment...Ch. 5.1 - 5.24 and 5.25 Draw the shear and bending-moment...Ch. 5.1 - 5.24 and 5.25 Draw the shear and bending-moment...Ch. 5.1 - Knowing that W = 12 kN, draw the shear and...Ch. 5.1 - Determine (a) the magnitude of the counterweight W...Ch. 5.1 - Determine (a) the distance a for which the...Ch. 5.1 - Knowing that P = Q = 480 N, determine (a) the...Ch. 5.1 - Solve Prob. 5.29, assuming that P = 480 N and Q =...Ch. 5.1 - Determine (a) the distance a for which the...Ch. 5.1 - A solid steel rod of diameter d is supported as...Ch. 5.1 - A solid steel bar has a square cross section of...Ch. 5.2 - Using the method of Sec. 5.2, solve Prob. 5.1a....Ch. 5.2 - Using the method of Sec. 5.2, solve Prob. 5.2a....Ch. 5.2 - Prob. 36PCh. 5.2 - Prob. 37PCh. 5.2 - Using the method of Sec. 5.2, solve Prob. 5.5a....Ch. 5.2 - Using the method of Sec. 5.2, solve Prob. 5.6a....Ch. 5.2 - Using the method of Sec. 5.2, solve Prob. 5.7. 5.7...Ch. 5.2 - Using the method of Sec. 5.2, solve Prob. 5.8. 5.7...Ch. 5.2 - Prob. 42PCh. 5.2 - Using the method of Sec. 5.2, solve Prob. 5.10....Ch. 5.2 - 5.44 and 5.45 Draw the shear and bending-moment...Ch. 5.2 - 5.44 and 5.45 Draw the shear and bending-moment...Ch. 5.2 - Prob. 46PCh. 5.2 - Prob. 47PCh. 5.2 - Prob. 48PCh. 5.2 - Using the method of Sec. 5.2, solve Prob. 5.20....Ch. 5.2 - 5.50 and 5.51 Determine (a) the equations of the...Ch. 5.2 - 5.50 and 5.51 Determine (a) the equations of the...Ch. 5.2 - 5.52 and 5.53 Determine (a) the equations of the...Ch. 5.2 - 5.52 and 5.53 Determine (a) the equations of the...Ch. 5.2 - 5.54 and 5.55 Draw the shear and bending-moment...Ch. 5.2 - 5.54 and 5.55 Draw the shear and bending-moment...Ch. 5.2 - 5.56 and 5.57 Draw the shear and bending-moment...Ch. 5.2 - 5.56 and 5.57 Draw the shear and bending-moment...Ch. 5.2 - 5.58 and 5.59 Draw the shear and bending-moment...Ch. 5.2 - 5.58 and 5.59 Draw the shear and bending-moment...Ch. 5.2 - Knowing that beam AB is in equilibrium under the...Ch. 5.2 - Knowing that beam AB is in equilibrium under the...Ch. 5.2 - The beam AB supports two concentrated loads P and...Ch. 5.2 - The beam AB supports a uniformly distributed load...Ch. 5.2 - Beam AB supports a uniformly distributed load of 2...Ch. 5.3 - 5.65 and 5.66 For the beam and loading shown,...Ch. 5.3 - 5.65 and 5.66 For the beam and loading shown,...Ch. 5.3 - 5.67 and 5.68 For the beam and loading shown,...Ch. 5.3 - 5.67 and 5.68 For the beam and loading shown,...Ch. 5.3 - 5.69 and 5.70 For the beam and loading shown,...Ch. 5.3 - 5.69 and 5.70 For the beam and loading shown,...Ch. 5.3 - 5.71 and 5.72 Knowing that the allowable normal...Ch. 5.3 - 5.71 and 5.72 Knowing that the allowable normal...Ch. 5.3 - 5.73 and 5.74 Knowing that the allowable normal...Ch. 5.3 - 5.73 and 5.74 Knowing that the allowable normal...Ch. 5.3 - 5.75 and 5.76 Knowing that the allowable normal...Ch. 5.3 - 5.75 and 5.76 Knowing that the allowable normal...Ch. 5.3 - 5.77 and 5.78 Knowing that the allowable normal...Ch. 5.3 - 5.77 and 5.78 Knowing that the allowable normal...Ch. 5.3 - A steel pipe of 100-mm diameter is to support the...Ch. 5.3 - Two metric rolled-steel channels are to be welded...Ch. 5.3 - Two rolled-steel channels are to be welded back to...Ch. 5.3 - Two L4 3 rolled-steel angles are bolted together...Ch. 5.3 - Assuming the upward reaction of the ground to be...Ch. 5.3 - Assuming the upward reaction of the ground to be...Ch. 5.3 - Determine the largest permissible distributed load...Ch. 5.3 - Solve Prob. 5.85, assuming that the cross section...Ch. 5.3 - Determine the largest permissible value of P for...Ch. 5.3 - Solve Prob. 5.87, assuming that the T-shaped beam...Ch. 5.3 - Beams AB, BC, and CD have the cross section shown...Ch. 5.3 - Beams AB, BC, and CD have the cross section shown...Ch. 5.3 - Each of the three rolled-steel beams shown...Ch. 5.3 - A 54-kip load is to be supported at the center of...Ch. 5.3 - A uniformly distributed load of 66 kN/m is to be...Ch. 5.3 - A roof structure consists of plywood and roofing...Ch. 5.3 - Solve Prob. 5.94, assuming that the 6-kN...Ch. 5.3 - Prob. 96PCh. 5.3 - Assuming that the front and rear axle loads remain...Ch. 5.4 - 5.98 through 5.100 (a) Using singularity...Ch. 5.4 - 5.98 through 5.100 (a) Using singularity...Ch. 5.4 - 5.98 through 5.100 (a) Using singularity...Ch. 5.4 - 5.101 through 5.103 (a) Using singularity...Ch. 5.4 - Prob. 102PCh. 5.4 - Prob. 103PCh. 5.4 - Prob. 104PCh. 5.4 - Prob. 105PCh. 5.4 - Prob. 106PCh. 5.4 - Prob. 107PCh. 5.4 - Prob. 108PCh. 5.4 - Prob. 109PCh. 5.4 - Prob. 110PCh. 5.4 - Prob. 111PCh. 5.4 - Prob. 112PCh. 5.4 - 5.112 and 5.113 (a) Using singularity functions,...Ch. 5.4 - Prob. 114PCh. 5.4 - 5.114 and 5.115 A beam is being designed to be...Ch. 5.4 - 5.116 and 5.117 A timber beam is being designed to...Ch. 5.4 - Prob. 117PCh. 5.4 - Prob. 118PCh. 5.4 - Prob. 119PCh. 5.4 - 5.118 through 5.121 Using a computer and step...Ch. 5.4 - Prob. 121PCh. 5.4 - 5.122 and 5.123 For the beam and loading shown and...Ch. 5.4 - 5.122 and 5.123 For the beam and loading shown and...Ch. 5.4 - 5.124 and 5.125 For the beam and loading shown and...Ch. 5.4 - Prob. 125PCh. 5.5 - 5.126 and 5.127 The beam AB, consisting of a...Ch. 5.5 - Prob. 127PCh. 5.5 - 5.128 and 5.129 The beam AB, consisting of a...Ch. 5.5 - 5.128 and 5.129 The beam AB, consisting of a...Ch. 5.5 - Prob. 130PCh. 5.5 - Prob. 131PCh. 5.5 - Prob. 132PCh. 5.5 - 5.132 and 5.133 A preliminary design on the use of...Ch. 5.5 - Prob. 134PCh. 5.5 - Prob. 135PCh. 5.5 - Prob. 136PCh. 5.5 - Prob. 137PCh. 5.5 - Prob. 138PCh. 5.5 - Prob. 139PCh. 5.5 - Assuming that the length and width of the cover...Ch. 5.5 - Two cover plates, each 12 in. thick, are welded to...Ch. 5.5 - Two cover plates, each 12 in. thick, are welded to...Ch. 5.5 - Prob. 143PCh. 5.5 - Prob. 144PCh. 5.5 - Two cover plates, each 7.5 mm thick, are welded to...Ch. 5.5 - Prob. 146PCh. 5.5 - Prob. 147PCh. 5.5 - For the tapered beam shown, determine (a) the...Ch. 5.5 - Prob. 149PCh. 5.5 - Prob. 150PCh. 5.5 - Prob. 151PCh. 5 - Draw the shear and bending-moment diagrams for the...Ch. 5 - Draw the shear and bending-moment diagrams for the...Ch. 5 - Determine (a) the distance a for which the...Ch. 5 - For the beam and loading shown, determine the...Ch. 5 - Draw the shear and bending-moment diagrams for the...Ch. 5 - Beam AB, of length L and square cross section of...Ch. 5 - Prob. 158RPCh. 5 - Knowing that the allowable normal stress for the...Ch. 5 - Prob. 160RPCh. 5 - (a) Using singularity functions, find the...Ch. 5 - Prob. 162RPCh. 5 - Prob. 163RP
Knowledge Booster
Learn more about
Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, mechanical-engineering and related others by exploring similar questions and additional content below.Similar questions
- using the theorem of three moments, find all the reactions and supports, I need concise calculations only. the answers are at the bottom, I need concise steps and minimal explanationsarrow_forwardIn an industrial facility, a counter-flow double-pipe heat exchanger uses superheated steam at a temperature of 155°C to heat feed water at 30°C. The superheated steam experiences a temperature drop of 70°C as it exits the heat exchanger. The water to be heated flows through the heat exchanger tube of negligible thickness at a constant rate of 3.47 kg/s. The convective heat transfer coefficient on the superheated steam and water side is 850 W/m²K and 1250 W/m²K, respectively. To account for the fouling due to chemical impurities that might be present in the feed water, assume a fouling factor of 0.00015 m² K/W for the water side. The specific heat of water is determined at an average temperature of (30+70)°C/2 = 50°C and is taken to be Cp J/kg-K. Water Steam Determine the heat exchanger area required to maintain the exit temperature of the water to a minimum of 70°C. The heat exchanger area required isarrow_forwardStress, ksi 160 72 150- 140 80 70 ༄ ྃ ༈ ཎྜ རྦ ༅ ཎྜ ྣཧྨ ➢ 130 120 110 100 90 2.0 2.8 3.6 4.4 5 Wire diameter, mm 6.0 6.8 2 7.6 8.4 Compression and extension springs. ASTM A227 Class II Light service Average service 0.020 0.060 0.100 0.140 0.180 0.220 0.260 0.300 0.340 0.380 0.420 0.460 0.500 Wire diameter, in Torsional stress due to initial tension, ksi 10 ४ 20 Preferred range 100 Stress, MPa 9.2 10.0 10.8 11.6 12.4 1100 1035 965 895 825 760 Severe service 690 620 550 50 150 3456789 10 11 12 13 14 15 16 Spring index, C = DJD FIGURE 18-21 Recommended torsional shear stress in an extension spring due to initial tension (Data from Associated Spring, Barnes Group, Inc.) 50 200 485 Stress, MPaarrow_forward
- Bolted Joint Design Bolted Frames Total Force due to door weight: P = 240 lb Number of Bolts: N = Distance to Bolt C/L: a = 4 N/A Bolt Material - Allowable shear stress of bolt material: T₂ = x Distance from Bolt centroid to bolt: x = y Distance from Bolt centroid to bolt: y = Degrees per Radian- Results y-Load on each bolt: F, = Moment resisted by bolt pattern: M = Radial distance from Bolt centroid to bolt: r = Sum squares of all radial distances: Σr² Force on each bolt to resist moment: F, - Angle for force composition: e= X-Force on each bolt to resist moment: F- y-Force on each bolt to resist moment: Fly Total y-Force on each bolt: Fy = Resultant force on bolt 1: R₁ = Required shear stress area for a bolt: A₂ = ASTM Grade A307 Steel 10,000 0 psi from Table 20-1 3.0 57.296 in degrees lb per bolt lb-in Formula FS-P/N M-Px XB r = (x² + y²)0.5 in² Σ 4r² Mr F₁ = Στ lb degrees lb lb lb Minimum Bolt Diameter: Din = Rounded up Bolt Diameter: D = 55 P. 1.5 in 2 in (3x) 1 in This bracket…arrow_forwardUniversity of Babylon Collage of Engineering/ Al-Musayab Department of Automobiles Final Examination/ Stage: 3rd Notes: Answer 4 questions only 2023-2202 Subject: Theory of vehicles Date: 2023\06\10-Saturday Time: Three Hours Course 2nd Attempt 1st Q1: A Hooke's coupling connects two shafts whose axes are inclined at 30°. The of the driven shaft? Find the maximum value of retardation or acceleration and driving shaft rotates uniformly at 600 rpm. What are the extreme angular velocities state the angle where both will occur. (12.5 Marks) Q2: Four masses, A, B, C, and D), revolve at equal radii and are equally spaced along a shaft. The mass B is 7 kg, and the radius of C and D make angles of 90° and 240°, respectively, with the radius of B. Find the magnitude of the masses A, C, and D and the angular position of A so that the system may be completely balanced. (12.5 Marks) Q3: A cam has straight worked faces that are tangential to a base circle of diameter 90 mm. The follower is a roller…arrow_forwardProblem 18-26 Added Extension Springs Spring Material ASTM A227 Modulus of Elasticity of the Material in Shear: G 1.150E+07 psi Average Service Max Operating Load: F₁ = 100 lb Max Length between attachment points: L₁ = 60.00 in 20.00 lb 26.00 1.400 Min Operating Load: F₁ = Min Length between attachment points: L₁ = Maximum Outside Diameter = in in Results Note: you select a wire diameter from the "US steel wire gage" column in table 18-2 Formula k = AF/AL k = (F0-F1)/(Lo - L₁) Spring Rate: k = lb/in Assumed Trial Outside Diameter: OD = Assumed Trial Mean: D ma Assumed Design Stress in Spring: Tda in 1.070 in 102,000 psi Assumed Wahl Factor: K = 1.2 Calculated Wire Diameter: Dwa Actual Wire Diameter: Dw Actual outer diameter: OD = Actual inner diameter: ID= Spring Index: C = See Figure 18-8 Dw= [8KF Dm πTd 1/3 in 5' 5' 5' 5' This corresponds to US Steel 9 wire gage ID = Dm - Dw C = Dm/Dw 4C - 1 0.615 K = + 4C - с Wahl Factor: K = 8KFDm 8KFC T = TD πD Stress in Spring at F = Fo: To psi…arrow_forward
- CHAIN DRIVE DESIGN Initial Input Data: Application: Garage Door Opener Drive type: AC Motor Driven machine Chain and Sprocket to pull the door up Degrees per Radian: 57.2958 degrees Sprocket Diameter: D = 1.690 in Number of strands: Chain number: 1 40 Service factor: 1.3 Table 7-10 No. of teeth Computed Data: Actual Motor Power Input: 0.000 hp Sprocket Speed (for sprocket attached to gear shaft) Design power: 0.00 rpm 0 hp 11 12 0.06 0.15 0.29 0.56 0.99 1.09 1.61 2.64 TABLE 7-7 Horsepower Ratings-Single Strand Roller Chain No. 40 0.500 inch pitch 10 25 50 100 180 200 300 500 700 900 1000 1: 0.06 0.14 0.27 0.52 0.91 1.00 1.48 2.42 3.34 4.25 4.70 ! 3.64 4.64 5.13 13 0.07 0.16 0.31 0.61 1.07 1.19 1.75 2.86 3.95 5.02 5.56 Design Decisions-Chain Type and Teeth Numbers: 14 Chain number: Use Table 7-7 Chain pitch: p = in 15 Number of Teeth: N = Per Table 7-7 16 0.08 0.20 0.39 0.75 1.32 1.46 2.15 3.52 0.07 0.17 0.34 0.66 1.15 1.28 1.88 3.08 0.08 0.19 0.36 0.70 1.24 1.37 2.02 3.30 4.55 5.80…arrow_forwardInput Data: Torque needed to overcome rolling friction in rollers, slides and other moving parts, except for Motor and Worm Gear the worm gear T₁ = Length of travel of door: Time for door to open or close: LD = 50 lb-in. 90 in t= 12.5 seconds Pitch diameter for chain sprocket: DPC 1.690 in Weight of Door: P = No. of worm threads: Nw = Worm Pitch diameter: Dw Diametral pitch: Pd Normal pressure angle: Degrees per Radian: Number of gear teeth: Calculated Data: Linear velocity of door and chain (in/sec): Linear velocity of door and chain (ft/min): Output Speed of Gear and Sprocket: Upward Force due to Weight of Door: Фо = = NG= 240 lb 2 1.250 in 12 14.5 degrees 57.2958 degrees 28 Vα= in/sec VC= ft/min NG = rpm FD lb Net Upward Force on Door: Fou lb Torque on gear ignoring rolling friction: TG = lb-in. Formula = FDU FD-2 x Fo (note: Fo is the Max Operating load of the extension springs). This is also the initial tension in the chain. TG = FDU X DPC/2 This is the also the torque on the…arrow_forwardQ5/A: A car with a track of 1.5 m and a wheelbase of 2.9 m has a steering gear mechanism of the Ackermann type. The distance between the front stub axle pivots is 1.3 m. The length of each track arm is 150 mm, and the length of the track rod is 1.2 m. Find the angle turned through by the outer wheel if the angle turned through by the inner wheel is 30°. (6 Marks) Q5/B: Write True on the correct sentences and False on the wrong sentences listed below:- 1- In automobiles, the power is transmitted from the gearbox to the differential through bevel gears. 2- The minimum radius circle drawn to the cam profile is called the base circle. 3- The Proell governor, compared to the Porter governor, has less lift at the same speed. 4- The balancing of rotating and reciprocating parts of an engine is necessary when it runs at a slow speed. (6.5 Marks) ***Best of Luck *** جامعة بابل UNIVERSITY OF BABYLON Examiner: Mohanad R. Hameed Head of Department: Dr. Dhyai H. Jawadarrow_forward
- University of Babylon Collage of Engineering/ Al-Musayab Department of Automobiles Mid Examination/ Stage: 3rd Subject: Theory of Vehicles Date: 14 \ 4 \2025 Time: 1.5 Hours 2025-2024 Q1: The arms of a Porter governor are 250 mm long. The upper arms are pivoted on the axis of revolution, but the lower arms are attached to a sleeve at a distance of 50 mm from the axis of rotation. The weight on the sleeve is 600 N and the weight of each ball is 80 N. Determine the equilibrium speed when the radius of rotation of the balls is 150 mm. If the friction is equivalent to a load of 25 N at the sleeve, determine the range of speed for this position. Q2: In a loaded Proell governor shown in Figure below each ball weighs 3 kg and the central sleeve weighs 25 kg. The arms are of 200 mm length and pivoted about axis displaced from the central axis of rotation by 38.5 mm, y=238 mm, x=303.5 mm, CE 85 mm, MD 142.5 mm. Determine the equilibrium speed. Fe mg E M 2 Q3: In a spring loaded Hartnell type…arrow_forwardusing the theorem of three moments, find all the reactions and supports, I need the calculations onlyarrow_forwardQ.5: (10 Marks) Select the correct answer (choose 10 only) 1. The forward whirling speed is ......... the static structure tilting speed. (a) Less than (b) Higher than (c) equal to 2. The divergence between the forward and backward whirling speeds increases as: (a) The rotating speed increase (b) the polar moment of inertia increases (c) Both (a) and (b) (d) do not change 3. Increasing the system natural frequency can be done by: (a) add masses (b) adding braces and supports (c) increase damping 4. The amplitude of vibration due to external force can be reduced by: (a) Increasing damping (b) Decreasing damping (c) Increasing mass 5. Tuned absorbers are used to: (a) Shift the natural frequency (b) increase damping (c) Increase stiffness 6. Accelerometers sensors contains: г (a) Piezoelectric materials (b) Magnet and coil (c) coil only 7. Increasing the stiffness of the system causes: (a) Less transmitted force (b) more transmitted force (c) Transmitted force does not change 8. The…arrow_forward
arrow_back_ios
SEE MORE QUESTIONS
arrow_forward_ios
Recommended textbooks for you
Elements Of ElectromagneticsMechanical EngineeringISBN:9780190698614Author:Sadiku, Matthew N. O.Publisher:Oxford University Press
Mechanics of Materials (10th Edition)Mechanical EngineeringISBN:9780134319650Author:Russell C. HibbelerPublisher:PEARSON
Thermodynamics: An Engineering ApproachMechanical EngineeringISBN:9781259822674Author:Yunus A. Cengel Dr., Michael A. BolesPublisher:McGraw-Hill Education
Control Systems EngineeringMechanical EngineeringISBN:9781118170519Author:Norman S. NisePublisher:WILEY
Mechanics of Materials (MindTap Course List)Mechanical EngineeringISBN:9781337093347Author:Barry J. Goodno, James M. GerePublisher:Cengage Learning
Engineering Mechanics: StaticsMechanical EngineeringISBN:9781118807330Author:James L. Meriam, L. G. Kraige, J. N. BoltonPublisher:WILEY
Elements Of Electromagnetics
Mechanical Engineering
ISBN:9780190698614
Author:Sadiku, Matthew N. O.
Publisher:Oxford University Press
Mechanics of Materials (10th Edition)
Mechanical Engineering
ISBN:9780134319650
Author:Russell C. Hibbeler
Publisher:PEARSON
Thermodynamics: An Engineering Approach
Mechanical Engineering
ISBN:9781259822674
Author:Yunus A. Cengel Dr., Michael A. Boles
Publisher:McGraw-Hill Education
Control Systems Engineering
Mechanical Engineering
ISBN:9781118170519
Author:Norman S. Nise
Publisher:WILEY
Mechanics of Materials (MindTap Course List)
Mechanical Engineering
ISBN:9781337093347
Author:Barry J. Goodno, James M. Gere
Publisher:Cengage Learning
Engineering Mechanics: Statics
Mechanical Engineering
ISBN:9781118807330
Author:James L. Meriam, L. G. Kraige, J. N. Bolton
Publisher:WILEY
Everything About COMBINED LOADING in 10 Minutes! Mechanics of Materials; Author: Less Boring Lectures;https://www.youtube.com/watch?v=N-PlI900hSg;License: Standard youtube license