
The beam AB is pin supported at A and supported by a cable BC. A separate cable CG is used to hold up the frame. If AB weighs 120 lb/ft and the column FC has a weight of 180 lb/ft, determine the resultant internal loadings acting on cross sections located at points D and E. Neglect the thickness of both the beam and column in the calculation.

Find the resultant internal loadings acting on cross sections located at D and E.
Answer to Problem 1RP
The resultant internal loadings at cross section at D are
The resultant internal loadings at cross section at E are
Explanation of Solution
Given information:
The beam AB is pin supported at A and supported by a cable BC.
The weight of the beam AB is
The weight of the column FC is
Calculation:
Find the loading at the center of the beam AB
Substitute
Convert the unit from lb to kip.
Sketch the Free Body Diagram of the beam AB shown in Figure 1.
Refer to Figure 1.
Find the angle of cable BC to the horizontal
Find the tension in cable BC as shown below.
Take moment about A is Equal to zero.
Find the support reaction at A as shown below.
Apply the Equations of Equilibrium as shown below.
Summation of forces along horizontal direction is Equal to zero.
Summation of forces along vertical direction is Equal to zero.
Find the loading at the center of the beam AD
Substitute
Convert the unit from lb to kip.
Sketch the Free Body Diagram of the section for point D as shown in Figure 2.
Refer to Figure 2.
Find the internal loadings as shown below.
Apply the Equations of Equilibrium as shown below.
Summation of forces along horizontal direction is Equal to zero.
Summation of forces along vertical direction is Equal to zero.
Take moment about D is Equal to zero.
Hence, the resultant internal loadings at cross section at D are
Find the loading at the center of the column FC
Substitute
Convert the unit from lb to kip.
Sketch the Free Body Diagram of the beam FC shown in Figure 3.
Refer to Figure 3.
Find the angle of cable CG to the horizontal.
Find the tension in cable CG as shown below.
Summation of forces along horizontal direction is Equal to zero.
Find the loading at the center of the column FE
Substitute
Convert the unit from lb to kip.
Sketch the Free Body Diagram of the section for point E as shown in Figure 4.
Refer to Figure 4.
Find the internal loadings as shown below.
Apply the Equations of Equilibrium as shown below.
Summation of forces along horizontal direction is Equal to zero.
Summation of forces along vertical direction is Equal to zero.
Take moment about E is Equal to zero.
Therefore, the resultant internal loadings at cross section at E are
Want to see more full solutions like this?
Chapter 7 Solutions
Statics and Mechanics of Materials (5th Edition)
Additional Engineering Textbook Solutions
Java: An Introduction to Problem Solving and Programming (8th Edition)
Fluid Mechanics: Fundamentals and Applications
Mechanics of Materials (10th Edition)
Elementary Surveying: An Introduction To Geomatics (15th Edition)
Starting Out with Programming Logic and Design (5th Edition) (What's New in Computer Science)
Computer Science: An Overview (13th Edition) (What's New in Computer Science)
- Hello I was going over the solution for this probem and I'm a bit confused on the last part. Can you please explain to me 1^4 was used for the Co of the tubular cross section? Thank you!arrow_forwardBlood (HD = 0.45 in large diameter tubes) is forced through hollow fiber tubes that are 20 µm in diameter.Equating the volumetric flowrate expressions from (1) assuming marginal zone theory and (2) using an apparentviscosity for the blood, estimate the marginal zone thickness at this diameter. The viscosity of plasma is 1.2 cParrow_forwardQ2: Find the shear load on bolt A for the connection shown in Figure 2. Dimensions are in mm Fig. 2 24 0-0 0-0 A 180kN (10 Markarrow_forward
- determine the direction and magnitude of angular velocity ω3 of link CD in the four-bar linkage using the relative velocity graphical methodarrow_forwardFour-bar linkage mechanism, AB=40mm, BC=60mm, CD=70mm, AD=80mm, =60°, w1=10rad/s. Determine the direction and magnitude of w3 using relative motion graphical method. A B 2 3 77777 477777arrow_forwardFour-bar linkage mechanism, AB=40mm, BC=60mm, CD=70mm, AD=80mm, =60°, w1=10rad/s. Determine the direction and magnitude of w3 using relative motion graphical method. A B 2 3 77777 477777arrow_forward
- The evaporator of a vapor compression refrigeration cycle utilizing R-123 as the refrigerant isbeing used to chill water. The evaporator is a shell and tube heat exchanger with the water flowingthrough the tubes. The water enters the heat exchanger at a temperature of 54°F. The approachtemperature difference of the evaporator is 3°R. The evaporating pressure of the refrigeration cycleis 4.8 psia and the condensing pressure is 75 psia. The refrigerant is flowing through the cycle witha flow rate of 18,000 lbm/hr. The R-123 leaves the evaporator as a saturated vapor and leaves thecondenser as a saturated liquid. Determine the following:a. The outlet temperature of the chilled waterb. The volumetric flow rate of the chilled water (gpm)c. The UA product of the evaporator (Btu/h-°F)d. The heat transfer rate between the refrigerant and the water (tons)arrow_forward(Read image) (Answer given)arrow_forwardProblem (17): water flowing in an open channel of a rectangular cross-section with width (b) transitions from a mild slope to a steep slope (i.e., from subcritical to supercritical flow) with normal water depths of (y₁) and (y2), respectively. Given the values of y₁ [m], y₂ [m], and b [m], calculate the discharge in the channel (Q) in [Lit/s]. Givens: y1 = 4.112 m y2 = 0.387 m b = 0.942 m Answers: ( 1 ) 1880.186 lit/s ( 2 ) 4042.945 lit/s ( 3 ) 2553.11 lit/s ( 4 ) 3130.448 lit/sarrow_forward
- Problem (14): A pump is being used to lift water from an underground tank through a pipe of diameter (d) at discharge (Q). The total head loss until the pump entrance can be calculated as (h₁ = K[V²/2g]), h where (V) is the flow velocity in the pipe. The elevation difference between the pump and tank surface is (h). Given the values of h [cm], d [cm], and K [-], calculate the maximum discharge Q [Lit/s] beyond which cavitation would take place at the pump entrance. Assume Turbulent flow conditions. Givens: h = 120.31 cm d = 14.455 cm K = 8.976 Q Answers: (1) 94.917 lit/s (2) 49.048 lit/s ( 3 ) 80.722 lit/s 68.588 lit/s 4arrow_forwardProblem (13): A pump is being used to lift water from the bottom tank to the top tank in a galvanized iron pipe at a discharge (Q). The length and diameter of the pipe section from the bottom tank to the pump are (L₁) and (d₁), respectively. The length and diameter of the pipe section from the pump to the top tank are (L2) and (d2), respectively. Given the values of Q [L/s], L₁ [m], d₁ [m], L₂ [m], d₂ [m], calculate total head loss due to friction (i.e., major loss) in the pipe (hmajor-loss) in [cm]. Givens: L₁,d₁ Pump L₂,d2 오 0.533 lit/s L1 = 6920.729 m d1 = 1.065 m L2 = 70.946 m d2 0.072 m Answers: (1) 3.069 cm (2) 3.914 cm ( 3 ) 2.519 cm ( 4 ) 1.855 cm TABLE 8.1 Equivalent Roughness for New Pipes Pipe Riveted steel Concrete Wood stave Cast iron Galvanized iron Equivalent Roughness, & Feet Millimeters 0.003-0.03 0.9-9.0 0.001-0.01 0.3-3.0 0.0006-0.003 0.18-0.9 0.00085 0.26 0.0005 0.15 0.045 0.000005 0.0015 0.0 (smooth) 0.0 (smooth) Commercial steel or wrought iron 0.00015 Drawn…arrow_forwardThe flow rate is 12.275 Liters/s and the diameter is 6.266 cm.arrow_forward
- Elements Of ElectromagneticsMechanical EngineeringISBN:9780190698614Author:Sadiku, Matthew N. O.Publisher:Oxford University PressMechanics of Materials (10th Edition)Mechanical EngineeringISBN:9780134319650Author:Russell C. HibbelerPublisher:PEARSONThermodynamics: An Engineering ApproachMechanical EngineeringISBN:9781259822674Author:Yunus A. Cengel Dr., Michael A. BolesPublisher:McGraw-Hill Education
- Control Systems EngineeringMechanical EngineeringISBN:9781118170519Author:Norman S. NisePublisher:WILEYMechanics of Materials (MindTap Course List)Mechanical EngineeringISBN:9781337093347Author:Barry J. Goodno, James M. GerePublisher:Cengage LearningEngineering Mechanics: StaticsMechanical EngineeringISBN:9781118807330Author:James L. Meriam, L. G. Kraige, J. N. BoltonPublisher:WILEY





