
Find the approximate axial forces, shears, and moments for the all members of the frames using cantilever method.

Explanation of Solution
Given information:
The axial force acting at point I
The axial force acting at point E
The vertical distance of the member AE, BF, CG, and DH
The vertical distance of the member EI, FJ, GK, and HL
The horizontal distance of the members AB, EF, and IJ
The horizontal distance of the members BC, FG, and JK
The horizontal distance of the members CD, GH, and KL
Take the counterclockwise moment is positive and clockwise moment is negative.
The axial force in horizontal direction, towards right is positive and towards left side is negative.
The axial force in vertical direction, towards upward is positive and towards downward is negative.
Calculation:
Insert the internal hinges at the midpoints of all the members of the given frame to obtain the simplified frame for approximate analysis.
Draw the simplified frame as in Figure (1).
For the calculation of column axial forces of story of the frame, pass an imaginary section aa through the internal hinges at the midheights of columns EI, FJ, GK, and HL, and pass an imaginary section bb through the internal hinges at the midheights of columns AE, BF, CG, and DH.
Draw the free body diagram of the frame portion with the passed imaginary lines as in Figure (2).
Column axial forces:
Above section aa:
Draw the free body diagram of the frame portion above the section aa as in Figure (3).
Refer Figure 3.
Determine the location of the centroid using the relation.
Substitute 0 m for
The given lateral load is acting on the frame to the right, therefore the axial force in column EI and FJ located to the left of the centroid, must be tensile, whereas the axial force in column HL and GK placed to the right of the centroid, must be compressive.
Consider the axial forces in the columns are to be linearly proportional to their distances from centroid.
Apply similar triangle rule.
Determine the relationship in column axial force between the member EI and FJ using the relation.
Substitute 11 m for
Determine the relationship in column axial force between the member EI and GK using the relation.
Substitute 8 m for
Determine the axial force in the column members EI, FJ, GK, and HL using equilibrium conditions.
Take moment about point M.
Substitute
Determine the axial force in the column members FJ.
Substitute 1.69 kN for
Determine the axial force in the column members GK.
Substitute 1.69 kN for
Determine the axial force in the column members HL.
Substitute 1.69 kN for
Draw the free body diagram of the frame portion above the section aa with the axial forces in the column members as in Figure (4).
Draw the free body diagram of the frame portion above the section bb as in Figure (5).
The given lateral load is acting on the frame to the right, therefore the axial force in column AE and BF located to the left of the centroid, must be tensile, whereas the axial force in column CG and DH placed to the right of the centroid, must be compressive.
Determine the relationship in column axial force between the member AE and BF using the relation.
Substitute 11 m for
Determine the relationship in column axial force between the member AE and CG using the relation.
Substitute 8 m for
Determine the axial force in the column members AE, BF, CG, and DH using equilibrium conditions.
Take moment about point N.
Substitute
Determine the axial force in the column members BF.
Substitute 11 kN for
Determine the axial force in the column members CG.
Substitute 11 kN for
Determine the axial force in the column members DH.
Substitute 11 kN for
Draw the free body diagram of the frame portion above the section bb with the axial forces in the column members as in Figure (6).
Girder shear and moments:
Consider girder IJ.
Determine the shear at upper left end joint I using equilibrium equation.
Substitute 1.69 kN for
Determine the shear at upper right end joint J using equilibrium equation.
Substitute 1.69 kN for
Determine the moment at left end of the girder IJ using equilibrium equations.
Substitute 1.69 kN for
Determine the moment at right end of the girder GH using equilibrium equations.
Take moment about point I.
Substitute
Consider girder JK.
Determine the shear at left end joint J using equilibrium equation.
Substitute 1.69 kN for
Determine the shear at right end joint K using equilibrium equation.
Substitute 2.15 kN for
Determine the moment at left end of the girder JK using equilibrium equations.
Substitute 2.15 kN for
Determine the moment at right end of the girder JK using equilibrium equations.
Take moment about point J.
Substitute
Consider girder KL.
Determine the shear at left end joint K using equilibrium equation.
Substitute 2.15 kN for
Determine the shear at right end joint L using equilibrium equation.
Substitute 1.69 kN for
Determine the moment at left end of the girder KL using equilibrium equations.
Substitute 1.69 kN for
Determine the moment at right end of the girder JK using equilibrium equations.
Take moment about point K.
Substitute
Consider girder EF.
Determine the shear at left end joint E using equilibrium equation.
Substitute 1.69 kN for
Determine the shear at right end joint F using equilibrium equation.
Substitute 9.31 kN for
Determine the moment at left end of the girder EF using equilibrium equations.
Substitute 9.31 kN for
Determine the moment at right end of the girder EF using equilibrium equations.
Take moment about point E.
Substitute
Consider girder FG.
Determine the shear at left end joint F using equilibrium equation.
Substitute 9.31 kN for
Determine the shear at right end joint G using equilibrium equation.
Substitute 11.85 kN for
Determine the moment at left end of the girder FG using equilibrium equations.
Substitute 11.85 kN for
Determine the moment at right end of the girder FG using equilibrium equations.
Take moment about point F.
Substitute
Consider girder GH.
Determine the shear at left end joint G using equilibrium equation.
Substitute 11.85 kN for
Determine the shear at right end joint H using equilibrium equation.
Substitute 9.31 kN for
Determine the moment at left end of the girder GH using equilibrium equations.
Substitute 9.31 kN for
Determine the moment at right end of the girder GH using equilibrium equations.
Take moment about point G.
Substitute
Column moments and shears:
Column moment for member EI, FJ, GK, and HL:
Determine the moment at the column member EI using moment equilibrium of joints.
Apply the moment equilibrium of joints at I.
Substitute
The moment at the column member EI is
Determine the moment at the column member FJ.
Apply the moment equilibrium of joints at J.
Substitute
The moment at the column member FJ is
Determine the moment at the column member GK.
Apply the moment equilibrium of joints at K.
Substitute
The moment at the column member GK is
Determine the moment at the column member HL using moment equilibrium of joints.
Apply the moment equilibrium of joints at L.
Substitute
The moment at the column member HL is
Column shear for member EI, FJ, GK, and HL:
Determine the shear at the end I in the column member EI using the relation.
Substitute
The shear at the column member EI must act towards right, so that it can produce Clockwise moment to balance the counterclockwise moment at joint I.
Determine the shear at the end of the column E using equilibrium conditions.
Substitute 3.38 kN for
Determine the shear at the end J in the column member FJ using the relation.
Substitute
The shear at the column member JF must act towards right, so that it can produce Clockwise moment to balance the counterclockwise moment at joint J.
Determine the shear at the end of the column F using equilibrium conditions.
Substitute 6.61 kN for
Determine the shear at the end K in the column member GK using the relation.
Substitute
The shear at the column member KG must act towards right, so that it can produce Clockwise moment to balance the counterclockwise moment at joint K.
Determine the shear at the end of the column G using equilibrium conditions.
Substitute 6.61 kN for
Determine the shear at the end L in the column member HL using the relation.
Substitute
The shear at the column member LH must act towards right, so that it can produce Clockwise moment to balance the counterclockwise moment at joint L.
Determine the shear at the end of the column H using equilibrium conditions.
Substitute 3.38 kN for
Column moment for member AE, BF, CG, and DH:
Determine the moment at the column member AE using moment equilibrium of joints.
Apply the moment equilibrium of joints at E.
Substitute
The moment at the column member AE is
Determine the moment at the column member BF.
Apply the moment equilibrium of joints at F.
Substitute
The moment at the column member BF is
Determine the moment at the column member CG using moment equilibrium of joints.
Apply the moment equilibrium of joints at C.
Substitute
The moment at the column member CG is
Determine the moment at the column member DH using moment equilibrium of joints.
Apply the moment equilibrium of joints at H.
Substitute
The moment at the column member DH is
Column shear for member AE, BF, CG, and DH:
Determine the shear at the end E in the column member AE using the relation.
Substitute
The shear at the column member EA must act towards right, so that it can produce Clockwise moment to balance the counterclockwise moment at joint E.
Determine the shear at the lower end of the column A using equilibrium conditions.
Substitute 10.16 kN for
Determine the shear at the end F in the column member BF using the relation.
Substitute
The shear at the column memer FB must act towards right, so that it can produce Clockwise moment to balance the counterclockwise moment at joint F.
Determine the shear at the lower end of the column B using equilibrium conditions.
Substitute 19.86 kN for
Determine the shear at the end G in the column member CG using the relation.
Substitute
The shear at the column member GC must act towards right, so that it can produce counterclockwise moment to balance the clockwise moment at joint G.
Determine the shear at the lower end of the column C using equilibrium conditions.
Substitute 19.86 kN for
Determine the shear at the end H in the column member DH using the relation.
Substitute
The shear at the column member HD must act towards right, so that it can produce Clockwise moment to balance the counterclockwise moment at joint H.
Determine the shear at the lower end of the column A using equilibrium conditions.
Substitute 10.16 kN for
Draw the free body diagram of frame with the column moments and shears for the portion EIJ as in Figure (7).
Girder axial forces:
Girder IJ.
Determine the girder end action at the upper left end joint I using the equilibrium condition.
Apply equilibrium condition at left end joint I.
Substitute3.38 kN for
The girder end action at joint I in the girder IJ is
Determine the girder end action at the upper right end joint H using the equilibrium condition.
Apply equilibrium condition at end joint J.
Substitute 16.62 kN for
Girder JK.
Determine the girder end action at the left end joint J for the girder JK using the relation.
Substitute 16.62 kN for
Determine the girder end action at the right end joint K.
Substitute 10.01 kN for
Girder KL.
Determine the girder end action at the left end joint K for the girder KL using the relation.
Substitute 10.01 kN for
Determine the girder end action at the right end joint L.
Substitute 3.4 kN for
Girder EF.
Apply equilibrium condition at left end joint E.
Substitute 10.16 kN for
The girder end action at joint E in the girder EF is
Determine the girder end action at the right end joint F using the equilibrium condition.
Apply equilibrium condition at left end joint F.
Substitute 33.22 kN for
Determine the girder end action at the left end joint F for the girder FG using equilibrium condition.
Substitute 33.22 kN for
Determine the girder end action at the right end joint G.
Substitute 19.97 kN for
Determine the girder end action at the left end joint G for the girder GH using equilibrium condition.
Substitute 19.97 kN for
Determine the girder end action at the right end joint H.
Substitute 6.72 kN for
Draw the freebody diagram of the member end forces and moments as in Figure (8).
Draw the freebody diagram of the frame with support reactions as in Figure (9).
Want to see more full solutions like this?
- 9.44 High-speed passenger trains are streamlined to reduce shear force. The cross section of a passenger car of one such train is shown. For a train 81 m long, estimate the shear force (a) for a speed of 81.1 km/hr and (b) for one of 204 km/hr. What power is required for just the shear force at these speeds? These two power calculations will be answers (c) and (d), respectively. Assume T = 10°C and that the boundary layer is tripped at the front of the train. 10 m Problem 9.44arrow_forwardA monitoring program for water flow in an unsaturated soil layer includes sensors to measure the volumetric water content and suction up to a depth of 3 m. The soil is a sand whose hydraulic properties are shown in the figures below. Using the drying curves, draw a quantitatively accurate set of vertical profiles of volumetric water content, pressure head, and total hydraulic head versus depth (with a datum at the base of the soil layer and an elevation head that is positive upward) expected for the following cases:A) The volumetric water content (moisture content) is 10% throughout the profile B) The pressure head is -150 cm throughout the profile C) The total hydraulic head is 100 cm throughout the profile (static no-flow case) Also, report the hydraulic gradient for each case. For parts (a) and (b), calculate the flow rates through the profile. For part (c), calculate the depth to the water table.arrow_forward9.16 Two vertical parallel plates are spaced 0.012 ft apart. If the pressure decreases at a rate of 100 psf/ft in the vertical z direction in the fluid between the plates, what is the maximum fluid velocity in the Z direction? The fluid has a viscosity of 10-3 Ibf s/ft² and a specific gravity of 0.80. .arrow_forward
- Please explain steps using software.arrow_forwardPlease explain steps for using softwarearrow_forwardDesign the reinforced masonry beam in the wall shown below. The wall is to be constructed of fully grouted hollow concrete masonry units in running bond. It is to carry its own weight plus a superimposed dead load of 2.5 kips/ft and a live load of 0.8 kip/ft. Determine the width of the masonry units (by trials), and the amounts of the longitudinal and shear reinforcement required using the strength design method of TMS 402-22. Show the layout of the reinforcements with diagrams. Use fm = 2,000 psi, Grade 60(60 ksi) steel, and Type S Portland cement mortar. Assume that the centroid of the bottom rebar is 3 inches from the bottom face of the beam. ( you may assume that the unit weight of fully grouted concrete masonry is 125 lbs per cubic foot.)arrow_forward
- 6. The easiest method to solve the beam shown in question number 14 is A. Force method B. Slope deflection method C. Moment distribution method D. Virtual work method E. Stiffness matrix method 17. The value of 8 caused by applying CW moment at A equal to 18. A. ML/2E1 B. ML/3E1 C. ML/4E1 D. ML/6EI E. None of the above For the beam shown below, the moment at A kN.m CCW. Assume P= 8 kN equals to ........ A. 20 B. 22.5 C. 25 D. 27.5 E. 30 M L A unlocked joint end pin P P P B A 1m 1m 2m 2m 19. The analysis of indeterminate non sway frames using moment distribution method does not need..... A. Finding stiffness factors of members B. Finding fix end moments C. Using compatibility equations D. Removing redundants E. Cand D 0. The frame shown is kinematically 6 kN/m indeterminate to ................ degree. A, C and D are fixed. E and B are pinned. A. First B. Second C. Third D. Fourth E. None of the above 6 m Sm 7 marrow_forward1. The moment at A using slope deflection method equals to 10 kN ..... kN. m CCW. A. 2.5 B. 5 C. 7.5 D. 10 E. None of the above 2m 2m B 10 kN + 2m + 2m 2. To solve the beam shown using slope deflection method,. ...... unknowns (s) 25 kN 15 kN/m should be selected. A. One B. Two fix C. Three D. Four E. None of the above magnitude of the rotation at B for the me shown using slope deflection method quals to El constant. A. -162/EI B. -162 El C. 40/El D. -40 El E. 0.3 radian B A 3 m 3 m -4 m- 4k/ft roller A fix 18 ft. To solve the beam shown using slope deflection method, should be fix selected as equilibrium equation (s). A. MAB+MBA = 0 B. MAB + MBA 0 and MBC=0 C. MBA+MBC = 0 D. MBA+MBC = 0 and MCB=0 E. None of the above B fix fix 9ft 20 kN/m 80 EN pin 9 m 3 m rollerarrow_forwardSolvearrow_forward
- 5. The number of unknowns for the frame shown using slope deflection method is... Assume A, B and D are fixed and interior hinge at C A. Two B. Four C. Six D. Eight E. None of the above 10 kN B Qc 4m A 3m + + 3m 3m 6. 7. The slope-deflection method was originally developed by Heinrich Manderla and Otto Mohr for the purpose of studying. A. secondary stresses in trusses B. secondary stresses in beams and frames C. Indeterminate beams and frames analysis D. Determinate beams and frames analysis E. None of the above In structures that have non-parallel end members, the displacement of the members will be..... A. Similar B. Different C. Proportional D. Zero E. None of the above. 8. The magnitude of the fix end moment at A 4k/ft using slope deflection method equals to pin exfix ...........k. ft. A. 25 B. -25 C. 40 D. -40 E. None of the above. A roller 15 ft- 12 f The magnitude of MBC for the frame shown in question number 3 using slope deflection method equals Assume El constant for all…arrow_forwardQ2. An isotropic rectangular slab (6 x 8) m is fixed at 3- edges and free at one edge as shown below. The reinforcement provides a positive yield moment of (10) kN.m/m and along the fixed edge a negative yield moment (m) of (14) kN.m/ m. Determine the collapse load if the slab carries a u.d.L. of (w) kN/m² including the slab own weight. W free C Gm fixed 8 m darrow_forwardReinforced Concrete Design 4 Second Monthly Exam 15/4/2025 Q1. A double T-concrete beam is prestressed with 2- tendons each of cross-sectional area of (600) mm² as shown below. Determine the allowable service load. Given: Span = 12 m, fse=1400 N/mm², fé= 50 N/mm², Ct = 163 mm, Cb =437 mm, I=7586 x 10 mm*. 10 KN/M * 25.00 x-500x 1500 +500 +100 163 不 -A 500 12m + 437 += 50 1 150 150 600mm 600mmarrow_forward
