--- title: "Beam AB with tie rod: AFD, SFD, BMD" subject: "Strength of Materials" chapter: 1 key: "semester_3/strength_of_materials:beam-tie-rod-afd-sfd-bmd" source: https://www.bctengineer.com/q/semester_3/strength_of_materials/beam-tie-rod-afd-sfd-bmd site: BCT Engineering --- # Beam AB with tie rod: AFD, SFD, BMD Strength of Materials · Chapter 1: Axial Forces, Shearing Forces and Bending Moments - Asked in: 2081 Bhadra, 10 marks; 2080 Baisakh, 12 marks (asked 2 times) - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_3/strength_of_materials/beam-tie-rod-afd-sfd-bmd - Whole chapter: https://www.bctengineer.com/md/semester_3/strength_of_materials/chapter-1 ## Question A beam AB carrying uniformly varying load of intensity 5 kN/m and a point load of 5 kN (at $60^\circ$) is hinged to a wall at end A and is connected by a tie rod DE which is hinged to the wall at end E and also hinged to a vertical bracket CD (of length 1 m) at end D as shown. Draw AFD, SFD and BMD for beam AB. [Figure: horizontal beam A-F-G-H-C-B with A hinged to the wall; segment lengths A to F 1 m, F to G 3 m, G to H 3 m, H to C 1 m, C to B 1 m; triangular loads of peak intensity 5 kN/m at F and at H, tapering to zero at G; bracket CD of length 1 m rises vertically from C and carries the tie rod DE; E is on the wall 4 m above A; 5 kN point load at B inclined at $60^\circ$.] ## Short answer Using equilibrium: tie force T = 26.43 kN, H_A = 22.24 kN, V_A = 10.05 kN. AFD: -22.24 kN left of C, 2.5 kN right. SFD varies from 10.05 to -4.95 kN, zero at 5.75 m. BMD peaks: sagging 28.18 kNm at 5.75 m, hogging -4.33 kNm just right of C. ## Answer **Assumptions (taken from the figure):** - Origin at A (hinge on the wall). Beam A–F–G–H–C–B: A to F = 1 m, F–G = 3 m, G–H = 3 m, H–C = 1 m, C–B = 1 m (B at x = 9 m). - Loads (downward): triangle FG rising from 0 at G to 5 kN/m at F; triangle GH with 0 at G and 5 kN/m at H. Each has a total load of $\tfrac12\times3\times5=7.5$ kN, acting 1 m from F and 1 m from H respectively. - The 5 kN load at B acts at 60° to the beam, downward and away from the wall: $F_x = 5\cos60^\circ = 2.5$ kN (tension), $F_y = 5\sin60^\circ = 4.33$ kN (down). - Bracket CD is rigid and vertical (D is 1 m above C, at x = 8 m); the tie rod DE joins D to E (0, 4 m), so its direction is $(-8, 3)/\sqrt{73}$. Sign convention: N is positive in tension; V is positive when the forces on the part before the section (travelling A → end) point along the left-hand normal of the member axis (upward on a left-to-right beam); M is quoted as a magnitude with the tension face named. Support reactions are drawn on the free-body of the whole frame. #### Support reactions The tie force $T$ acts at D, 1 m above the beam: $T_x = \frac{8}{\sqrt{73}}T$ (to the left) and $T_y = \frac{3}{\sqrt{73}}T$ (upward). Moments about A: $8T_y + 1\cdot T_x = 7.5(2) + 7.5(6) + 4.33(9) = 98.97$, so $3.745\,T = 98.97$. - $H_A$ = 22.24 kN → - $V_A$ = 10.05 kN ↑ - $T$ = 26.43 kN (tension in the rod; components 24.74 kN ← and 9.28 kN ↑ on the beam) Check: the three equilibrium equations of the complete frame are satisfied to rounding. #### Axial force, shear force and bending moment at the key sections | Section | N (kN) | V (kN) | M (kN·m) | |---|---|---|---| | AB at A | -22.24 | 10.05 | 0.00 | | AB at B | 2.50 | 4.33 | 0.00 | | At F (x = 1 m) | -22.24 | 10.05 | 10.05 (bottom) | | At G (x = 4 m) | -22.24 | 2.55 | 25.21 (bottom) | | At H (x = 7 m) | -22.24 | -4.95 | 25.36 (bottom) | | Just left of C | -22.24 | -4.95 | 20.42 (bottom) | | Just right of C | 2.50 | 4.33 | -4.33 (top) | The jump in M at C ($20.42 + 4.33 = 24.75$ kN·m) is the couple $T_x\times1$ m that the bracket CD transfers to the beam. #### Shape of the diagrams - Between loads V is constant; under a UDL it varies linearly and under a triangular load parabolically. M is one degree higher than V (linear, parabolic, cubic respectively). V jumps at point loads and reactions; M jumps at couples. - AFD: constant between loads on a member, changing only where a load component acts along the member. - The BMD is drawn on the tension side of each member; it is continuous at rigid joints (moment is transferred) and zero at hinges and at free ends. #### Salient features - Maximum positive (sagging) M = 28.18 kN·m at x = 5.75 m - Maximum negative (hogging) M = -4.33 kN·m just right of C (x = 8.00 m) - V = 0 at 5.75 m from A along AB: M = 28.18 kN·m (bottom), the local maximum #### Sketches (sagging +ve above the line) SFD: ```text ******** *** ***** ****** ************** | ------------------------------*****----------------- **** | *******| ``` BMD: ```text ******************* ********** *****| ****** | **** | **---------------------------------------------***** * ``` ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Why is the tie force T found by taking moments about the hinge A? A. Because the moments of H_A and V_A about A are zero, so T is the only unknown left in the equation B. Because T acts at A C. Because the beam is in pure bending D. Because moments about A are always equal to the shear force Answer: A. Both hinge reactions pass through A and have no moment about it, so the moment equation contains only T and the known loads. (Key point: Moment equation about A) 2. D is 8 m along and 1 m above the beam, and E is on the wall 4 m above A. What is the direction of the tie rod DE used to split T into components? A. (-8, 3)/sqrt(73), so T_x = 8T/sqrt(73) to the left and T_y = 3T/sqrt(73) upward B. (-8, 4)/sqrt(80), so T_x is to the left and T_y is downward C. (-3, 8)/sqrt(73), so T_y is larger than T_x D. Horizontal, so T has no vertical component Answer: A. D is at (8, 1 m above beam) and E is at (0, 4), so the rod rises 3 m over 8 m horizontally, giving the direction (-8, 3)/sqrt(73). (Key point: Tie rod components) 3. The 5 kN load at B acts at 60 degrees to the beam, away from the wall. What are its axial and transverse components? A. F_x = 5 cos60 = 2.5 kN (tension), F_y = 5 sin60 = 4.33 kN (down) B. F_x = 4.33 kN, F_y = 2.5 kN C. F_x = 5 kN, F_y = 0 D. F_x = 2.5 kN (compression), F_y = 2.5 kN Answer: A. The component along the beam is 5 cos60 = 2.5 kN pulling away from the wall (tension), and the component perpendicular to it is 5 sin60 = 4.33 kN downward. (Key point: Inclined load at B) 4. V is +2.55 kN at G and -4.95 kN at H. What does this tell you about the maximum bending moment? A. It lies between G and H, at the point where V = 0 (x = 5.75 m, M = 28.18 kN·m) B. It occurs at B C. It occurs at A D. It cannot be found because V changes sign Answer: A. M has zero slope where V = 0. V changes sign between G and H, so the maximum sagging moment occurs there. 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